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EUCLIDEAN GEOMETRY AND TRANSFORMATIONS Clayton W. Dodge University of Maine DOVER PUBLICATIONS, INC. Mineola, New York
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Euclidean Geometry and Transformations

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Page 1: Euclidean Geometry and Transformations

EUCLIDEANGEOMETRYANDTRANSFORMATIONS

ClaytonW.DodgeUniversityofMaine

DOVERPUBLICATIONS,INC.Mineola,NewYork

Page 2: Euclidean Geometry and Transformations

CopyrightCopyright©1972byClaytonW.DodgeAllrightsreserved.

BibliographicalNoteThis Dover edition, first published in 2004, is an unabridged, corrected

republicationoftheworkoriginallypublishedbyAddison-WesleyPublishingCompany, Inc., Reading, Massachusetts, in 1972. Some minor correctionshavebeenmadewithinthetextandaSupplementtopages112–113hasbeenaddedonpage296.

LibraryofCongressCataloging-in-PublicationDataDodge,ClaytonW.

Euclideangeometryandtransformations/ClaytonW.Dodge.p.cm.

Originallypublished:Reading,Mass.:Addison-WesleyPub.Co.,1972,inseries:Addison-Wesleyseriesinmathematics.Includesbibliographicalreferencesandindex.eISBN13:978-0-486-13842-81.Geometry.2.Transformations(Mathematics)I.Title.

QA453.D672004516.2—dc22

2004041357ManufacturedintheUnitedStatesofAmerica

DoverPublications,Inc.,31East2ndStreet,Mineola,N.Y.11501

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Toourdaughter,KathyAlthoughwesometimesgoaroundincircles,westillreflecteachother.

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PREFACEJustasanalyticgeometryisrecognizedtodayasanimportanttoolingeometry,so also are isometries and similarities important geometric tools. It is wellknownthatEuclideangeometryisthestudyofthosepropertiesofpointsthatareinvariant under isometries and similarities, but just how such properties areexhibited using these transformations has not been widely discussed intextbooks.Aprimary purpose of this book is to provide a source for both thetheory and the practical application to geometry of these transformations forcollege students of mathematics in general, and for teachers and prospectiveteachersofgeometryinparticular.

ThespiritofmodernelementarygeometryisalsopresentedwithtopicssuchasMenelaus’ andCeva’s theorems,Euclideanconstructions, and thegeometryof special lines and points associated with a triangle, thereby reviewing andrefreshingthereader’smemoryforhighschoolgeometryandpreparinghimtodogeometry.Thehighschoolgeometryteacherwhohasmasteredthistextcanbe confident that he is prepared tohandle thegeometryproblems that arise inhighschoolclasses.

Prerequisites for this material include high school algebra, geometry, andelementary trigonometry. In addition, some familiarity with the concept offunctionwillprovehelpful.

The primary goal of this book is to prepare the reader to do Euclideangeometry. Hence much of it is written in the style of the classic CollegeGeometry by N. A. Court. The reader is given many opportunities to workexercises,forsuchisthekeytounderstandingmathematics.Itissuggestedthatthe reader pause amoment after reading the statement of each theorem in thetext, draw an appropriate figure, and attempt a proof of the theorem beforereading further.Compare the attemptedproofwith theproofgiven in the text.Work an abundance of exercises. Look first in the section of “Hints” whenunable to obtain a solution, then look at the “Answers” section only as a lastresort.Steadyprogresstowardgenuineunderstandingwillresult.

Geometry,whenunderstood,isindeedafascinatingstudy.Eachchapterbeginswithasectionofhistoryorcommentarywhichneednot

be assigned for formal class study. Although exercises are provided for thesesections, their purposes are towhet the appetite of the student and to providesomeenrichmentmaterial.

FollowingthecommentarysectionineachofChapters2,3,4,and6,oneortwo sections entitled “Introduction.. inform the reader of the theory that isdevelopedin thesectionsthatfollow.Thecasualreadermaywishtoskipover

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the formal development at that time and look directly at the “Applications”sections,returninglatertofillintheoreticgaps.

Hintsforthesolutionsofabouthalfoftheexercisesareprovidedinthebackofthebook,followedbyasectionofanswerstoalternatepartsofallmulti-partexercises and to all other odd-numbered exercises. The bibliography, whichprecedesthe“Hints”section,containsfullinformationonallbooksreferredtointhetextandonotherselectedsources.

Those items preceded by a solid triangle ( )within a section, or sectionsprecededbysolidtriangles,maybeomittedwithoutlossofcontinuity.

Although it isdivided intosixchapters, thebook isnumberedaccording tosectionsanditemsorparagraphswithinsections.ThusDefinition15.3referstothe thirdnumberedparagraph inSection15,and thatparagraph isadefinition.Similarly,15.4referstothefourthnumberedparagraphinSection15.Exercise15.3 is the third exercise in Exercise Set 15, which follows Section 15. Suchdouble numbers always refer to text items unless the word “Exercise” isspecifically stated. Furthermore, please note that the index lists item numbersinsteadofpagenumbers.Thereadershouldfind iteasierandfaster touse thisindexthanapageindex.

Except for Section 43 and a small portion of Section 44,which are easilyomitted,Chapters4,5,and6areindependentofoneanother.Thusareasonableone-semester (45-hour) course for students with little or no background incollegegeometrymight includeChapters1 to3,omittingSections21,22,and30, covered at the rate of about two sections each three hours. Enough timeshouldremaintostudyoneofChapters4,5,and6.Historicalsectionsmaybeassignedasoutsidereading.

Theauthorextendshisdeep thanks toProfessorsHenrikBresinsky,GeorgeCunningham,andHowardEvesfortheirinspirationandkindwordsofadvice,to34students in threeclasseswhoaided theauthor inclass-testing thismaterial,andtothestaffatAddison-Wesleyfortheirpatientunderstandingofanauthor’sidiosyncrasies.

C.W.D.Orono,MaineJanuary1972

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CONTENTSCHAPTER1MODERNELEMENTARYGEOMETRY

1TheBeginningsofGeometry2Directedsegmentsandangles3Idealpointsandratios4ThetheoremofMenelaus5Ceva’stheorem6Somegeometryofthetriangle7Moregeometryofthetriangle8Geometricconstructions

CHAPTER2ISOMETRIESINTHEPLANE9TheAmazingGreeks10Introductiontotranslations,rotations,andreflections11Introductiontoisometries12Transformationtheory13Isometriesasproductsofreflections14Translationsandrotations15Halfturns16Productsofreflections17Productsofisometrie;asummary18Applicationsofisometriestoelementarygeometry19Furtherelementaryapplications20Advancedapplications21Analyticrepresentationsofdirectisometries22Analyticrepresentationsofoppositeisometries

CHAPTER3SIMILARITIESINTHEPLANE23Therebirthofmathematicalthinking24Introductiontosimilarities25Homothety26Similarity27Applicationsofsimilaritiestoelementarygeometry28Furtherelementaryapplications29Advancedapplications30Analyticrepresentationsofsimilarities

CHAPTER4VECTORSANDCOMPLEXNUMBERSINGEOMETRY31Thesearchforthemeaningofcomplexnumbers32Introductiontocomplexnumbers33Vectors

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34Vectorsmultiplication35Vectorsandcomplexnumbers36TrianglesintheGaussplane37LinesintheGaussplane38Thecircle39IsometriesandsimilaritiesintheGaussplane

CHAPTER5INVERSION40Matchlessmodernmathematics41Inversion42Progressions,ratios,andPeaucellier’sCell43Inversionandcomplexgeometry44Applicationsofinversion

CHAPTER6ISOMETRIESINSPACE45Whatnext?46Introductiontothreedimensions47Reflectioninaplane48Basicspaceisometries49Morespaceisometries50Someapplications51AnalyticrepresentationsAppendixesA.ASummaryofBookIofEuclid’SElementsB.BasicRulerandCompassConstructionsBibliographyHintsforSelectedExercisesAnswersIndex

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1MODERNELEMENTARYGEOMETRY

SECTION1 THEBEGINNINGSOFGEOMETRY

1.1Thefirstsectionineachchapterofthisbookisdevotedtoadiscussionofthehistoryofgeometry,specificallyahistoryofthetypeofmaterialcoveredbythistext. These sections, although they contain a few exercises appropriate to thehistory discussed, are not an integral part of the general textmaterial, so theymaybereadatanyconvenienttime.

WiththeexceptionofSections31and45inChapters5and6,thesehistoricalsections progress chronologically, so reading them in their given order issuggested.Section45,whichislesshistoricalandmoreeditorialinform,mayberead at any time, butwill bemoremeaningful if the student reads it after hestudiesthecontentsofChapter2.1.2 The geometry, indeed all the mathematics, which has come down to usthroughEuropehaditsoriginsinthepracticalengineeringandagricultureoftheancientBabyloniansandEgyptiansfromabout5000to2000B.C.Theseearliest“practicalmathematicians”wereconcernedonlywiththesolutionstoproblems:howmuchgrainacertaingranarycanhold,howmuchareainafarmer’slandfortaxpurposes,etc.Theheightof thisearlymathematicalskill isquitevisible inthe great Egyptian pyramids and other structures. The pyramid of Gizeh, forexample, was built about 2900 B.C., using about two million huge stones, asheavyas54tonseach,hauled600milesandcuttoanaccuracygreaterthanonepart in ten thousand!Greatadmiration isdue thesehard-workingearlypeoplesforsuchmagnificentstructures.Ofcourse,theheavymanuallaborwasdonebyasmany as 100,000 slavesworking for as long as 30 years, butmuch carefulmathematicalthoughtcertainlyprecededsuchprojects.1.3IntheRhindpapyrus,decipheredin1877andcopiedabout1700B.C.bythescribeAhmesfromanearlierworkofabout3400B.C.,wefind“Directions forObtainingKnowledgeofallDarkThings.”Heretheareaofanisoscelestriangleofside10andbase4istakenas20;thatis,halfthebasetimestheside.Theareaof a circle is given as the square of eight-ninths of the diameter, a goodapproximationwhichassumesthatπ=3.1604….Theareaofaquadrilateral isgivenas(a+c)(b+d)/4,whichiscorrectforarectangle,buttoomuchforanyotherquadrilateral.1.4Manycorrectformulasweregiven,suchastheareasofatrapezoidandofatriangle,andthevolumeofarightcircularcylinder.Mostamazingofallisthecorrectformula

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for the volume of the frustum of a square pyramid of (lower) base edge B,summit (upperbase)edgeb,andaltitudeh,given in theMoscowpapyrus (ca.1850 B.C.). “The greatest Egyptian pyramid” is how E. T. Bell refers to theEgyptian’s knowledge of this formula. It is surely curious that the Egyptiansshould have known this formula and not a correct formula for the area of aquadrilateral.1.5Mathematics inEgyptdeclinedafterabout2000B.C.Poornotationand thecomplete lack of any evidence of logical reasoning seem the most probablecauses for this stagnation.Although theyusedaknotted rope to forma3–4–5triangletoobtaintheirrightangles,thereisnoevidencewhateverthattheywereawareofevenoneinstanceofthePythagoreantheorem.1.6ThemathematicsofancientChinawasverysimilar to thatofEgypt,but itdid continue to develop over the succeeding centuries to bring forth anoccasionaltheorem–suchasHorner’smethodforreducingeachoftherootsofapolynomialequationbyaconstant–afull500yearsbeforeitwasdiscoveredintheWest.1.7 TheBabylonianswere bettermathematicians, if the term “mathematician”canreallybeappliedtoanyoftheseearlypeoples.ItwastheBabylonianswhodividedthecircleinto360parts.Theyknewthatthealtitudefromthebaseofanisosceles triangle bisects the base, that an angle inscribed in a semicircle is aright angle; they knew the Pythagorean theorem, and that the sides of similartriangles areproportional. Invariousplaces theyhaveπ equal to3 and to3 .TheBible(IKings7:23andIIChronicles4:2)alsogivestheapproximationπ=3.1.8 By constructing a table of values forn3 +n2, theywere enabled to solvecubicequationsoftheformn3+n2=c.PerhapsthemostadvancedtableofallisthatknownasPlimpton322, dating fromabout1800B.C.Thisclay tablet listsPythagoreantriplesandthevaluesofsec2θobtainedfromthemforanglesfrom45° to 31°, with amazingly regular increments in the function values. SuchcalculationsindicateafairlyadvancedunderstandingoftrigonometryandofthePythagoreantheorem.1.9 TheBabylonians never discovered the correct volume of the frustum of apyramid.Byanalogytheysaidthatitshouldbehalfthesumoftheareasofthebases times thealtitude, since that is the right idea for theareaofa trapezoid.Manymathematicianslivingathousandormoreyearslaterhavefallenintothesame trap: Because a formula holds for a certain two-dimensional figure, thesameformulaisassumedforthecorrespondingthree-dimensionalfigure.

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1.10 All mathematics recorded prior to about 600 B.C. was very practical innature,lackingingeneralizations,andlackinginlogicalstructure.Eachspecialcase was treated separately. Several numerical examples would be given,followed by a statement to the effect that “such is the procedure.”The readerwastodeducetheformulafromthemanyexamples.Therearetimeswhenoneistempted to question whether our teaching today has, in many cases, reallyimprovedoverthelast2000to4000years,sincethemethodmentionedaboveisusedsooftenbothintheclassroomandintheliterature.Again,therecentfloodofmathematics textbooks includesmanyworks of truly superb quality, clear,concise, accurate, and readable.But alas! there is also a glut ofmediocre andevenvenomouswritings thatuseall the“right”words,butaremisleading,andevencontaindownrightlies.Sotheteachermustbemostcarefulinselectingthetexts forhiscourses.Letushope that futurehistoriansofmathematicswillbekindenoughtojudgeusbyourbestandnotbyourworst.ExerciseSet11.Findthecorrectareaofanisoscelestrianglewithbase4andside10.2.Showthatwhenonetakestheareaofacircleasthesquareofeight-ninthsofthediameter,thenoneistakingπ=3.1604….

3.Showthat(a+c)(b+d)/4isgreaterthantheareaofanonrectangularquadrilateralwhosesuccessivesideshavelengthsa,b,c,d.Findacorrectformulaforthisarea.

4.DerivetheformulaV= (B2+Bb+b2)hforthevolumeofthefrustumofasquarepyramidofbaseedgeB,summitedgeb,andheighth.

5.Showhowtheaccompanyingfiguremaybeusedtoprovethata3–4–5triangleisarighttriangle.

Exercise1.56.LookupIKings7:23andIIChronicles4:2intheBible.Drawandlabelafiguretoshowwhatvalueofπisassumedthere.

7.Constructatableofvaluesforn3+n2forn=1,2,…,12.Thenusethistabletofindarootforeachoftheseequations:a)x3+x2–1452=0,b)x6+

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2x5+x4–22,500,c)2x3+x2=468,d)x3+3x2=2160.SECTION2 DIRECTEDSEGMENTSANDANGLES

2.1 We begin with more theorems in high school geometry, one purpose ofwhichistohelpeasethereaderbackintogeometricthinking.ThushighschoolEuclideangeometryisassumed,andnoaxiomsorpostulatesarestatedhere.ThereadermayfindithelpfultoreadthecontentsofBookIofEuclid’sElements,assummarizedinAppendixA.Thetheoremslistedthereinwillprovideasufficientbasisforthegeometryofthistext.Abasicknowledgeofalgebraicmanipulationandofthesineandcosinefunctionsisalsoassumed.Thepurposeoftheentirefirstchapteristorefreshthereader’smemoryabouthighschoolgeometryandtoleadhimbackontothepathofgeometricalthinking.Thissectionintroducestheconceptofadirectedsegmentorangle,anideamostusefulinmoderngeometry,aswillbeseenespeciallyinthenextthreesections.Theorems2.17and2.19willbeofparticularvaluetousinthelaterdevelopment.2.2 Throughout this book, unless otherwise stated, we shall always write thecorresponding members of congruent or similar figures in the same orderrelativetooneanother.Thus,whenwewriteΔABC≅ΔDEF(triangleABCiscongruenttotriangleDEE),werequirethat A≅ D, B≅ E,AB≅DEetc.Thecarefulstudentofgeometrywillbesure toobserve thisconvention inhisownwriting.

Theunderstandingofthisconventionmakesclear,forexample,theintentinthe following proof. The reader is urged to draw a figure to illustrate thistheorem.2.3TheoremThebaseanglesofanisoscelestrianglearecongruent.LetAB≅ACintriangleABC.Wehave BAC≅ CABbyidentity.Since

alsoAB≅ACandAC≅AB,wehaveΔBAC≅ΔCABbySAS(twosidesandtheincludedangleofonetriangleeachcongruenttothecorrespondingpartsofthe other triangle). Now B≅ C since they are corresponding parts ofcongruentfigures. *2.4DefinitionAlineisproperlyanundefinedterm,butwetakethewordlinetomeanastraightlinewithoutbeginningorendpoints,infiniteinlength.IfpointCisbetweenpointsAandB,thenthesethreepointsaredistinctandtheyalllieonaline.Conversely,ifA,B,Carethreedistinctpointsonaline,thenexactlyoneof these points is between the other two. A segment AB is the set of pointsconsistingofpointsAandBandallpointsbetweenAandB.2.5Sincealineorasegmentisasetofpoints,weusethenotationsP∈mandQ∉mtodenotethatpointPliesonlinemandpointQdoesnotlieonlinem.Ofcourse, a line is also a generalizationof thephysical concept of the edgeof a

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tableorofasheetofpaper.Similarly,apointistheidealizationofadotoraspotor a location. In fact, Euclidean geometry is basically the idealized study ofcertainpropertiesascribedtotherealphysicalworld.2.6DefinitionPointslyingonalinearecalledcollinear,andtheyformarangeofpointswith the lineasbase.Lines thatallpass throughonepointarecalledconcurrent,and thispoint iscalled theirvertex.Lines thatallconcurorareallparallelaresaidtoformapenciloflines(seeFig.2.6*).

Figure2.62.7DefinitionWe denote byAB either the line on the pointsA andB or thesegment terminated by A and B. The context will make clear which use isintended.Themeasure (length)ofsegmentABwillbedenotedbym(AB). IfAandBcoincide,wewriteA=Borm(AB)=0.Similarly BACdenotestheangleformedbytheraysABandAC.Thenotation Aalsowillbeusedfor BACwhennoconfusionarises.Themeasure(generallyindegrees)of BACor Awillbedenotedbym( BAC)orm( A).2.8 FromDefinition 2.7, it follows thatwewrite A = B onlywhen theseanglescoincide.Iftheseangleshavethesamemeasure,wewrite A≅ B(A is congruent to B). Similarly,m(AB) =m(CD) orAB≅CD means thatsegmentsAB andCD have the same length, whereasAB =CD indicates thatthesesegments(orlines)coincide.2.9 Definition Choose a direction along line m as positive. We define thedirectedlengthfromAtoB,denotedbyd(AB),by

ifthedirectionfromAtoBispositive,and

ifthedirectionfromBtoAispositive.(SeeFig.2.9.)

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Figure2.92.10TheoremForanytwopointsAandB,

2.11TheoremIfA,B,Careanythreecollinearpoints,then

2.12 Theorem IfO, A, B are three collinear points, then the midpointM ofsegmentABsatisfiestherelation

2.13DefinitionLet thedirectedmeasureof BAC,denotedbyd( BAC),bedefinedby

when BACismeasuredcounterclockwise(acounterclockwiserotationcarriesrayABintorayAC),and

when BACismeasuredclockwise.(SeeFig.2.13.)

Figure2.13

2.14TheoremForanyangleBAC,d( BAC)+d( CAB)=0.

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2.15 Agreement Since directed distances and directed angles are used quiteextensively in thisbook,weshalldenote these sensedmagnitudesbyboldfacetype,andundirectedmagnitudesbylightfaceitalictype,informulasinwhichitisclearthatdistancesareimplied.Inallothercasesthemanddnotationswillbeused.*Thus,informulas,wewrite

2.16TheoremEuler’sTheorem.IfA,B,C,Dareanyfourcollinearpoints,then

ByTheorem2.11,write

Thenthegivenexpressionbecomes

2.17TheoremTheareaKoftriangleABCisgivenby

thatis,theareaofatriangleishalftheproductofanytwosidesandthesineoftheangleincludedbetweenthem.

2.18Forconvenienceintheformulasthatfollow,weagreethata/b=c/dshallbe termed truewheneverad=bc is true,whetherornotb=0ord=0.Thisconvention will prove useful when we are using the algebraic expressions inMenelaus’ and Ceva’s theorems in Sections 4 and 5. It eliminates manyawkwardspecialcases,treatingallpossibilitiesatonce.

2.19TheoremLetABCbeanytriangleandletLbeanypointonlineBC.Then

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FirstnotethatlengthsABandCAarenotdirected,butallothermeasuresinthisformulaaredirected.

Leth denote the lengthof the altitude fromvertexA in triangleABC. (SeeFig.2.19.)TheareasK1andK2oftrianglesABLandALCaregivenby

fromwhichweobtain,providedL≠C,

Figure2.19

Now BL/LC and (sin BAL)/(sin LAC) are both positive or bothnegativeaccordingasLliesbetweenBandCoroutsidesegmentBC.IfL=B,then both numerators are zero. Thus in all cases it follows that these twofractionshavethesamesign,sothetheoremfollowswhenL/C.

IfL=C,thetheoremfollows,by2.18,sincebothdenominatorsarezero.

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ExerciseSet2

1.ProvethatifA,B,Careanythreecollinearpoints,thend(AB)+d(BC)+d(CA)=0.

2.ProveTheorem2.10.3.ProveTheorem2.11.4.ProveTheorem2.17.5.Provethataninternalanglebisectorinatriangledividestheoppositesideintosegmentsproportionaltotheadjacentsides.

6.ProveTheorem2.12.7.LetA,B,C,Dbecollinearpoints.IfMandNarethemidpointsofABandCD,showthat2MN=AC+BD=AD+BC.

8.IfA,B,C,DarecollinearpointsandifthemidpointsofABandCDcoincide,showthatd(AC)=d(DB).

9.IfA,B,C,Dareanyfourcollinearpoints,thenprovethat

10.Stewart’stheorem.ProvethattheformulaofExercise2.9holdsevenwhenpointDdoesnotlieonlineABC.

11.ProvethatifA,B,C,Darecollinearpointssuchthatd(AC)=d(DB),thenthemidpointsofABandCDcoincide.

12.UseExercise2.10tofindthelengthsofthemediansofatriangle.13.UseExercise2.10tofindthelengthsoftheinternalanglebisectorsofa

triangle.14.LetthelengthsofthesidesoftriangleDBCbea,b,cforsidesBC,CD,DB.

LetaltitudeDAhavelengthh,andletd(BA)=dandd(AC)=esothata=d+e.Thenc2=d2+h2.Seetheaccompanyingfigure.UsetheserelationsalongwithExercise2.10toshowthat

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Exercise2.14

15.UseExercise2.14toproveHeron’sformulafortheareaKofatrianglewithsidesa,b,candsemiperimeters=(a+b+c)/2:

Thisalsoshowsthatthealtitudehtosideaisgivenby

16.UsetheaccompanyingfiguretoproveEuler’stheorem,Theorem2.16.

Exercise2.15

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SECTION3 IDEALPOINTSANDRATIOS

3.1 Although we shall work primarily in the Euclidean plane, we shalloccasionally use Euclidean space of three dimensions. The next definitionpermits both considerations. No picture of these ideal elements will bepresented, since they simply do not appear in ordinaryEuclidean figures. Thereader is urged to answer carefully and completely Exercises 3.1 and 3.2 toreinforcetheconceptofidealelements.

ManypropertiesofEuclideangeometryhaveratherunfortunatespecialcases.For example, two distinct points always determine exactly one line (passingthroughthetwopoints),buttwodistinctcoplanarlines(lineslyinginthesameplane) determine exactly one point (of intersection) only when they are notparallel. This deficiency can be remedied by imagining a point at infinity atwhich the two parallel linesmeet. Definition 3.2 provides the details of suchinfiniteelements.Ratiosofdivisionofasegmentthenservetotietogetherboththeideasofinfiniteelementsanddirectedmeasures.Sections4and5willusealltheseideas.

3.2DefinitionToeachEuclideanline,hereaftercalledanordinaryline,weaddoneidealpoint(orpointatinfinity)havingthefollowingproperties.1.Parallelordinarylinessharethesameidealpoint.2.Skeworintersectingordinarylineshavedistinctidealpoints.3.AlltheidealpointsbelongingtotheordinarylinesinagivenEuclideanplane,hereaftercalledanordinaryplane,formtheideallineofthatplane.4.Parallelordinaryplanessharethesameidealline.5.Intersectingordinaryplaneshavedistinctideallines.6.Alltheidealpoints(andideallines)inspaceformtheidealplane.7.Everyidealpointisconsideredtobeinfinitelyfarremovedfromeveryother(ordinaryorideal)point.

The Euclidean plane thus augmented is called the extended plane andEuclideanspacethusaugmentediscalledextendedspace.Points,aswellaslinesandplanes,thatarenotidealarecalledordinary.

3.3DefinitionLetAandBbeordinarypointsand letPbeanypointcollinearwithAandB.WedefinetheratiorinwhichPdividessegmentABby

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IfPisbetweenAandB,thenPissaidtodivideABinternally;ifP=AorP=B,thenPdividesABimproperly;otherwisePdividesABexternally.Inallcaseswewriter=AP/PB.

3.4 It follows that the ratio r inwhichP divides segmentAB can be any realnumber;r=1ifPisthemidpointofAB,forexample.IfPliesbetweenAandB,thenr>0;0<r<1 ifP is closer toA; andr>1 ifP is closer toB. IfB isbetweenAandP,thenr<–1,andr→–∞asP→B.IfAisbetweenBandP,then–1<r<0.IfPisideal,thenr=–1.IfP=A,r=0,andifP=B,r=∞.TheratiosofdivisionofsegmentABareindicatedforseveralpointsinFig.3.4.

Figure3.4

3.5Theorem IfAP/PB=AQ/QB,whereA andB aredistinctordinarypointsandPandQlieonlineAB,thenP=Q.

Fromthegivenequationwehave

Since the first and last fractions have equal nonzero numerators, theirdenominatorsareequaltoo.ThusPB=QB,soP=Q.

3.6DefinitionThecross ratio of four collinearpointsA,B,C,D, denotedby(AB,CD),isdefinedby

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3.7 Definition The cross ratio of four concurrent lines VA, VB, VC, VD,denotedbyV(AB,CD),isdefinedby

3.8TheoremIfanordinarytransversalmcutsfourconcurrentlinesVA,VB,VC,VDinthefourpointsA,B,C,D,thenthecrossratioofthefourlinesisequaltothecrossratioofthefourpoints.Thatis,V(AB,CD)=(AB,CD).

3.9TheoremIfA,B,C,Darecollinearordinarypoints,then

3.10DefinitionTheequationsofTheorem3.9areused todefine (AB,CD) incaseAorB isan idealpointandCandDaredistinctordinarypoints.That is,(AB,CD)=(CD,AB).

3.11Definition If (AB,CD) = –1, then pointsC andD are said todivideABharmonically, and D is called the harmonic conjugate of C with respect tosegmentAB.

3.12 Theorem IfC andD divideAB harmonically, thenA andB divideCDharmonically.

3.13TheoremTheharmonicconjugateofthemidpointofasegmentistheidealpointonthatline.

ExerciseSet3

1.Decidewhethereachstatementistrueorfalseinextendedspace.a)Eachlinehasexactlyoneidealpoint.b)Eachplanehasexactlyoneidealline.c)Eachtwodistinctplanesmeetinjustoneline.d)Thereareinfinitelymanyideallines.

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e)Thereareinfinitelymanyidealplanes.f)Thereareinfinitelymanyidealpoints.

2.Decidewhethereachstatementistrueorfalseintheextendedplane.a)Eachlinehasexactlyoneidealpoint.b)Eachlinehasatleastoneidealpoint.c)Eachtwodistinctlinesmeet(haveapointincommon).d)ThroughagivenpointPnotonagivenlinemtherepassesexactlyonelinemeetinglineminanidealpoint(aparallelline).

e)Iflinesmandnmeetatanidealpoint,andlinesnandpmeetatanidealpoint,thenlinesmandpmeetatanidealpoint.

f)Thereareinfinitelymanyidealpoints.g)Thereareinfinitelymanyideallines.

3.Drawatrianglehavingexactly:a)zeroidealvertices,b)oneidealvertex,c)twoidealvertices,d)threeidealvertices.

4.LocatethepointPthatdividessegmentABintheindicatedratio,whereAandBarepointswithCartesiancoordinates(0,0)and(1,0).?a)1b)2c)0d)∞e)–1f)–2g)h)–i)j)–

5.Verifythestatementsin3.4.6.Findtwoequivalentfractionswithequalnumeratorsandunequaldenominators.DoesthisinvalidatetheproofofTheorem3.5?

7.a)ProvethatiftwopointsdividesidesABandACoftriangleABCinthesameratio,thenthelinejoiningthesepointsisparalleltosideBC.

b)Isthetheoremstilltrueif“AC”isreplacedby“C4”?8.ProveTheorem3.8.9.ProveTheorem3.9.10.ProvethatifDistheharmonicconjugateofCwithrespecttoAB,thenCis

theharmonicconjugateofDwithrespecttoAB.

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11.a)Provethatif(AB,CD)=(AB,CE),thenD=E.b)Provethattheharmonicconjugateofagivenpointwithrespecttoagivensegmentcollinearwiththegivenpointisunique.

12.ProveTheorem3.12.13.ProveTheorem3.13.14.Thereare24differentwaysoftakingthecrossratiooffourdistinctpoints

A,B,C,D,fourofwhicharedisplayedinTheorem3.9.Showthatthese24arrangementsyieldonly6differentcrossratios,andthat,if(AB,CD)=r,theotherfivevaluesare1–r,1/r,r/(r–1),1/(1–r),and(r–1)/r.

15.Given(AB,CD)=r,showthat:a)interchangingthefirstandsecondpairsofpoints,orinterchangingthepointswithinthefirstpairandalsothosewithinthesecondpair,doesnotchangethevalueofthecrossratio,b)interchangingthepointswithinthefirstpaironly,orwithinthesecondpaironly,changesthevalueofthecrossratioto1/r,c)interchangingthefirstandfourthpointsonly,orthesecondandthirdpointsonly,changesthevalueofthecrossratioto1–r,andd)interchangingthefirstandthirdpointsonly,orthesecondandfourthpointsonly,changesthevalueofthecrossratiotor/(r–1).

16.IntheCartesianplane,letA(0,0)andB(1,0).ShowthatthepointPthatdividessegmentABintheratiorhascoordinates(r/(r+1),0).

SECTION4 THETHEOREMOFMENELAUS

4.1DefinitionLetABCbeatriangle.ApointPlyingonthelinedeterminedbyasideofthetriangleiscalledaMenelauspointforthatside.Ifthepointisnotavertexofthetriangle,itisaproperMenelauspoint.

4.2TheoremMenelaus’theorem.LetL,M,NbeMenelauspointsforsidesBC,CA,ABofordinarytriangleABC.ThenpointsL,M,Narecollineariff*

First, suppose thatL,M,N are properMenelaus points collinear on a baselinem.DropperpendicularsAR,BS,CToflengthsr,s,t,tolinemfrompointsA,B, C, as shown in Fig. 4.2. The following pairs of right triangles are similar

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becausc they either share an acute angle or have a pair of vertical angles forcorrespondingacuteangles:

Figure4.2

Hence

Since either one or three of the divisions must be external, this product isnegative.

Toestablishtheconverse,letL,M,NbeproperMenelauspointssuchthat

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Let the line joiningpointsL andM cut lineAB atpointN′.ThenL,M,N′ arethree collinear Menelaus points for triangle ABC, so by the first part of thisproof,

NowAN/NB=AN′/N′B,soN=N′byTheorem3.5.

4.3 Theorem Trigonometric form of Menelaus’ theorem. Let L, M, N beMenelauspointsforsidesBC,CA,ABofordinarytriangleABC.ThenpointsL,M,Narecollineariff

ThistheoremfollowsquitereadilyfromTheorems4.2and2.19.

As examples of the application of Menelaus’ theorem, the followingtheoremsaregiven.

4.4TheoremThelinejoiningthemidpointsoftwosidesofatriangleisparalleltothethirdside.

LetMandNbethemidpointsofsidesABandCAoftriangleABC.(SeeFig.4.4.)LetlineMNmeetsideBCatpointL.ByMenelaus’theorem,

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Figure4.4

ButCM/MA=AN/NB=1,sinceMandNaremidpoints.ThusBL/LC=–1,soLisanidealpoint;thatis,linesBCandMNareparallel.

4.5TheoremThelinestangenttothecircumcircleofatriangleatitsverticescuttheoppositesidesinthreecollinearpoints.

LetthetangenttothecircumcircleatAmeetlineBCatLasinFig.4.5.ThenBAL≅ C, sinceeachangle ismeasuredbyhalfofarcAB.Alsowehave

that LAC=180°– ABC, since theseanglesaremeasuredbyhalvesof thetwooppositearcsAC.Then

byTheorem2.19andthelawofsines.Sincethedivisionisclearlyexternal,theminussignholds.Similarly,

where M and N are the corresponding intersections of the tangents to thecircumcircleatBandCwiththelinesCAandAB.Now

sothetheoremfollowsfromMenelaus’theorem.

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Figure4.5

4.6DefinitionTwotrianglesABCandA′B′C′aresaidtobecopolariffthethreelinesAA′BB′CC′joiningcorrespondingverticesareconcurrent(inanordinaryor idealpoint) ; theyarecoaxial iff the threepointsL,M,Nof intersectionofcorrespondingsidesBCandB′C′,CAandC′A′,ABandA′B′arecollinear(onanordinaryoridealline).(SeeFig.4.6.)

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Figure4.6

4.7 Theorem Desargues’ two-triangle theorem. If two triangles are copolar,then they are coaxial ; conversely, if two triangles are coaxial, then they arecopolar.

Let triangles ABC and A′B′C′ be copolar at O. (See Fig. 4.7.) ApplyingMenelaus’theoremtotriangleOBCwithcollinearMenelauspointsL,C′,B′,totriangleOCAwithpointsM,A′,C′, and to triangleOABwithpointsN,B′,A′,obtain

Figure4.7

Multiplythesethreeequationssideforsidetoget

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Thus, sinceL,M, N areMenelaus points for triangleABC, thenL,M,N arecollinear;thatis,trianglesABCandA′B′C′arecoaxial.

Conversely,supposetrianglesABCandA′B′C′arecoaxialinpointsL,M,N.LetBB′ andCC′ meet atO. Now trianglesMCC′ andNBB′ are copolar atL.Hence,bythefirstpartofthisproof,thesetrianglesarecoaxial;thatis,pointsA,A′,Oarecollinear.ThustrianglesABCandA′B′C′arecopolaratO.

4.8 TheoremPappus’ theorem. If hexagonABCDEF has its verticesA, C, Elyingonone lineand itsverticesB,D,F lyingonanother line, then the threepointsL,M,NofintersectionofpairsofoppositesidesABandDE,BCandEF,CDandFAarecollinear.

LetABmeetCDatPandEFatR,andletCDandEFmeetatQ,asshowninFig.4.8.ApplyingMenelaus’theoremtotrianglePQRcutinturnbylinesFAN,MBC,andELD,obtain

Figure4.8

Now multiply these three equations side for side, and regroup the factors to

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obtain

SinceF,B,DandE,A,CformtwosetsofcollinearMenelauspointsfortrianglePQR,theneachofthelasttwotripleproductsinparenthesesisequalto–1.Thusthisequationreducesto

establishing, by the converse part of Menelaus’ theorem, that L, M, N arecollinear.

4.9Wehave seen in this section amost convenient test for the collinearityofthreepoints.Inthenextsectionweshallfindasimilartestfortheconcurrenceofthreelines.Theseideasarecalleddualsofoneanother.Thatis,twostatementsare called duals of one another if each is transformed into the other by theinterchangeofthewords“point”and“line,”and,ofcourse,alsosuchassociatedterms as “collinear” and “concurrent.” A striking illustration of duality is theDesarguestheorem,forcopolartrianglesandcoaxialtrianglesaredualconcepts.The theorem states that these two dual concepts are equivalent, provided, ofcourse,thatwepermitidealelements.

4.10 Inprojectivegeometry, inwhich theDesargues andPappus theoremsarebasic, extended space is always used, and the theorem of duality, a theoremabout theorems, states that the dual of any projective theorem is anotherprojective theorem. Observe that the Desargues theorem is its own dual. WeclosethissectionbystatingwithoutproofthedualofPappus’theorem.Compareits statement word for word with that of Pappus’ theorem, noting how eachconceptisdualized.

4.11TheoremThedualofPappus’theorem.Ifahexagon(withsides)a,b,c,d,e,fhasitssidesa,c,econcurrentinonepointPanditssidesb,d,fconcurrentinanotherpointQ, then the three lines l,m,n joiningoppositevertices (at theintersectionsof)abandde,beandefcdandfaareconcurrent(atapointR).(SeeFig.4.11.)

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Figure4.11

4.12 Now that Theorem 4.11 has been stated and illustrated, we find that itswordingmaybeimproved.Thisfactdoesnotatalldetractfromtheusefulnessofduality,forwehaveobtainedatleastonestatementofthetheorem,andthereareveryfewlongsentenceswhosewordingcannotbeclarified.Werestate thetheoremasfollows:Ifthesidesofahexagonpassalternatelythroughtwopoints,thenthethreediagonalsthatjoinoppositeverticesareconcurrent.

ExerciseSet4

1.TheproofofMenelaus’theoremassumedthattheMenelauspointswereproper.CompletetheproofofthistheorembyestablishingthetheoremforthecaseinwhichoneormoreMenelauspointsarenotproper.

2.RestateandproveMenelaus’theoremforatrianglehavingoneidealvertex.3.CanMenelaus’theorembeappliedtoatrianglehavingtwoorthreeidealvertices?

4.ProveTheorem4.3.5.GiventhatPandQarepointsonlinesABandACoftriangleABC,provethatPQisparalleltoBCiffAP/PB=AQ/QC.

6.ChoosepointPonsideABoftriangleABC.LettheparalleltoBCthroughPmeetsideACinQ,theparalleltoABthroughQmeetBCinR,theparalleltoACthroughRmeetABinS,theparalleltoBCthroughSmeetACinT,

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andtheparalleltoABthroughTmeetBCinU.ProvethatPUisparalleltoAC.

7.ProveDesargues’two-triangletheorem(Theorem4.7)forthecaseinwhichOisanidealpoint.

8.Provethatthethreeexternalanglebisectorsofatrianglemeettheoppositesidesinthreecollinearpoints.

9.Provethattwointernalanglebisectorsofatriangleandtheexternalbisectorofthethirdvertexmeettheoppositesidesinthreecollinearpoints.

10.PointsPandQareisotomicconjugateswithrespecttosegmentABiffA,B,P,Qarecollinearandd(AP)=–d(BQ);thatis,iffPandQlieonlineABandaresymmetricwithrespecttothemidpointofsegmentAB.ProvethatwhenL,M,NarecollinearMenelauspointsfortriangleABC,thentheirisotomicconjugatesL′,M′,N′arecollinear,too.

11.LinesAPandAQareisogonalconjugateswithrespecttoangleBACiffAPandAQaresymmetricwithrespecttothebisectorof BAC;thatis,iffd(BAP)=–d( CAQ).LetAL′,BM′,CN′betheisogonalconjugatesofAL,BM,CN,withLandL′,MandM′,NandN′lyingonlinesBC,CA,AB.ProvethatifL,M,Narecollinear,thenL′,M′,N′arecollinear.

12.GeneralizetheconceptofaMenelauspointtoapplytoaquadrilateral.ThenprovethatMenelauspointsL,M,N,OforsidesAB,BC,CD,DAofquadrilateralABCDarecollinearonlywhen

Istheconversetrue?13.Lettheincircle(circleinscribed)fortriangleABCtouchsidesBC,CA,AB

inpointsX,Y,Z,andextendYZtocutBCatK.ProvethatBX/XC=–BK/KC.

14.LetPbethemidpointofmedianAA′intriangleABC,andletCPcutABatQ.ProvethatAQ/QB= .

15.Surveyorshavetheproblemofdrawingalinebetweentwopointshavinganobstruction(suchasabuildingoramountain)intervening.ShowhowDesargues’two-triangletheoremcanbeusedtolocateotherpointsonthelinedeterminedbytwosuchpoints.

16.Theε-rulerproblem.ShowhowtodrawlineABwhenpointsAandBare

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muchfartherapartthanthelengthoftheavailablestraightedge.17.ShowhowtodrawalinethroughapointPandthroughtheinaccessible

pointofintersectionoftwogivenlinesmandn.

SECTION5 CEVA’STHEOREM

5.1DefinitionA line throughavertexofa triangle iscalledaCevian for thatvertex of the triangle. If it does not coincide with a side of the triangle, theCevian is called a proper Cevian. We agree that when stating that AL is aCevian,we imply thatA is thevertex andL is thepoint of intersectionof theCevianwiththeoppositesideofthetriangle.

5.2 TheoremCevcCs theorem. Three CeviansAL, BM, CN for triangleABCconcuriff

SupposeproperCeviansAL,BM,CNmeetatO,asinFig.5.2.ThenC,O,NandB,O,Mare triosofcollinearMenelauspointsfor trianglesABLandALC,respectively.So,byMenelaus’theorem,

Figure5.2

Now,multiplyingtheseequationssideforside,obtain

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Theconverse to this theorem is established in the samemanner aswas theconversetoMenelaus’theorem.

5.3TheoremTrigonometric form ofCeva’s theorem. ThreeCeviansAL, BM,CNfortriangleABCconcuriff

5.4TheoremLettheincirclefor(thecircleinscribedin)triangleABCtouchsidesBC,CA,ABatpointsX,Y,Z.ThenCeviansAX,BY,CZconcurinapointcalledtheGergonnepointforthetriangle.

SinceBXandBZaretangentsfromapointtoacircle(seeFig.5.4),thenZB=BX.SimilarlyXC=CYandYA=AZ.Then

The plus sign holds since points X, Y, Z divide the triangle internally. ThetheoremfollowsfromCeva’stheorem.

Figure5.4

5.5TheoremThethreealtitudesofatriangleconcur.

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ReferringtoFig.5.5,letD,E,FbethefeetofthealtitudesfromverticesA,B,CoftriangleABC.FromrighttriangleBAD,

Figure5.5

theminussignisusedwhen Bisacute,theplussignwhen Bisobtuse.Thussin BAD = ±cos B. Similar relations occurring in the other five righttriangleshavingasaleganaltitudeoftriangleABCyield

The plus sign holds, since either none or two of these feet D, E, F divideexternallythesidesonwhichtheylie.

5.6TheoremIfAL,BM,CNarethreeconcurrentCeviansfortriangleABC,and ifL′,M′N′ are isotomic conjugates (see Exercise 4.10) ofL,M, N withrespect to sides BC,CA, AB, then AL′, BM′,CN′ concur. The two points ofconcurrencearesaidtobeisotomicconjugatepointsfortriangleABC.

SinceLandL′areequidistantfromthemidpointofsideBC,thenBL=L′CandBL′=LC.(SeeFig.5.6.)SimilarrelationsholdforMandM′andforNandN′.Fromtheserelayionsitfollowsthat

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byCeva’stheorem,sinceAL,BM,CMconcur.HenceAL′,BM′,CN′concurbyCeva’stheorem.

Figure5.6

5.7 In the last section, Menelaus’ theorem was proved by general syntheticmeans, then in thissectionCeva’s theoremwasshown tobeaconsequenceofMenelaus’ theorem. The process may be reversed. In the exercises, you areaskedtoproveCeva’stheoremsynthetically.So,assumingCeva’stheoremhasbeen so established, let usproveMenelaus’ theorem. Its converseneednot beprovedagain.

5.8TheoremCeva’stheoremimpliesMenelaus’theorem.Let L,M, N be three collinear Menelaus points for sides BC, CA, AB of

triangleABC,asinFig.5.8a.LetCNmeetBMinPandALinQ,andletALandBMmeetin

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Figure5.8a

R.WeapplyCeva’stheoremtoeachofthesixtriangleslistedinFig.5.8b.Bymultiplyingthesesixequationssideforside,weobtain

from which the theorem follows, since either one or three of the collinearMenelauspointsdividethesidesexternally.

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Figure5.8b

5.9TheoremIfAL,BM,CNarethreeconcurrentCeviansfortriangleABCandifMNmeetsBCatL′then(BC,LL′)=–1;thatis,

Applying Ceva’s theorem to triangle ABC of Fig. 5.9 and the concurrentCevians AL, BM, CN, and applying Menelaus’ theorem to triangle ABC andcollinearMenelauspointsL′M,N,obtain

Thetheoremfollowswhenwedividethesetwoequationssideforside.

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Figure5.9

5.10 Now that we have stated, proved, and examined Ceva’s theorem, it isappropriate to remind you of the remarks in 4.9, that Ceva’s and Menelaus’theorems are duals. First, a triangle itself is a self-dual figure: It is the figureformed by three noncollinear points and the three lines joining pairs of thepoints; it is also the figure formed by three nonconcurrent lines and the threepointsofintersectionofpairsofthelines.Thusatriangleisalsoatrilateral.

WhereMenelaus’ theorem discusses the collinearity of threepoints on thelines (sides) of a triangle, Ceva’s theorem considers the concurrence of threelineson(through)thepoints(vertices)ofatriangle.Soeachisdualtotheother.It is odd that, although Menelaus’ theorem was known by 100 A.D., Ceva’stheoremescapeddetectionuntil1678,morethan1500yearslater.

ExerciseSet5

1.ProvetheconversetoCeva’stheorem.2.TheproofofCeva’stheoremgiveninthetextassumedthattheCevianswereproper.Provethetheoremwiththisrestrictionremoved.

3.ProveTheorem5.3.4.ProveCeva’stheoremsynthetically.[Hint:DrawalinethroughAparalleltoBCtomeetBMinSandCNinT.Thenusesimilartriangles.]

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5.ProveTheorem5.5forthecaseinwhich B=90°.6.ShowhowtoconstructtheharmonicconjugateofapointClyingonlineABwithrespecttosegmentAB,7.LetAL,BM,CNbeCeviansfortriangleABC.LetL′M′N′bethepointsofintersectionofMNandBC,NLandCA,LMandAB.ProvethatL′M′N′arecollineariffAL,BM,CNconcur.

8.Showthatthemediansofatriangleconcur.Theirpointofconcurrenceiscalledthecentroidofthetriangle.

9.ShowthatthecentroidistheonlypointwhoseCeviansdividethesidesofthetriangleinequalratios.

10.ShowthatthecentroidistheonlypointwhichdividesitsCeviansinequalratios.

11.Showthatthethreeinternalanglebisectorsofatriangleconcur.12.Showthatoneinternalandtwoexternalanglebisectorsofatriangleconcur.13.ShowthatifthreeCeviansconcur,thentheirisogonalconjugatesconcur

(seeExercise4.11).14.AcircleintersectssidesBC,CA,ABoftriangleABCinpointsLandL′M

andM′NandN′.ShowthatAL,BM,CNconcuriffAL′BM′CN′concur.

SECTION6 SOMEGEOMETRYOFTHETRIANGLE

6.1It isconvenienttointroducesomestandardterminologyandsymbolismforcertain points, lines, and other objects related to a given triangle ABC. Theexistence of some of these objects will be established later in this chapter.Following Definition 6.2 and continuing through Section 7, various theoremsconcerning these defined elements are presented. The purpose of thisdevelopment is to familiarize the reader with some of the basic properties oftriangles andwith themethods of proving them. Perhaps the item of greatestinterestinSections6and7isthatwhichdescribesthefamousninepointcircle.TherelevantinformationiscontainedinDefinition6.2andTheorems7.13and7.14,andsummarizedin7.15.

6.2 Definition In triangle ABC the measures of the angles A, B, C will bedenotedbyA,B,C.Thesidesoppositetheseverticeswillhavelengthsa,b,c,respectively,andthesemiperimeter(a+b+c)/2isdenotedbys.Thetermsideof a triangle will refer to either the segment or the entire line on which thesegmentlies.Thecontextwillmaketheuseclear.

Themidpointsofthesidesa,b,caredenotedbyA′,B′C′.ThelinesAA′BB′,

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CC′arecalledmediansandhavelengthsma,mb,mc.TriangleA′B′C′iscalledthemedial triangle.Thefeetof thealtitudesaredenotedbyD,E,F.ThealtitudesAD,BE,CFhavelengthsha,hb,hc.TriangleDEF iscalledtheorthictriangle.TheinternalanglebisectorsmeettheoppositesidesatU,V,W,andthelengthsofthebisectorsAU,BV,CWareta,tb,tc.

ThecircleinsidetriangleABCandtangenttoitssidesistheinscribedcircle(orincircle)withcenterIandradiusr.ThethreecirclesexteriortotriangleABCandtangenttoitssides(seeFig.6.2)arecalledexcirclesandhavecentersIa,Ib,Icandradiira,rb,rc.Theincircle touchesthesidesof triangleABCatX,Y,Z,andtheexcircleIjtouchesthesidesatXj,Yj,Zj(forj=a,b,c).Thesefourcirclesarealsocalledequicircles.

The circle through the vertices A, B, C is called the circumcircle and itscenterandradiusaredenotedbyOandR.

We useG to denote the centroid (the meeting of the medians),H for theorthocenter (themeeting of the altitudes), andN for theninepoint center (thecenterofthecirclethroughthemidpointsofthesides).TheareaoftriangleABCisdenotedbyKorbyKABC.

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Figure6.2

6.3 Theorem Themedial triangle of a given triangle (assumed to be triangleABC)hassidesparalleltoandhalfthelengthofthesidesofthegiventriangle.

SinceAC′=AB/2andAB′=AC/2and C′AB′= BAC(seeFig.6.3).thenΔBAC ~ (is similar to) ΔC′AB′ by SAS (two pairs of corresponding sidesproportional and the included angles congruent). ThusC′B′≅ BC/2≅ BA′.SimilarlyA′B′≅BC′,soBA′B′C′isaparallelogram.Thetheoremfollows.

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%

Figure6.3

6.4 Corollary The medial triangle partitions the given triangle into fourcongruenttriangles.

6.5CorollaryTheninepointradiusishalfthecircumradiusofthetriangle.

6.6 Theorem The three medians concur at a point (the centroid) that trisectseachmedian.

LetGdenotetheintersectionofmediansBB′andCC′,asshowninFig.6.6,andletPandQbethemidpointsofsegmentsBGandCG.SincesegmentsB′C′andPQ are each parallel to and half the length ofBC by Theorem 6.3, thenPQB′C′isaparallelogram,soitsdiagonalsPB′andQC′bisecteachother.Thatis,PG=GB′andQG=GC′.SincealsoBP=PGandCQ=QG,itfollowsthatBB′ andCC′ trisect eachother.By symmetry,AA′ is also trisectedbyG. (SeeExercise5.8.)

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Figure6.6

6.7Theorem Themedians of a triangle, as vectors, form a trianglewhoseareaisthree-fourthsoftheareaofthegiventriangle.

ExtendmedianAA′ itsownlengthtopointPtoformΔPCB≅ΔABC.(SeeFig. 6.7.) LetB″ be the midpoint ofPC. NowBB″CC is a parallelogram, sovector

Figure6.7

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CC′ is equal to vectorB″B. Since segmentB′B″ is parallel to and has lengthequaltohalfofAP, thenvectorB′B″equalsvectorAA′.ThustriangleBB′B″ isformedfromvectorsequaltothemediansoftriangleABC.Theproofoftherestofthetheoremisleftasanexercise.

6.8TheoremAtriangleanditsmedialtrianglehavethesamecentroid.

6.9TheoremIftwomediansofatrianglearecongruent,thenthetriangleisisosceles.

Supposethatmb=mc.ThenuseTheorem6.6toobtainΔBGC′≅ΔCGB′.

6.10TheoremTheinternalanglebisectorsofatriangleconcurattheincenterofthetriangle.

Let I denote the intersection of the internal bisectors of anglesA andB intriangleABC,asinFig.6.10.DropperpendicularsIX,IY,IZfromItothesidesBC,CA,AB.SinceBIisthebisectorofangleB,thentrianglesBIXandBIZarecongruent

Figure6.10

(byAAS),soIX≅IZ.SimilarlyIZ≅IY.ThenI isequidistant fromthe threesides, so I is indeed the incenter of the triangle.SinceΔCIY≅ΔCIX byHL(hypotenuseandlegofonerighttrianglecongruenttothecorrespondingpartsofasecondrighttriangle),thenCIbisectsangleC.

6.11TheoremTwoexternalanglebisectorsandtheinternalbisectorofthethird

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angleofatriangleconcurateachexcenterofthetriangle.

6.12TheoremTheexternalandinternalbisectorsofagivenangleofatrianglecut the circumcircle again at the extremities of that circumdiameterperpendiculartotheoppositesideofthetriangle.

If triangle ABC is isosceles with vertex A, then bisector AU is thecircumdiameter perpendicular to BC, and the external bisector of angle A istangenttothecircumcircle.Sothetheoremistruewhenthepointoftangencyisconsideredtobethesecondpointofintersectionoftheexternalbisectorandthecircumcircle.

SoassumeAB≅AC.Let the internalandexternalbisectorsofangleAcutthecircumcircleagainatSandT.(SeeFig.6.12.)SinceASbisectsangleBAC,italso bisects arc BC. Thus the circumdiameter through S is the perpendicularbisectorofsideBC.WeneedonlyshowthenthatSTisadiameter.Thisistrue,since SAT=90°.

Figure6.12

6.13CorollaryTheincenterofatriangleistheorthocenterofthetriangleIaIbIcformedbythethreeexcentersofthegiventriangleandtheverticesofthegiventrianglearethefeetofthealtitudesoftriangleIaIbIc.

6.14CorollaryEach internalanglebisectorofa trianglebisects thearcof thecircumcirclecutoffbytheoppositesideofthetriangle.Hencethisbisectorand

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theperpendicularbisectorof thisoppositesidemeeton thecircumcircleof thetriangle.

6.15TheoremTheareaofatriangleisequaltotheproductofitsinradiusandsemiperimeter;thatis,K=rs.

Figure6.15

In triangleABC with incenter I (see Fig. 6.15), the areas of trianglesBCI,CAI,andABIaregivenby

Thetheoremfollowswhenoneaddsthesethreequantities.

6.16 Theorem The area of a triangle is equal to the product of a givenexradiusrjandthesemiperimeterdiminishedbysidej.Thatis,

6.17TheoremHeron’sformula.TheareaoftriangleABCisgivenby

IntriangleABC,letx=m(BD),asshowninFig.6.17.Then

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fromwhichoneobtains

Figure6.17

Now

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Fromthisequationweobtain

6.18 For a synthetic geometric proof of Heron’s formula, see H. Eves’ AnIntroductiontotheHistoryofMathematics,thirdedition,pages171–172.

6.19CorollaryTheareaofatriangleisgivenby

6.20TheoremTheproductoftwosidesofatriangleisequaltotheproductofthecircumdiameterandthealtitudetothethirdside.

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LetcircumdiameterAOoftriangleABCcutthecircumcircleagainatP.(SeeFig.6.20.)Then ABC≅ APC,sinceeachangleismeasuredbyhalfofarcAC.

Figure6.20

ThusrighttrianglesABDandAPCaresimilar,so

Thatis,cb=2Rha.

6.21CorollaryTheareaoftriangleABCisgivenby

6.22 Theorem The midpoint of a side of a triangle is the midpoint of thesegmentonthatsidewhoseextremitiesarethepointsofcontactof theincircleandtheexcirclenamedforthatside.InFig.6.22,sincetwotangentsfromapointtoacirclearecongruent,wehave

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Figure6.22

ButAZa=AYa,so

ThusCXa=CYa=AYa–AC=s–b.AlsoBX=BZ=AB–AZ=c–AYandBX=BC–XC=a–YC,so

whence

fromwhichthetheoremfollows.

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ExerciseSet6

1.ReviewthetheoremsinBookIofEuclid’sElements.2.ProveCorollary6.4.3.ProveCorollary6.5.4.Provethatthediagonalsofaparallelogrambisecteachother.5.CompletetheproofofTheorem6.7.6.ProveTheorem6.8.7.ProveTheorem6.9.8.ProveTheorem6.11.9.Provethattheinternalandexternalbisectorsofagivenangleareperpendicular.

10.ProveCorollary6.13.11.ProveCorollary6.14.12.ProveTheorem6.16.13.ProveCorollary6.19.14.ProveCorollary6.21.15.Thediameterofasemicircleisdividedintotwosegmentsaandbbyits

pointofcontactwithaninscribedcircleasintheaccompanyingfigure.Showthatthediameterdoftheinscribedcircleistheharmonicmeanofaandb;thatis,d=2ab/(a+b).(PiMuEpsilonJournal,spring1970,page237.)

Exercise6.15

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16.Giventhatacentralangleinacircleandthearcitinterceptshavethesamemeasure,provethesetheorems.a)Anangleinscribedinacircleortheanglebetweenachordandatangentismeasuredbyhalfitsinterceptedarc.

b)Theanglebetweentwotangents,twosecants,oratangentandasecanttoacircleismeasuredbyhalfthedifferencebetweentheinterceptedarcs.

c)Theanglebetweentwochordsinacircleismeasuredbyhalfthesumofthearcsinterceptedbyitanditsverticalangle.

17.Provethatthetwotangentsfromapointtoacirclearecongruent.18.Provethattheproductofthewholesecantanditsexternalsegmentdrawn

fromapointtoacircleisequaltothesquareofthetangentfromthatpoint.19.Provethatiftwochordsinacircleintersect,thentheproductofthe

segmentsintowhichonechordisdividedisequaltotheproductofthesegmentsoftheother.

20.Provethat1/r=1/ra+1/rb+1/rc.

21.Provethatrbrc+rcra+rarb=s2.22.a)Provethatinanytrianglesin A=a/2R.

b)Deducethelawofsines:a/sin A=b/sin B=c/sin C.23.Provethatthebisectorofanangleofatrianglealsobisectstheangle

betweenthealtitudeandthecircumdiameterissuingfromthatvertex.

SECTION7 MOREGEOMETRYOFTHETRIANGLE

7.1TheoremTheperpendicularbisectorsofthesidesofatriangleconcuratthecircumcenterofthetriangle.

Since the circumcenter O is equidistant from all three vertices, it isequidistant from the endpoints of any side of the triangle. Thus it lies on theperpendicularbisectorofeachside.

7.2TheoremInanytriangle,R=(ra+rb+rc–r)/4.

InFig.7.2,letIA′meetIaXaatP,andletthecircumdiameterthroughA′meetthecircumcircleandtheinternalandexternalbisectorsofangleAatSandT(byTheorem6.12).SinceA′bisectsXXa(byTheorem6.22)andIXandPXaareboth

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Figure7.2

perpendiculartoXXa,thenIXPXaisaparallelogram,soIX=PXa,andIaP=ra–r.SinceA′S isperpendicular toatA′ it follows thatA′S joins themidpointsofsidesIPandIIaoftriangleIPIa.HenceA′S=(IaP)/2=(ra–r)/2.

ByCorollary6.13, IcBIb= IcCIb=90°,sothecircleonIcIbasdiameterpassesthroughBandC.ItfollowsthatTisthemidpointofIcIb.FromtrapezoidIcXcXbIb,wehave

Nowwehavefromwhichthetheoremfollows.

7.3CorollaryThecircleonanytwoexcentersasdiameterpassesthroughthe

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twovertices of the triangle that are not collinearwith the excenters. (SeeFig.7.2.) 7.4CorollaryThemidpointofasideofatriangleisthemidpointofthesegmentcutofffromthatsidebythepointsoftangencyofthetwoexcirclesnotnamedfor thatside. (SeeFig.7.2.) 7.5CorollaryThesegment joining twoexcenters of a triangle is bisected by the circumcircle of the triangle. If thetriangle is not isosceles when viewed from the vertex collinear with theequicenters,thenthatmidpointisnotavertex.(SeeFig.7.2.)7.6TheoremThealtitudesofatrianglearetheanglebisectorsoftheorthictriangle.

ThecircleonsideBCoftriangleABCasdiameterpassesthroughthefeetEandFofthealtitudesBEandCF,since BFC= BEC=90°.(SeeFig.7.6.)Thus

Figure7.6

BCEFisacyclicquadrilateral;thatis,itcanbeinscribedinacircle,so FEC+B=180°.Thus FEC DEA, sinceBDEA is alsocyclic, and sinceBE is

perpendicular to AC, then BE bisects angle DEF. Similarly the other twoaltitudesbisecttheotheranglesoftheorthictriangle.

7.7CorollaryThecircleonthesideofatriangleasdiameterpassesthroughtheoppositeverticesoftheorthictriangle.

7.8 Corollary The sides of the orthic triangle form with those of the giventrianglethreetrianglessimilartothegiventriangle.

InFig.7.6, B=180°– FEC= FEA.HenceΔABC~ΔAEFbyAA(two angles of one triangle congruent to the corresponding two angles of the

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secondtriangle).SimilarlyΔABC~ΔDBF~ΔDEC.

7.9 Theorem The tangent to the circumcircle of a triangle at a vertex of thetriangleisparalleltotheoppositesideoftheorthictriangle.

ThistheoremfollowsfromtheproofofTheorem4.5andfromCorollary7.8.

7.10CorollaryTheanglemadeatanendpointofonesideoftheorthictrianglewith the side of the given triangle onwhich it terminates is congruent to theanglebetweentheothertwosidesofthetriangle.

TheresultfollowsfromTheorem7.8.

7.11CorollaryThealtitudesofatriangleconcur.ByTheorem6.10appliedtotheorthictriangle.

7.12 Theorem The product of the segments into which the orthocenterdivideseachaltitudeisaconstantforagiventriangle.

Any two altitudes, sayAD andBE (see Fig. 7.6), are chords of the samecircle, by Corollary 7.7. Hence the products of the segments into which theydivideeachotherareequal,byExercise6.19.Thatis,AH·HD=BH·HE.Bysymmetry,itfollowsthatthiscommonvalueisalsoequaltoCH·HF.

7.13TheoremThefeetofthealtitudesofatrianglelieonitsninepointcircle.

ReferringtoFig.7.13,AC′≅C′D,sinceC′isthemidpointofthehypotenuseof right triangleABD. Thus triangleC′DA is isosceles, and sinceC′B′A′B is aparellelogram, it follows thatC′B′A′D is an isosceles trapezoid, so the circlethroughC′B′andA′alsopassesthroughD.SimilarlyEandFlieonthiscircle,theninepointcircle.

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Figure7.13

7.14TheoremThemidpointsNa,Nb,Ncof thesegmentsAH,BH,CH joiningtheverticestotheorthocenterlieontheninepointcircleoftriangleABC.

InFig.7.13,C′NajoinsthemidpointsofsidesABandAHoftriangleABH,soC′NaisparalleltoBE,henceperpendiculartoACandtoA′C′.Thatis, NaC′A′=90°.SimilarlyNaB′isparalleltoCH,so NaB′A′=90°.HencethecircleonNaA′asdiameterpassesthroughB′andC′,soitistheninepointcircle.SimilarlytheninepointcirclepassesthroughNbandNc,too.

7.15Weseethattheninepointsontheninepointcirclearethemidpointsofthesides, the feet of the altitudes, and themidpoints of the segments joining theorthocentertoeachofthevertices.

7.16CorollaryTheninepoint center liesmidwaybetween theorthocenter andthecircumcenter.Infact,anysegmentjoiningtheorthocentertoapointonthecircumcircleisbisectedbytheninepointcircle.

SincetheninepointcenterAlliesontheperpendicularbisectorsofDA′andofB′E,thefirstpartofthiscorollaryfollows.(SeeFig.7.16.)

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%

Figure7.16

7.17TheoremLetHbetheorthocenterfortriangleABC.ThenA,B,CaretheorthocentersfortrianglesHBC,AHC,ABH,respectively.

7.18 Definition Four points such that each is the orthocenter of the triangleformedbytheotherthreearesaidtoformanorthocentricquadrangle.

7.19TheoremThe four trianglesof anorthocentricquadranglehave the sameorthic triangle, thesameninepointcircle,andequalcircumradii. (SeeFigs.7.6and 5.5.) 7.20 Theorem The circumcenters of the four triangles of anorthocentric quadrangle formanother orthocentric quadrangle having the sameninepoint circle. Furthermore, the four points of either quadrangle are thecircumcentersofthetrianglesoftheotherquadrangle.

Let thecircumcentersof trianglesABC,HBC,AHC,ABHbedenotedbyO,Oa,Ob,OcasinFig.7.20.SinceObOcistheperpendicularbisectorofAH,thenit isparallel toBC.SinceOOa isperpendiculartoBC,O liesonthealtitudetovertexOa of triangleOa Ob Oc. It follows thatO is the orthocenter of thistriangle.Hencethefourcircumcentersformanorthocentricquadrangle.

Since N bisects HO (by Corollary 7.16), then N also bisects AOa, thecorrespondingsegmentfortriangleHBC.ItfollowsthatquadrangleOOaObOcis

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symmetric to quadrangle HABC in point N. Now, since ObOc is theperpendicular bisector ofAH, thenBC is the perpendicular bisector ofOa O.ThusA′,B′,C′arethemidpointsofthesegmentsjoiningtheverticesOa,Ob,OctotheorthocenterOoftriangleOaObOc.Thetheoremfollows.

Figure7.20

7.21Herewefindeighttrianglesallhavingthesameninepointcircle.Toshowjust how rich such a figure is,we commentwithout proof that the nineteenth-century German mathematician Karl Wilhelm Feuerbach proved that theninepoint circle is tangent to eachof the four equicircles of the given triangle(seeTheorem44.6). FromTheorem7.19, the four triangles of an orthocentricquadrangleprovideatotalof16equicircles,alltangenttotheninepointcircleofthe quadrangle. (Try drawing all these circles.) By Theorem 7.20, itscircumcentersprovidefourmoretriangleshavingthissameninepointcircle,sotheyinturnprovideanother16equicirclestangenttoit.Hencefromonetriangleweobtainagrandtotalof32equicircles,alltangenttotheninepointcircle!

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7.22Weconcludethissectionwithafaulty“proof”ofafalse“theorem.”Recallthe theoremswe have proved in this and the preceding sections to locate thefallacyintheargument.Itshouldhelptodrawamostaccuratefigure,orseveralaccuratefigures.

7.23False“Theorem”Alltrianglesareisosceles.

IfAB≅AC,wearedone.SosupposeitisnotgiventhatAB≅AC.Letthebisector of angleA and the perpendicular bisector ofBCmeet atP. (See Fig.7.23.)DrawBPandCPanddropperpendicularsPQandPRtosidesCAandAB.ThenΔAPR≅ΔAPQbyAAS.,soAR≅AQandPR≅PQ.AlsoΔBPA′≅ΔCPA′bySAS,soBP≅CP.NowΔBPR≅ΔCPQbyHL,soRB£QC.Then

sotriangleABCisisosceles. (?)

Figure7.23

ExerciseSet7

1.ProveCorollary7.3.2.ProveCorollary7.4.3.ProveCorollary7.5.4.Provethattheoppositeanglesofaconvexcyclicquadrilateralaresupplementary.

5.ProveCorollary7.7.

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6.ProveCorollary7.11.7.Provethatthemidpointofthehypotenuseofarighttriangleisequidistantfromitsthreevertices.

8.ProvethesecondpartofCorollary7.16.9.ProveTheorem7.17.10.ProveTheorem7.19.11.Findthefallacyinthe“proof”of“Theorem”7.23.12.DrawthediametersPAandPBintwocirclesintersectingatP,asshownin

theaccompanyingfigure.LetABcutthetwocirclesinRandSasshown.Then PRA=90°and PSB=90°,sinceeachangleisinscribedinasemicircle.ThusPRandPSaretwoperpendicularsfromapointPtoalineAB.Findthefallacyinthisargument.

Exercise7.12

13.ProvethatONaandAA′intriangleABCbisecteachother.

14.Provethat(AH)2+(BC)2=4(AO)2intriangleABC.15.Provethat,intriangleABC,HA′andAOconcuronthecircurncircle.16.Provethatthefourcentroidsofthetrianglesofanorthocentricquadrangle

formasimilarorthocentricquadrangle.17.Provethatthefourverticesofanorthocentricquadranglearethecentroids

ofanothersimilarorthocentricquadrangle.18.Provethattheequicentersofatriangleformanorthocentricquadrangle

whoseninepointcircleisthecircumcircleofthegiventriangle.19.Provethattheradiiofthecircumcircletotheverticesofatriangleare

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perpendiculartotheoppositesidesoftheorthictriangle.20.ProveTheorem7.2algebraicallyusingTheorems6.15through6.21.

SECTION8 GEOMETRICCONSTRUCTIONS

8.1OneoftheoldestgamesintheworldisthegameofEuclideanconstructions.Much energy has been expended discovering just what quantities can beconstructedusingonly a compass and straightedge (anunmarked single-edgedruler).

8.2Equallylargeamountsofefforthaveshownthatcertainquantitiescannotbeconstructed—thatit isimpossible toconstructthem.Noconstructionofafinitenumberofstepswilleverbefound,usingjustEuclideantools, thatwillenableone to trisect every general angle (trisect an angle), to draw a square havingexactlythesameareaasagivencircle(squareacircle),ortoconstructtheedgeofacubehavingjusttwicethevolumeofacubewhoseedgeisgiven(duplicateacube).Thesearethethreeso-calledclassicalconstructionproblems.

Amateurgeometersnotfamiliarwiththecontentofsuchimpossibilityproofsstillinventconstructionsthatappeartosolvetheseancientproblems.Athoroughexamination of each such construction will uncover errors. Any suchconstruction is either only an approximation (and some are excellentapproximations) or else it is incorrect in that it uses the Euclidean toolsimproperlyormakesuseofothertools.

8.3 It can be shown that, given any set of numbers (line lengths), one canconstruct thesum,difference,product,andquotientofany twonumbers in theset, hence any rational combination of these numbers.Also one can constructsquare roots of numbers (see 8.11). And these are the only constructiblemagnitudes.Hencecuberootsorfifthrootsortranscendentalnumberscannotbeconstructed generally. Each of the classical construction problems involvesfindingsuchaninconstructiblenumber.Theexercisespursuethismatterfurther,andacompletediscussionisfoundinthereferenceslistedinExercise40.7.

8.4NotationWedenote thecirclewithcenterA and radiusr=BC byA(r)orA(BC).ThecirclewithcenterAandpassingthroughpointPisdenotedbyA(P).

8.5Weassumethatthereaderisfamiliarwiththebasicconstructionofthesumand difference of given lengths, of angles congruent to given angles, and ofperpendiculars and parallels to given lines. The details of such basic

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constructionsappearinAppendixB.

8.6 Our first construction involves another ancient problem. The first threepostulates of Euclid’s Elements describe the uses of the straightedge andcompass.Theystate:1.Alinesegmentcanbedrawnbetweenanytwopoints.2.Alinesegmentcanbeextendedindefinitely.3.Acirclecanbedrawnhavinganygivenpointascenterandpassingthroughanyothergivenpoint.

Thefirsttwopostulatesindicatehowthestraightedgeistobeusedandthethirddescribestheuseofthecompass.(AcompletelistofEuclid’spostulatesappearsinAppendixA.)8.7According to the third postulate, Euclid’s compass couldnotbeusedtotransferdistances.Hiscompasscouldbesetontwopointsandanarcswung,buttheliftingofeitherpointfromthepaperwouldcausethecompassto lose its setting, to collapse. The modern compass can be used to transferdistances.That is, onemay draw the circleA(BC)with amodern compass bysettingitspointsonBandC,thenliftingthecompassandmovingittopointA.

ThequestionariseswhethermoderncompassescandomorethancollapsingEuclidean compasses. Certainly they can do no less. To show that the twocompassesareindeedequivalent,wepresentthefollowingtheorem.

8.8TheoremThemodernandEuclideancompassesareequivalent.

WeshallconstructthecircleA(BC)usingaEuclideancompass.Notethattheconstructiondoesnotrequiretheaidofastraightedge.

GivenpointsA,B,C,asinFig.8.8,drawcirclesA(B)andB(A)tointersectatPandQ.NowdrawcirclesP(C)andQ(C)tomeetagainatD.CircleA(D)isthedesiredcircleA(BC).

%

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Figure8.8

TheconstructionisprovedbyobservingthatPandQlieontheperpendicularbisectorsofsegmentsABandCD.ThusAandDarethereflectionsofBandCinlinePQasmirror,soAD≅BC.

8.9AsshownintheproofofTheorem8.8,aconstructionconsistsoftwoparts;first,theexplanationofthestepsonetakestofindthedesiredobject;andsecond,a proof that the constructed object is precisely thatwhichwas desired. In theremainderofthissectiontheproofsareleftforthereadertosupply.

8.10ProblemConstructthemeanproportionalbetweentwosegments.

Wearegiven twosegmentsof lengthsa andb andweare required to findtheirmeanproportional;thatis,weneedtofindalengthxsothata/x=x/b;thatis, x = . On a line (see Fig. 8.10) mark AP = a and PB = b. Draw asemicircle onAB as diameter and let the perpendicular toAB atP meet thissemicircleatpointX.NowPX=x.

Figure8.10

8.11Toconstruct ,weuse theconstructionofProblem8.10, takingb=1.Henceitisnecessarytohaveaunitsegmentavailablewhentakingsquarerootsoflengths.Aunitsegmentisrequiredalsoformultiplicationandfordivision,asProblem8.12shows.

8.12ProblemFindtheproductoftwosegments.

Given segments of lengths a, b, and 1, we shall use proportion in similartrianglestoconstructalengthx=ab.Onaline,wemarkPU=1andUB=b.OnanyotherraythroughP,wemarkPA=a(seeFig.8.12).Welettheparallelto

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AUthroughBcutPAatX.ThenAXhaslengthx=ab.

Figure8.12

8.13ProblemInscribeasquareinagivensemicircle.FromthesolvedproblemasshowninFig.8.13a, let thesquarePQRShave

sidesandthesemicircleAOBhaveradiusr.Bysymmetry,OQ=s/2.SinceOR=randQR=s,then(s/2)2+s2=r2,sos=2r/ .Hencewemustconstruct2r/

.GiventhesemicircleAOB,drawrighttriangleEFGhavinglegEGtwicethe

lengthoflegFG.(SeeFig.8.13b.)OnhypotenuseEF,markEH=r.ThenEJ=s,whereJ is the foot of the perpendicular dropped fromH toEG.Nowmarklengths/2=m(HJ)oneithersideofpointOonlineAB togetpointsPandQ.Thesquareiseasilycompleted.

Figure8.13a

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Figure8.13b

8.14ProblemOnagivensegmentashypotenuse,constructarighttriangleequal(inarea)toagiventriangle.

LetthesegmentbePQandthetriangleABC,asinFig.8.14a.DrawaltitudeAD.Thenconstructx=AD·BC/PQ,asshowninFig.8.14b.Let theparallel toPQ, x units distant fromPQ, cut in pointsR andR′ the semicircle onPQ asdiameter.ThenPQRisthedesiredtriangle.

Figure8.14a

Figure8.14b

8.15ProblemConstructtriangleABC,giventa,hb,c.

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AtapointEonabaselinem,erectaperpendicularhbunitsinlengthtopointB,asinFig.8.15.DrawcircleB(c)tocutmatA.BisectbothanglesatAandontheseisectorsmarkAU1andAU2,eachoflengthta.LetBU1andBU2cutmatC1andC2.TheneachoftrianglesABC1andABC2isasolution.

Figure8.15

8.16DefinitionA rusty compass is a compasswhoseopening is fixed andcannotbealtered.

8.17 Many interesting problems occur when the permissible uses of theEuclideantoolsarealteredslightly.Considertheproblemofperformingvariousconstructions with a compass whose opening cannot be changed (perhapsbecause it has rusted). These so-called rusty compass problems can be quitechallenging. It has been shown that every construction possiblewith compassandstraightedgecanbedonewithrustycompassandstraightedge(insofarasthedesiredobjectsarepointsandlines).Foracompletediscussion,seeEves’SurveyofGeometry(Vol.1,Section4.5).

8.18 Problem Construct a perpendicular at a point on a line using a rustycompass.

Given pointP on linem and letting the fixed compass opening be r, drawcircleP(r)tocutmatAandB.DrawcirclesA(r)andB(r)tocutsemicircleP(r)atCandD,asshowninFig.8.18.NowletcirclesC(r)andD(r)meetatQ(theyalsomeetatP).ThenPQisthedesiredperpendicular.

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Figure8.18

8.19ProblemUsingarustycompass,constructasegmentofagivenlengthABatpointPonalinem.

DrawAP. Construct a linen throughP parallel toAB as follows (see Fig.8.19). Erect a perpendicular p to line AB so that p passes within a compasslength ofP. Then drop a perpendicular n fromP to line p. Similarly draw aparallel to AP through B to meet line n atC. Now PC≅ AB. Draw line rbisectingtheanglebetweenlinesmandn,andlettheperpendicularfromCtorcutmatapointQ.ThenPQ≅AB.

Figure8.19

ExerciseSet8

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1.Completeeachproblembyprovingthatthegivenconstructioniscorrect,a)Problem8.10b)Problem8.12c)Problem8.13d)Problem8.14e)Problem8.15f)Problem8.18g)Problem8.19

2.Constructthequotientoftwosegments.3.Showhowtobisectananglegreaterthan120°usingarustycompass.4.Atagivenpointonagivenlineconstructananglecongruenttoagivenangleusingstraightedgeandrustycompass.

5.Showhowtofindsquarerootswithastraightedgeandarustycompass.6.Constructatrianglesimilartoagiventriangleandfour-ninthsaslarge(inarea).

7.Constructatrapezoidsimilartoagiventrapezoidandtwo-thirdsaslarge.8.Onagivensegmentasbase,constructarectangleequaltoagivenparallelogram.

9.Constructanisoscelestriangleonagivenlinesegmentasbaseandequaltoagiventriangle.

10.Constructanisoscelestriangleequaltoagivenparallelogram,giventhelengthofitsequalsides.Istheconstructionalwayspossible?

11.ConstructtriangleABC,givena)B,tb,hbb)ha,ma,Bc)b,hb,cd)A,C,tce)b,ma,Cf)C,A,hc

12.“Givenacircle,takeone-fourthofthecircumferenceasthesideofasquare.Sincethesquareandthecirclenowhaveequalperimeters,theyhaveequalareas.Thissquaresthecircle.”Whatiswrongwiththisargument?

13.GivenangleAOB,extendBOanddrawanycirclewithcenterOcuttingthesidesoftheangleatAandB.(Seetheaccompanyingfigure.)Onastraightedge,marksegmentCDequaltoOA.Slidethestraightedgealong

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pointAwithpointDlyingonlineBOuntilpointCliesonthecircle.Drawthisline.ShowthatangleADBtrisectsangleAOB.Doesthisconstructionsolvetheancienttrisectionproblem?

Exercise8.13

14.An“angletrisection”appearshere(seefigure),simplifiedsomewhatfromitsoriginalformintheFebruary1966issueofMechanixIllustrated.Findthefallacyintheargument.DrawacirclewithcenterOtocutthesidesofthegivenangleinpointsAandB.ExtendAOandBOtocutthecircleagainatCandDasintheaccompanyingfigure.LetthebisectorofangleAOBcutchordCDatGandthecircleatR,asshown.MarkHonORsothatHR=RG.LetcircleG(H)cutBCandADatC′andD′Then C′OD′=(AOB)/3.

Exercise8.14

Proof.ThecirclewithradiusequaltoOAandpassingthroughC′andD′asshowncutsthefirstcircleatpointsMandNwhereOD′andOC′meetthatcircle.Then RON= NC′C= NCC′=( NOB)/2,showingthat RON

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trisects ROB.Thetheoremfollows, (?)15.Inthe“trisection”ofExercise8.14,take AOB=120°andfind C′OD′usingtrigonometry.

16.Provethestatementslistedbelow,fromwhichitfollowsthat cannotbeconstructed,sothattheduplicationofacubeisimpossible.Sinceτistranscendental,then isalsotranscendental,sothecirclecannotbesquared.Finally,sincesin20°isarootoftheequation8x3–6x–1=0,whichhasnorationalroots,thensin20°cannotbeconstructed,soa60·anglecannotbetrisected.a)Showhowtoconstructthesum,difference,product,andquotientoftwogivenlengths,andthesquarerootofagivenlength,inthepresenceofaunitsegment.

b)GiventheunitsegmentOU,inwhichO(0,0)andU(1,0)arepointsintheCartesianplane,showhowtoconstructanypointhavingrationalcoordinates.

c)Showthattheequationofalinethroughtwopointswithrationalcoordinateshasrationalcoefficients.

d)Showthatthepointofintersectionoftwolineswhoseequationshaverationalcoefficientshasrationalcoordinates.

e)Showthat,givenastraightedgeandpointswithrationalcoordinates,thenonecanconstructonlypointswithrationalcoordinates.

f)ShowthattheequationofcircleA(B),whereAandBarepointswithrationalcoordinates,hasrationalcoefficients.

g)Showthatthepointsofintersectionoftwocirclesorofacircleandaline,whoseequationshaverationalcoefficients,havecoordinatesthatarerationalorthatinvolveonlysquarerootsofrationalnumbers.

h)Showthatastraightedgeandcompasscanconstructonlypointswhosecoordinatescanbeobtainedbyfinitelymanyapplicationsoftheoperationsofaddition,subtraction,multiplication,division,andsquarerootappliedtothecoordinatesofgivenpoints.

i)Showthatthecommentsatthebeginningofthisexercisenowfollow.

* TheHalmossymbol indicatestheendofaproof.* Figures arenotnumberedconsecutively. Instead, theyare identifiedby thenumberof the item theyaccompany.* Sinceitisquitedifficulttowriteinboldface,itissuggestedthatanoverbarbeusedfordirectedlengthsinwrittenformulas.ThentheformulaofTheorem2.11wouldbehandwrittenas .* Thewordiffmeans“ifandonlyif,”andissoread.

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2 ISOMETRIESINTHEPLANE

SECTION9 THEAMAZINGGREEKS

9.1About600B.C.markedthestartofanewerainmathematics.Peoplestartedtoask“Why?”andtodemandlogicalanswers.Moreleisuretimeandasocietywhichplacedincreasingemphasisonlogicalreasoningcausedthethinkingaboutmathematicstochangefromthepracticaltothemoretheoretical.Theresultwastheso-called“materialaxiomatics”whichdominatedmathematicalreasoningformore than2000years. Inmaterial axiomatics one considers a bodyof appliedmathematics,suchasthegeometryoftherealworld.Certainobviousfactsaboutthisstudyareassumedtruewithoutproofasaxiomsorpostulates.Thenallotherfactsarelogicallydeducedfromthepostulates.9.2Themethodofmaterialaxiomatics reached fullmaturity in just300years,blossomingforthbrilliantlyinEuclid’sElementsabout300B.C.9.3ThefirstmantoprovesimpletheoremswasthemerchantThales(640–546B.C.) of the island ofMeletus. In his travels about theMediterranean Sea, helearnedofthemathematicsofEgyptandotherEasterncountries.Thenheprovedsuch simple theorems as, for example, that a diameter of a circle bisects thecircleandthatthebaseanglesofanisoscelestrianglearecongruent.9.4While residing in Egypt, Thales is said to have amazed King Amasis bycalculatingtheheightofapyramidbymeasuringitsshadowandtheshadowofavertical stick of known height and then using simple proportion. Very likelyThalesdidcalculate theheightof thepyramidusing shadowsandproportions,butitisabsurdtoimaginethathecouldusethemethoddescribed,foronecouldnotpossiblymeasuredirectlythelengthofthepyramid’sshadow.

Bypredictingasolareclipsein585B.C.heattainedgreatfame.Itissaidthathe fell into a ditch one night while studying the stars during a stroll. An oldwomanwhohelpedhimoutprotested,“Howcanyoupossiblyseeanything intheheavenswhenyoucannotevenbeholdwhatisatyourfeet?”9.5Clothedinmuchmysticismandcloudedwithmanylegends,Pythagoras(572?–500? B.C.), of the Aegean island of Samos, and his followers contributedgreatly to mathematics. The Pythagorean Society which he founded, a secretfraternitywhich valued highly both brotherhood andmathematics, credited alldiscoveriestoPythagoras,soitisnoteasytotellwhathehimselfdevelopedasagainstwhathisfollowersdeduced.9.6 It is felt,nonetheless, thatPythagorasdidprove the theorem thatbearshisname.Itissaidthathewassodelightedwiththisdiscoverythathesacrificedahundred oxen. Undoubtedly this story is not true, since the Pythagoreans

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believedintransmigrationofthesoul.Infact,onedayPythagoraswasreportedtohavecomeuponamanbeatingadog.Hecriedforthemantostop,sincehecould recognize the voice of a departed friend in the dog’s howls.And in thenextlife,hetoldtheman,thetablesmightbereversed,sothathemightbethedumbanimal.Soeffectivewere thesewords that themanfell tohiskneesandbeggedforgivenessfromthedog.9.7 One must not overlook the great Academy of Plato (429–348 B.C.), overwhosedoorwasthemotto,“Letnooneunversedingeometryenterhere.”Whenasked what occupied the Deity, Plato immediately replied, “God geometrizescontinually.” Eudoxus (408–355 B.C.), the most brilliant of the earlymathematicians,inventedatheoryofproportionthattookirrationalnumbersintoaccount.Twothousandyears laterRichardDedekind(1831–1916)showedjusthowaccurateEudoxushadbeeninhistheory.Aristotle(384–322B.C.),althoughknown primarily for his systematization of deductive logic, improved somegeometricdefinitionsanddiscussedcontinuity.9.8Themostfamousscientifictreatiseofallantiquity,whichhasbeenpublishedinmoreeditionsthananyotherbookexcepttheBible,istheElementsbyEuclid(365 ?–300? B.C.), written about 325 B.C. This basic mathematics textbook soovershadowed all earlier writings that no trace of its predecessors remains.IndeeditisdifficulttofindmuchinformationaboutanymathematicianpriortoEuclid.ThethirteenbooksoftheElementsembracenotonlybasicgeometrybutalso number theory and elementary algebra. How much of the material wasEuclid’s ownwork is difficult to ascertain, but it is felt that he contributed atleastmanynewproofs.9.9LittleisknownofEuclid’slife,butheisreportedtobethefirstprofessorofmathematics at the greatUniversity ofAlexandria.When askedbyPtolemy iftherewereaneasierwaytomastergeometry,hereplied,“Thereisnoroyalroadto geometry.” Perhaps geometry treated by algebraic means—using analyticgeometry,isometries,andsimilarities—isactuallythat“royalroad”thatdidnotexistinEuclid’sday.9.10ThegreatestmathematicianofallantiquitywasArchimedes(272–212B.C.)ofSyracuse.Hisaccomplishments ingeometryandmechanicswerelegion.Helocatedπbetween and byinscribingandcircumscribingpolygonsof96sides in and about a circle.Hiswondrouswarmachines forestalled the fall ofSyracusetotheRomans.Whenthecitywasfinallyconqueredduringafestivalwhen its guard was down, a Roman soldier came upon the aged Archimedesworkingatageometric figure inasand tray;hisadmonitionnot todisturb theworkenragedthemanofwarandheranArchimedesthroughwithhissword.It

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is worthy of note here that no ancient Roman ever achieved fame inmathematics.9.11Many lessermathematicians followed, eachcontributinghisbit, butnonebringingbackthegreatglorythatreacheditszenithinArchimedes.Eratosthenes(ca.230B.C. ),Apollonius (262?–190?B.C. ),Hipparchus (ca.140B.C.),Heron(ca.110B.C.),Menelaus(ca.100B.C.),ClaudiusPtolemy(85?–165?),Pappus(ca.340),TheonofAlexandria(ca.390),andhisdaughterHypatia(375–415),thefirstrecordedwomanmathematician,arebutafew.9.12Awomanofoutstandingbeautyandanexcellentteacher,Hypatiawasalsoadevoutandoutspokenpagan in thisearlyChristianera.Somuchso thatoneday in March, 415, a band of these gentle Christians dragged her from herchariot,stonedhertodeathwithclamshells,dismemberedher,andthenburnedher body, just to make sure the job was well done with proper Christianaffection.9.13ThefamousUniversityofAlexandria, foundedbyPtolemyabout300B.C.andthecenterofallknowledgeformorethan900years,nowbegantofadeasRoman society started to break up, its glory becoming just a shadow of itsformerself.Finally,in641,theArabsconqueredAlexandria,burningwhathadnotbeendestroyedbytheChristians.NowGreecewasdead.AndtheDarkAgesovertookEurope.ExerciseSet91.ExplainwhyThalescouldnothavecalculatedtheheightofapyramidasindicatedin9.4,andshowhowhecouldhavecalculateditsheightbysimpleproportion,usingtwoshadowobservationsatdifferenttimesofday.

2.FindseveraldifferentproofsofthePythagoreantheorem.3.a)Showthatwhenxisthesideofaregularpolygoninscribedinacircleof

radiusr,then

isthelengthofthesideofaregularpolygonhavingtwicethenumberofsidesandinscribedinthesamecircle.

b)Showthattheperimeterofaregularpolygonof96sidesinscribedinacircleofdiameter1leadstotheformula

4.FindtheformulascorrespondingtothoseofExercise9.3forcircumscribedpolygons.

SECTION10 INTRODUCTIONTOTRANSLATIONS,ROTATIONS,ANDREFLECTIONS

10.1 If a first triangleABC is congruent to a second triangleA′B′C′, then it ispossibletomovethefirsttriangleandperhapsturnitoversothatitwillcoincide

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with the second triangle. Such a motion is called an isometry. That is, anisometryisarigidmotion(ormaportransformation)ofthepointsoftheplane.The defining property of an isometry is that it preserves distances. If theisometryamapsPandQtoP′andQ′thenPQ≅P′Q′.Itfollowsthateachpointmapstojustonepoint(itsimage)andeachpointP′istheimageofjustonepointP(thepreimageofP′).Wewriteα(P)=P′todenotethatamapsPtoP′.10.2IftheisometryαmapstriangleABCtocongruenttriangleA′B′C′andifthelengthsofthesegmentsAA′,BB′,CC′areequalandthesesegmentsareparallelto one another and similarly directed, then α is called a translation throughvector* .(SeeFig.10.2a.)Ourpointofviewisthatatranslationmovesallthepointsofthe

Figure10.2aplane.Thatis,atranslationthroughvector mapseachpointPintheplanetoapointP′suchthat and areparallel,equalinlength,andhavethesamedirection;thatis,theyareequalvectors(seeFig.10.2b).Thus,iflinesAA′andPP′

Figure10.2b

donotcoincide,thenAA′P′Pisaparallelogram.Intheeventthattheselinesdocoincide,we shall still call the figure they form a (degenerate) parallelogram.Observe that triangle ABC translates to triangle A′B′C′ iff AB, CA, BC are

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similarlyoriented toandparallel toor lieon the same lineasA′B′,C′A′,B′C′,respectively.

10.3LetObea fixedpoint in theplaneand letθbea fixeddirectedangle.ArotationaboutpointO (called thecenterof therotation) throughangleθ isanisometrythatmapseachpointPintoapointP′sothatOP≅OP′and POP′=θ,measuredcounterclockwise.Figure10.3showsarotationappliedtoatriangleABC.PointO is itsownimage.Anypoint that is itsownimageunderagivenisometryiscalledafixedpoint(oraninvariantpoint)forthatisometry.Thusthecenter of a rotation is a fixed point, and no other point is fixed by a rotationthroughananglenotamultipleof360°.Atranslationthroughanonzerovectorhasnofixedpoints.Note that,althougharotationpreservesdistances, it isnottrue that vectors are each equal to vectorswhentriangleABCisrotatedintotriangleA′B′C′throughananglenotamultipleof 360°. That is, each side of the triangle, as a vector, is not equal to thecorrespondingsideofitsimage.

%

Figure10.3

10.4Thethirdisometryweconsiderinthissectionisperhapsthemostbasicandmostimportantisometryofall.AreflectioninalinemmapseachpointPoftheplane into its“mirror image”P′with linemasmirror.Figure10.4shows this

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reflection,alongwithatriangleABCanditsimageunderthereflection.ThusP′isthereflectionofPinlinemiffmistheperpendicularbisectorofPP′orPandP′coincideonlinem.

Figure10.4

Henceallpointsofmarefixedpoints.IfonereadstheverticesoftriangleABC(in Fig. 10.4) in cyclic order, that is,“A–B–C–A” one is traveling around thetriangleinthecounterclockwisedirection.Ontheotherhand,triangleA′B′C′ isclockwise.Thusareflectionreversesthesenseofatriangle.Anyisometrythatreverses the sense of a triangle is called opposite. A reflection is an oppositeisometry. An isometry that preserves the sense of a triangle is called direct.Translationsandrotationsaredirectisometries.

10.5Thereflectionisthebasicbuildingblockofisometries.Eachtranslationandeach rotation can be written as a product of two reflections. Denoting thereflectioninlinembyσm,ifσmmapsPtoP′andthenσnmapsP′toP″,thentheproductofthereflectionsinlinesmandn,inthatorder,mapsPtoP″,andwewrite

Notecarefullythatσnσmmeansfirstperformσm,thenperformσnontheresult.This definition of “product” holds for transformations in general; it is not

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restrictedjusttoreflections.

10.6Letmandnbetwoparallellinesandletvdenotethevectorfromlinemtolinen(measuredperpendiculartotheselines).LetPbeanypointwithσm(P)=P′andσn(P′)=P″,let2xdenotethevectorfromPtoP′and2ythevectorfromP′ toP″. (See Fig. 10.6.) Then x + y = v, so 2x + 2y = 2v. That is, σn σmtranslates each point P through vector 2v. It follows that the product of tworeflectionsσnσminparallellinesmandnisatranslationthroughtwicethevectorfrom the first line to the second line. Conversely, each translation through avector2vcanbefactoredintoaproductoftworeflectionsinlinesmandn,linem being chosen arbitrarily as any line perpendicular to the direction ofv, andthenlinenisthatlineparalleltomsothatthedirecteddistancefrommtonisv.

Figure10.6

10.7 It deserves repeating that we write βα to denote the product of thetransformationsαandβinthatorder.Inthistext,productsoftransformationsareperformedfromright to left.Sincesomeauthorsprefer towriteproducts fromleft to right,besure toobserve thenotationconventionsof theparticularbookyouareusing.

10.8NowletmandnbetwolinesintersectingatapointOsothatthedirectedanglefromlinemtolinenisθ.LetPbeanypoint,letσm(P)=P′andσn(P′)=P″.LetthedirectedanglesPOP′andP′OP″be2xand2y.(SeeFig.10.8.)Thenx+y=θ,sothedirectedanglePOP″is2x+2y=2θ.Furthermore,OP≅OP′≅OP″,soσnσmisarotationaboutpointOinangle2θ.Conversely,everyrotationaboutapointO inanangle2θ canbe factored intoaproductof reflections intwolines,thefirstonembeingchosenarbitrarilythroughO,thenthesecondonenisthatlinethroughOsuchthatthedirectedanglefromlinemtolinenisθ.

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Figure10.8

ExerciseSet10

1.Bydefinition,anisometrypreserveslengths.Showthatanisometryalsopreservesangles.

2.IfABB′A′isarectangle,whichtwoisometriesmapsegmentABtoA′B′?3.IntheCartesianplane,ifoneeffectsatranslationbymovingthecoordinateaxesinsteadofthepointsoftheplane,howshouldtheaxesbemovedforthetranslationthroughvector ?

4.Showthattheproductoftwotranslationsisatranslation.

5.Showthatanytwodirectlycongruenttrianglesarerelatedtoeachothereitherbyatranslationorarotation.

6.Giventwodirectlycongruenttriangles,findthevectoroftranslationorthecenterandangleofrotationthatmapsonetotheother.

7.Displaytwocongruenttriangleswithcorrespondingsidesparallelthatarerelatedbyarotationandnotatranslation.Whatistheangleofrotation?

8.Useareflectiontoprovethatthebaseanglesofanisoscelestrianglearecongruent.

9.Showthataproductofreflectionsinthreeparallellinesisequaltoareflectioninanotherlineofthispencil.

10.Showthataproductofreflectionsinthreeconcurrentlinesisequaltoareflectioninanotherlineofthispencil.

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SECTION11 INTRODUCTIONTOISOMETRIES

11.1An isometryα is completely determined by any three points that form atriangle and their images, because any pointP in the plane can be located bygiving its distancesm(AP),m(BP),m(CP) from these three points. Hence itsimageP′isdeterminedsincetheisometrypreservesthesedistances.Figure13.6showsthisidea.

11.2 It follows that each isometry in the plane is the product of atmost threereflections,foritwouldtakeatmostthreereflectionstomapatriangleABCtoacongruenttriangleA′B′C′,essentiallyonereflectionpervertex,asshowninFig.13.13. Thus each direct isometry, being a product of an even number ofreflections,isaproductoftworeflections,henceitisatranslationorarotation.Eachoppositeisometryiseitherareflectionoraproductofthreereflections.

11.3Letιdenotethatisometrythatleaveseachpointfixed;thatis,ι(P)=PforeverypointP.We call ι (lower-caseGreek iota) the identitymap. Ifm is anyline,thenσmσm=ιforFig.10.4showsthatwheneveralinemreflectsapointPtoapointP′,thenitreflectsP′toP;thatis,thereflectionofthereflectionofapointisthepointitself.Furthermore,ifmandnareanytwolines,then

since transformation multiplication is associative. It is clear that ι is a directisometry,sinceitcanbewrittenasaproductofanevennumberofreflections.

11.4 Ifα andβ are transformations such thatαβ=βα= ι, thenβ is called theinversetransformationtoα.Ifα(P)=Q,thenβ(Q)=P.Suchaninversealwaysexists for transformations and is unique, sowewriteα–1 for the inverse toα.Sinceσmσm= ι, thenareflection is itsowninverse;σm–1=σm.The letterσ isreservedforself-inverseisometries.Writingβ2forαα,atransformationα,otherthantheidentityι,thatisself-inverse(α2=ι)iscalledinvolutoric.Rotationsaregenerally not involutoric, and translations through nonzero vectors are neverinvolutoric.

11.5Theinverseoftherotationα=σnσmisgivenbyα–1=σmσnandisarotationaboutthesamepointthroughtheoppositeangle.Theinverseofthetranslationα

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=σnσmisagaingivenbyα–1=σmσnandisatranslationintheoppositedirectionthrough the same distance. Figures 10.6 and 10.8 show that if the order ofreflectionisreversedfromσnσmtoσmσn,thenpointP″ismappedfirsttoP′andthentoP,justreversingthetranslationorrotation,henceproducingitsinverse.

11.6There is just one rotation that is involutoric, the rotation about a pointAthrough180°(orthroughanoddmultipleof180°),calledahalfturnabout(orareflectionin)pointAanddenotedbyσA.Figure11.6showstrianglePQRanditsimage triangle P′Q′R′ in a halfturn about point A. Note that involutoricisometries with lower-case subscripts (as in σm) represent reflections in lines,whereas those with upper-case subscripts (as in σA) represent halfturns aboutpoints. In fact, throughout this book, unless otherwise stated,we shall alwaysdenotepointsbyupper-case italic letters, linesby lower-case italic letters, andtransformations by lower-case Greek letters. Later, when needed, we shalldenoteplanesbyupper-caseGreekletters.Thisconventionissubjectedtosomeminorviolationinthechapteroncomplexnumbers,butthatneednotconcernushere.

Figure11.6

11.7 A halfturn about point A is the product of two reflections in any twoperpendicular linesmandn throughA (seeFig.11.7a).ThusσA=σnσm.SincealsonandmaretwoperpendicularlinesthroughA,itfollowsthatσA=σmσn,too.Equating these two expressions forσA,we obtainσnσm =σmσn. In fact,σsσr =σrσsiffr=sorrandsareperpendicular.

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Figure11.7a

The product σBσA of two halfturns is a translation through vector 2 .Figure 11.7b shows that if σA(P) = P′ and σB(P′) = P″, then A and B aremidpointsofsidesPP′andP′P″oftriangleP′PP″.ItfollowsthatPP″isparalleltoandtwiceaslongasAB.NowforgivenpointsA,B,andC, takepointDsothatABCDisaparallelogram(Fig.11.7c).ThenσDσCσBσA=ι,sinceσDσCandσBσAareinversetrans;ations,so

Figure11.7b

that is, the product of threehalfturns (σcσBσA) is a halfturn (σD)in the fourthvertex of the parallelogram the first three vertices ofwhich are the first three

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pointsintheirgivenorder.

Figure11.7c

11.8Wenextturnourattentiontoproductsofreflectionsinlinesm,n,andp.Ifm,n,andpallconcurinanordinarypointA,letrbethelinethroughAsothattherotationsσnσmandσpσrareequal(seeFig.11.8a).Then

Similarly, if m,n,p are parallel, let r be a line parallel to them so that thetranslationsσnσmandσpσrareequal(seeFig.11.8b).Hereagainwehave

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Figure11.8a

So,ifthethreelinesm,n,pfromapencil,thentheproductσpσnσmofreflectionsintheselinesreducestoareflectioninsomelineofthesamepencil.

Figure11.8b

11.9 Any other opposite isometry α can be factored into a product of threereflectionsα=σpσnσmwithmparallelton,andpperpendiculartobothmandn(seeTheorem16.4).Thisisometryiscalledaglide-reflection;andistheproductof the translation σn σm followed by the reflection σp in line p parallel to thedirectionofthetranslation.Notethat

sincepisperpendiculartolinesmandn.Theinverseofthisglide-reflectionα=σpσnσmcanbewrittenas

and in its last form α–1 is seen to be another glide-reflection, namely thetranslation inverse to the given glide, and the reflection equal to the givenreflection.Thus,inFig.11.9,α=σpσnσmcarriespointAtopointA′whileα–1carriesA′whileα–1carriesA′back(dottedlines)topointA.

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Figure11.9

11.10 Principal results on isometries are summarized in the table of Theorem16.14.OthergeometricpropertiesaregiveninthetheoremsofSection17.

ExerciseSet11

1.ForthegiventriangleABCinthecoordinateplanewithA(0,0),B(0,0),andC(0,2),findcoordinatesfortheverticesoftheimagestriangleA′B′C′undereachisometry:a)atranslationof5unitsinthepositivex-directionb)atranslationof unitsat45°intothefirstquadrantc)arotationof90°aboutthepointP(3,0)d)arotationof–45°aboutthepointP(3,0)e)arotationof–90°aboutthepointQ(2,–3)f)areflectioninthemirrory=0g)areflectioninthemirrory=xh)areflectioninthemirrorx+y=4i)ahalfturnaboutP(3,0)j)ahalfturnaboutQ(2,–3)k)aglide-reflectionof5unitsinthepositivex-directionwithmirrory=2

1)aglide-reflectionof unitsat45°intothefirstquadrantwithmirrorx–y=4

m)theidentityι2.WriteeachisometryofExercise11.1asareflectionorasaproductofreflections.

3.Notranslationorglide-reflection(withnonzerovector)canmaparectangleintoitself.Findallrotations,halfturns,andreflectionsthatmapagivenrectangleintoitself.

4.RepeatExercise11.3forthefollowing:a)squareb)parallelogramc)

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rhombusd)equilateraltrianglee)regularhexagonf)regularpentagon5.WritetheinverseofeachisometryinExercise11.1.

6.DecidewhichoftheisometriesinExercise11.1areinvolutoric.

7.DecideforeachisometryαofExercise11.1whetherthereisapositiveintegernsuchthatαn=ι.Ifso,findthesmallestsuchn.Wecallthatntheorderoftheisometry.Involutoricisometrieshaveorder2,ιhasorder1.Translationsandglide-reflectionsthroughnonzerovectorsandmanyrotationshavenosuchpositiven,sotheorderofsuchisometriesisdefinedtobezero.

8.Giventhelinesm,n,pwithequationsy=0,y=2x,x=0,findlineqsothata)σq=σpσnσmb)σq=σmσnσpc)σq=σmσpσnd)σq=σpσmσn

9.RepeatExercise11.8takingy=0,y=2,y=3asequationsforthelinesm,n,p.

10.Showthatatranslationcanbewrittenasaproductoftworotations.Furthermore,showthatonecenterofrotationandoneangleofrotation(≠0°)maybechosenarbitrarily.

11.Showthatthemidpointofthehypotenuseofarighttriangleisequidistantfromthethreevertices;dosobyusingahalfturnaboutthatmidpoint.

SECTION12 TRANSFORMATIONTHEORY

12.1Inthissectiononewillfindaformaltreatmentofthetransformationtheoryrequired fora formal treatmentof isometries. It isassumed that the readerhasread theoverviewgiven inSections10 and11, so that he is familiarwith thedirectionthisdevelopmentistotake.

12.2 Definition A transformation of the plane, or more simply, atransformation,αisaone-to-onemaporfunctionofthepointsoftheplaneontothemselves.Thatis,–mapseachpointoftheplanetoauniqueimagepoint,andeachpointintheplaneistheimageofexactlyonepoint.

12.3 Definition If α and β are transformations, we define their product (orcomposition)βαby(βα)(P)=β(α(P))foreachpointP.Wealsowriteα2forαα,

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α3forαα2,etc.

12.4Sincea transformation isa function, it follows that twotransformationsαandβ areequal iff theyare identical, that is, iff foreachpointP in theplane,12.5TheoremIfαandβaretransformations,thenβαisatransformation.

12.6TheoremTransformationmultiplicationisassociative;thatis,ifα,β,γaretransformations,then(γβ)α=γ(βα).

ForagivenpointP,letα(P)=Q,β(Q)=R,andγ(R)=S.Then

and

Hence(γβ)α=γ(βα).

12.7DefinitionFortransformationαandpointP,ifα(P)=P,thenPiscalledafixedpoint(oraninvariantpointordoublepoint)fortransformationα.

12.8DefinitionThetransformationι,definedbyι(P)=PforeachpointPintheplane,iscalledtheidentitytransformationortheidentitymap.Thatis,ιisthattransformationthatleaveseachpointfixed.

12.9TheoremIfαisatransformation,thenαι=ια=α.

12.10DefinitionIfαandβaretransformationsandβα=ι,thenβissaidtobeinversetoα.

12.11TheoremIfβisinversetoα,thenαisinversetoβ.Letβ(P)=Qandα(Q)=RforagivenpointP.Sinceβα=ι,then

Sinceβisatransformation,itisone-to-one;thatis,sinceβ(R)=β(P),thenR=P.Now

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so–β=ι;thatisinversetoβ.

12.12 Theorem If β and γ are both inverse to α, then β = γ. That is, atransformationhasatmostoneinverse.

Forβ=βι=β(αγ)=(βα)γ=ιγ=ιγ=γbyTheorems12.6,12.9,and12.11.

12.13TheoremEachtransformationhasaninversetransformation.Given transformationα and anypointP in theplane, letα(Q)=P.This is

alwayspossiblesinceαisanontomap.Nowdefineβbyβ(P)=Q.Thenβisatransformationbecauseaisatransformation,andβα=ι.

12.14DefinitionIfαisatransformation,denoteitsuniqueinversebyα–l.

12.15DefinitionAsetG,alongwithabinaryoperation*,iscalledagroup iffthefollowingfourpostulatesaresatisfied:G1:whenevera,b∈G,thena*b∈G,G2:ifa,b,c∈G,then(a*b)*c=a*(b*c),G3:thereisanelementiinGsuchthat,ifa∈G,thena*i=i*a=a,andG4:foreachainG,thereisanelementa–1inGsuchthat=a–1*a=i.

12.16DefinitionAgroupiscalledabelian iffitsatisfiesthecommutativelaw:G5:foreacha,b∈G,a*b=b*a.

12.17 Theorem The set T of all transformations of the plane, along withtransformationmultiplication,isagroup.

ByTheorems12.5,12.6,12.9,and12.13.

12.18ThatthisgroupofTheorem12.17isnotabelianwillbeshownlater(seeExercise12.3).

12.19 Geometry can be defined by means of its transformation groups.Euclidean geometry is the study of those properties (congruence of figures,equality of lengths, parallelism, etc.) which are invariant under thetransformations of the transformation group containing all reflections andproductsofreflections(translations,rotationsandglide-reflections).

12.20 High school geometry courses also study similar figures. This study iscalled plane equiform geometry, the study of properties invariant under the

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groupofallsimilarities,namelyreflections,productsofreflections,homotheties(or uniform stretches of the plane), and all products of these transformations.Thesetransformationsarestudiedinthenextchapter.

12.21If,inthetransformationgroup,oneincludesallprojectionsofthepointsofthe plane (projections from one plane onto another), the resulting study isprojective geometry. Every theorem of projective geometry also holds inequiform and Euclidean geometries since every property preserved by all thetransformationsofprojectivegeometrywillcertainlybepreservedbyanysubsetofthesetransformations.Similarly,everytheoremofequiformgeometryistrueinEuclideangeometry.

12.22 In1872FelixKlein(1849–1925)conceived thedefinitionofageometryas the study of those properties of a set of points that are invariant under thetransformationsofsometransformationgroup.Hehad justacceptedachaironthefacultyattheUniversityofErlangen,andthisdefinitionwaspresentedinhisinauguraladdress,nowknownashisErlangerprogramm.Thus,whenstudyingthe Euclidean geometry determined by isometries, we are concerned withcongruence, lengths, measures of angles, similarity, concurrence of lines,collinearityofpoints,etc. Inplaneequiformgeometry,congruenceandlengthsarenolongerpreserved.Inprojectivegeometry,onlycollinearityofpointsandconcurrenceof linesare left invariant in thelistabove.Eachtimethegroupoftransformationsisenlarged,thefieldofstudyisnarrowed,forfewerpropertiesare preserved when more transformations are considered. Conversely, allproperties preserved under a given group of transformations will certainly bepreservedunderanysubgroupofthesetransformations.

12.23Euclid assumed (seeAppendixA) that “thingswhich coincidewith oneanother are equal.” Itwashis intent to “pickup andmove” a figure so that itcouldbesuperimposeduponanother figureasa test forcongruence.Since thelogicalfoundationsofthismethodofsuperpositionarenotmadeclear,moderntreatments assume the SAS condition for congruence of triangles. Wheresuperpositionishazy(howcanyoupickupagenuineline,andwhathappenstoit if you do manage to pick it up ?), the SAS postulate is precise. Now, anisometryisexactlythemotionEuclidhadinmind,andwethereforedevelopthetheoryofisometriestoclarifyEuclideantransformations.Soamoderntreatmentofsuperpositionwouldstatethat“twotrianglesarecongruentifareflectionoraproduct of reflectionsmaps one triangle into the other,” thereby avoiding thevague“pickupandmove”idea.Studentsbeginninghighschoolgeometryhave

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the background to base their sophomore geometry course on transformations,providedthematerialiswellwrittenforthatlevel.Thetextthatyouuseinsucha course shoulddomore thanpaymere lip service to transformations, but theproperties of triangles and circles, etc., integrated with the correspondingpropertiesofprismsandspheres,etc.,mustreceivefullcoverage.

ExerciseSet12

1.ProveTheorem12.5.

2.ProveTheorem12.9.

3.Findtransformationsαandβsuchthatαβ≠βα.4.Provethatιisunique.

5.Provethat(α–1)–1=α.

6.Provethat(βα)–1=α–1β–1.

7.Ifα2=α,thentransformationαissaidtobeidempotent.a)Showthatifαisidempotent,thenα=ι.b)Showthattherearetransformationsotherthanιsuchthatα3=α.

8.Forgiventransformationsαandβ,showthatthereareuniquetransformationsγandδsuchthatγα=βandαδ=β.Findexpressionsforγandδintermsofαandβ.

9.Showthatifγα=γα,orifγα=γβ,fortransformationsα,β,γ,thenα=β.10.Showthatifα2β2=(αβ)2fortransformationsαandβ,thenαβ=βα11.a)Findtransformationsα,β,γsuchthatγα=βγandα≠β.

b)Ifαandγaregiven,findanexpressionforβ.c)Cansuchβalwaysbefoundforgivenαandγ?

12.ShowthatanysetSoftransformationsformsagroupundertransformationmultiplicationwhen:1)Sisnonempty,2)wheneverα∈S,thenα–1∈S,and3)wheneverα,β∈S,thenβα∈S.

13.ShowthatthethreeconditionsofExercise12.12canbereplacedbythetwoconditions:1)Sisnonempty,and2*)ifα,β∈S,thenβ–1α∈S.

14.ShowthateachsetofisometriesfoundinExercises11.3and11.4isagroup.

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SECTION13 ISOMETRIESASPRODUCTSOFREFLECTIONS

13.1 To give us a starting point, we shall postulate that two triangles arecongruentiftheysatisfytheSAScondition;thatis,iftwosidesandtheincludedangleofonetriangleareeachcongruenttothecorrespondingpartsofthesecondtriangle.Other sufficient conditions for congruenceof triangleswillbeprovedlaterinthischapter.

13.2DefinitionAn isometry isamapof thepointsof theplane thatpreservesdistance;thatis,αisanisometryiffforeachpairofpointsPandQintheplane,ifα(P)=P′andα(Q)=Q′thenPQ≅P′Q′.

13.3TheoremAnisometrymapslinesegmentsintocongruentlinesegments.

LetαbeanisometryandPQasegment.Letα(P)=P′andα(Q)=Q′.ThenPQ≅P′Q′.TakeanypointAonsegmentPQ.ThenPA+AQ=PQ.So,ifα(A)=A′, thenP′A′ +A′Q′ =P′Q′ since α preserves distances. HenceA′ lies onsegmentP′Q′andthetheoremfollows.

13.4CorollaryAnisometrymapslinesintolinesandcirclesintocircles.

13.5DefinitionIfSisapointsetandαisatransformation,letα(S)denotethesetofallimagesofpointsofSunderthemapα;thatis,P∈Siffα(P)∈α(S).

13.6TheoremThereisatmostonepointXatgivendistancesfromthreegivennoncollinearpointsP,Q,R.

Thereisacircle(whichmayreducetoasinglepoint)ofpointsYsuchthatPX≅PYandacircleofpointsYsuchthatRX≅RY.Thesetwocirclesmeetinatmost twopointsY1andY2.SincePQR isa triangle,Qdoesnot lieonPR, theperpendicular bisector ofY1Y2.HenceQY1 QY2, so atmost one of thesesegments can be congruent toQX. Thus there is atmost one pointX at givendistancesfromthreenoncollinearpointsP,Q,R.(SeeFig.13.6.)

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Figure13.6

13.7 Corollary If a map of the points of the plane to the plane preservesdistance,thenitisone-to-oneandonto;thatis,anisometryisatransformationoftheplane.

13.8TheoremIfPQRisatriangleandαandβareisometriessuchthatα(P)=β(P),α(Q)=β(Q),andα(R)=β(R),thenα=β.

ByTheorem13.6,foreachpointXintheplane,α(X)=sinceαandβpreservedistance.Soα=β.

13.9DefinitionAplanemapαiscalledareflectioninlinem,orareflection,iffwheneverB=α(A) forpointsAandB, thenm is theperpendicularbisectorofsegmentAB,orelseA=BandA∈m.Thereflectioninlinemisdenotedbyσmandlinemiscalleditsmirror.

13.10TheoremAreflectionisanisometry.SupposeAandBareanytwopoints,andletA′=σm(A)andB′=σm(B).LetA

A′andBB′cutminpointsFandG,andsupposeA,B,andA′arenotcollinear,asshowninFig.13.10.ThentrianglesAGFandA′GFarecongruentbySASsince

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Figure13.10

AF≅A′F,FG=FG,and AFG≅ A′FG=90°.ThusAG≅A′G.AlsoAGB≅ A′GB′bysubtractionandBG≅B′G.HencetrianglesABGandA′B′Gare congruent by SAS. Now AB≅ A′B′. The case when A, B, and A′ arecollinear is not difficult. Thus σm preserves distances, and every point has animage.Thusσmisamapthatpreservesdistances.Henceitisanisometry.

13.11CorollaryEveryproductofreflectionsisanisometry.

13.12TheoremForeachreflectionσm,σm–1=σm.

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13.13TheoremEachisometryisaproductofatmostthreereflections.LettheisometryαmaptriangleABCtoA′B′C′.

Case1.IfA=A′,B=B′andC=C′thenα=ι,thesquareofanygivenreflectionσm

Case 2. If A = A′ and B = B′ butC ≠C′ as in Fig. 13.13a, then AB is theperpendicularbisectorofCC′sinceCandC′areequidistantfromAandfromB.HenceareflectioninlineABmapstriangleABCtoA′B′C′.

Figure13.13a

Figure13.13b

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Figure13.13c

Case3.IfA=A′butB≠B′asinFig.13.13b,thenAliesonm,theperpendicularbisectorofBB′.ThenσmmapsBtoB′andleavesAfixed,reducingthiscasetooneofthetwocasesabove.

Case 4. If A ≠ A′ (see Fig. 13.13c), reflect triangle ABC in line m, theperpendicularbisectorofAA′sothatσm(A)=A′.Nowthiscasereducestooneofthethreecasesabove.

13.14 Definition Each isometry that is a product of an even number ofreflectionsiscalleddirect.Anoppositeisometryistheproductofanoddnumberofreflections.

13.15DefinitionThesenseofatriangleABCisclockwiseiffonetravelsintheclockwise direction when reading the vertices in cyclic order A-B-C-A; it iscounterclockwiseiffonegoesinthecounterclockwisedirectionwhenreadingA-B-C-A.When trianglesABC andA′B′C′ arecongruentandbothare read in thesame sense, they are said to be directly congruent; if their senses are not thesame,theyareoppositelycongruent(Fig.13.15).

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Figure13.15

13.16 Theorem A reflection maps a triangle into a congruent triangle.Furthermore,thecongruenceisopposite.

We must first prove, assuming only the SAS condition for congruence oftriangles, that ifσm(ABC)=A′B′C′, then∆ABC≅∆A′B′C′.Wedothis inthreecases.

Case1.SupposeAandBbothlieonthemirrorm,andletCC′cutmatR(seeFig.13.16a).Nowanytwocorrespondingsegmentsarecongruentsinceσmisanisometry.SinceCC′ isperpendiculartom, then∆ACR=∆AC′Rand∆BCR≅∆BC′RbySAS.Atleastoneofthesetwotrianglesisnotdegenerate,soatleastoneofthetwostatements BAC≅ BAC′and ABC≅ ABC′mustbetrue.NowtrianglesABCandABC′arecongruentbySAS.

Figure13.16a

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Figure13.16b

Figure13.16c

Case2.IfAliesonthemirrorm,butBandCdonotlieonm,thenAcannotlieonbothBB′andCC′.SupposeAdoesnotlieonCC′,andletCC′cutmatRasinFig.13.16b.ByCase1,∆CAR≅∆C′ARand∆BAR≅∆B′AR,so CAR≅C′ARand BAR≅ B′AR.Then BAC≅ B′AC′bysubtraction.Hence∆ABC≅∆AB′CbySAS.

Case 3. In any other case, at most one side, sayAB, of triangleABC can beparalleltothemirrorm,sosupposeCAcutsmatPandBCcutsmatQ.ByCase1,∆PCQ≅∆PC′Q,so ACB≅ A′C′B′.Now∆ABC≅∆A′B′C′bySAS.

Figure10.4givesthereaderabasisforaproofofthesecondsentenceofthistheorem.

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13.17 Corollary An opposite isometry maps a triangle into an oppositelycongruenttriangle.

13.18 Corollary A direct isometry maps a triangle into a directly congruenttriangle.

13.19TheoremNo isometry is both direct and opposite, but each isometry iseitherdirectoropposite.

13.20TheoremThereareexactly twoisometries,onedirectandoneopposite,thatcarryagivensegmentAB intoacongruent segmentA′B′ andmapA toA′andBtoB′.

LetCbeanyotherpointsothatABCisatriangle.ThereareonsegmentA′B′exactlytwotrianglesA′B′C′andA′B′C″congruenttotriangleABC.Furthermore,A′B′ istheperpendicularbisectorofC′C″sotriangleA′B′C″isthereflectioninlineA′B′oftriangleA′B′C′.Thusthesetwotrianglesareoppositelycongruent,soone of themmust be directly congruent and the other oppositely congruent totriangleABC.ThustheisometriesthatmaptriangleABCintothesetwotriangleswillbedirectandopposite,respectively.

ExerciseSet13

1.ProveCorollary13.4.

2.ProveCorollary13.7.

3.ProveTheorem13.10forthecasewhenA,B,andA′arecollinear.4.UsemathematicalinductiontoproveCorollary13.11.

5.ProveTheorem13.12.

6.IndicatethenumbersofreflectionspossibleforthecasesofTheorem13.13.

7.CompletetheproofofTheorem13.16.

8.ProveCorollary13.17.

9.ProveCorollary13.18.10.ProveTheorem13.19.11.UsemathematicalinductionandCorollaries13.4,13.17,and13.18toprove

thatanisometrymapsanypolygonintoacongruentpolygon.

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12.LetP(a,0),Q(b,0),R(0,c)beCartesianpointswitha≠bandc≠0.a)ShowthattherearejusttwopointsXsuchthatPX=sandQX=t,where|s–t|<PQ<s+t.GiventhatoneofthesepointsXhascoordinates(u,v),findthecoordinatesfortheotherpointX.

b)Findm(RX)forbothpointsX,andshowthatthetwoofthesedistancesareneverequal.

13.FindimagesforeachofthepointsA(0,0),B(1,0),C(0,3),D(1,1),E(2,3),F(15,12),G(–3,–5)inthereflectioninthemirror:a)y=0b)x=0c)y=2d)x=–1e)y=xf)x+y=2

14.Drawtwocongruenttriangleswithnocorrespondingverticescoincidingandconstructthereflectionsthatcarryonetotheother.Findsuchapairoftrianglesthatrequireexactly:a)onereflectionb)tworeflectionsc)threereflectionsd)fourreflections

SECTION14 TRANSLATIONSANDROTATIONS

14.1DefinitionAplanemapαiscalledatranslationthroughvector iffforeach point A, α(A) = B where ; that is, AB and PQ are parallel,congruent,similarly-directedsegments.

14.2TheoremIflinesmandnareparallelandifthenormalvectorfrommtonisv,thenσnσmisatranslationthroughvector2v.

14.3 In Theorem 14.2, note that vector 2v is perpendicular tom and to n, isdirectedfrommtowardsn,andisinlengthtwicethedistancebetweenmandn.(See Fig. 10.6.) 14.4 Theorem Every translation is an isometry and can befactored into a product of two reflections in parallelmirrorsm and n, one oftheselinesbeingarbitrarilychosenperpendiculartothevector2voftranslation,then the other mirror is parallel to the first and placed so that the directeddistancefrommtonisv.

14.5 Theorem The inverse of the translation through vector is thetranslationthroughvector .Ifthetranslationcanbewrittenasσnσm,thenits

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inverseisσmσn

14.6TheoremTranslationscommute.Byvectormethods,2v+2w=2w+2v,sinceineithercasethisvectorisa

diagonal of the parallelogramdetermined by the vectors 2v and 2w. (SeeFig.14.6.)

Figure14.6

14.7Theorem14.6canalsobeprovedalgebraically.SuchaproofisleftforthereaderinExercise15.11.

14.8DefinitionLetObeapointandθthemeasureofadirectedangle.Aplanemapα iscalledarotationaboutpointO throughangleθ iff foreachpointA,α(A)=BwhereA=B=OorOB≅OAandd( AOB)=θ.PointOiscalledthecenteroftherotation.

14.9TheoremIflinesmandnintersectatpointOwiththedirectedanglefrommtonofmeasureθ,thenσnσmisarotationaboutpointOthroughangle2θ.

Letσm(P)=P′andσn(P′)=Q. IfP∉m, thenOPP′ isanisoscelestrianglewithm thebisectorof itsvertexanglePOP′.Similarly, ifP′∉n, thenn is thebisectorof thevertexangleP′OQ of the isosceles triangleOP′Q. (SeeFig.14.9.) It follows that POQ= POP′+ P′OQ, so POQ is twice theanglefrom linem to linen.Also, since trianglesOPP′ andOP′Q areboth isosceles,thenOP≅OQ.

ThecasewhenP∈morwhenP′∈nissimpler.Thetheoremfollows.

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Figure14.9

14.10TheoremEachrotationisanisometryandcanbefactoredintoaproductoftworeflectionsinintersectinglinesmandn.Furthermore,oneofthemirrorsmaybearbitrarilychosenthroughthecenterOofrotation,thentheothermirroristhatlinethroughO,suchthatthedirectedanglefrommtonishalftheangleofrotation.

14.11TheoremEachdirectisometryisarotationoratranslation.

14.12TheoremTheproductoftwotranslationsisatranslation.Geometrically, the proof is trivial. The vector of the product of the two

translationsisthevectorsumofthevectorsofthetwogiventranslations.Hencetheproductisatranslation.

Alternatively, letusgiveaproofutilizingTheorems14.2and14.4.Let thetranslations beα andβ and letα =σnσm andβ =σqσp. Ifm,n,p,q are allparallelorcoincident,thenletrbeanotherlineofthispencilsuchthatσpσr=σmσm.(SeeFig.14.11a.)Now

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Figure14.11a

atranslationsinceitistheproductoftworeflectionsinparallelmirrors.Ifnandparenotparallelorcoincident,letthemmeetatA.(SeeFig.14.11b.)

Letlinesn′andp′passthroughAsothatn′isperpendiculartomandσp′σn′=σpσn.Thenp′isperpendiculartoqbyTheorems14.9and14.10.(SeeFig.14.11c.)Letmandn′meetatB,andp′andqatC.LetrdenotelineBCandletm′andn′be the perpendiculars to r at B and C, respectively. (See Fig. 14.11d.) ByTheorems14.9and14.10again,σqσp′=σq′σrandσrσm′=σn′σmThen

atranslationsincem′andq′areparallelmirrors.

Figure14.11b

Figure14.11c

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Figure14.11d

14.13TheoremTheinverseofarotationaboutOthroughangleθistherotationaboutOthroughangle–θ.Iftherotationcanbewrittenasσnσm,thenitsinverseisσmσn(seeFig.14.9).

14.14 Theorem The product of two rotations about the same center O is arotationaboutO,andtheproductcommutes.

Letαandβbetherotationswithanglesθ1andθ2.ByTheorem14.10,writeα=σnσmandβ=σnwherem,n,parethreeproperlychosenlinesthroughO(seeFig.14.14).Then

arotationaboutOwithangleθ1+θ2.Sinceθ1+θ2=d2+θ2+θ1itfollowsthattherotationscommute.

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Figure14.14

14.15TheoremTheproductoftworotationsisarotationoratranslationandis,ingeneral,notcommutative.

If thecentersof therotationsare thesame, thenTheorem14.14applies.SoassumethecentersAandBarenotthesame.LetpdenotethelineAB.(SeeFig.14.15.)ByTheorem14.10,therearelinesmandnthroughAandBrespectivelysothatthetworotationsαandβaregivenby

Nowwehave

atranslationifmisparallelton,orarotationifmandnmeetinanordinarypoint.

Figure14.15

14.16 Definition A glide-reflection is the product of a translation and areflectioninamirrorparalleltothedirectionofthetranslation(seeFig.11.9).

14.17Theglide-reflectioncompletesourlistofoppositeisometries.Inthenextsectionweshallshowthateveryoppositeisometryiseitheraglide-reflectionor,asa specialglide-reflection,a reflection.Sinceeachdirect isometry iseitherarotation or a translation, we have introduced enough basic isometries.Nonetheless,inthenextsectionweshallintroduce—forconvenience—onemoredirectisometrytocompleteourlistofisometries.

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14.18TheoremThetranslationandreflectionofaglide-reflectioncommute.

ExerciseSet14

1.ProveTheorem14.2.

2.ProveTheorem14.4

3.ProveTheorem14.5.

4.ProveTheorem14.10.

5.ProveTheorem14.11.

6.ProveTheorem14.13.

7.Showthattworotationsaboutdifferentcentersdonot,ingeneral,commutebyexaminingthenatureoftheproductsαβandααwhereαandβare90°rotationsaboutpointsA(–1,0)andB(1,0).

8.Whatmustbetrueoftworotationsαandβiftheirproductβαisatranslation?

9.Iftheproductβαoftworotationsαandβisatranslation,whatcanbesaidabouttheproductαβ?

10.ProveTheorem14.18.11.Explainhowareflectionisaspecialglide-reflection.Isittruethata

translationisalsoaspecialglide-reflection?12.Provethataglide-reflectionisanisometry.Isitdirectoropposite?13.Findtheinverseofaglide-reflection.14.Whatsortofisometrymighttheproductoftwoglide-reflectionsbe?Could

itbeaglide-reflection?Explain.15.Istheproductoftworeflectionseverareflection?Explain.16.FindimagesforeachofthepointsA(0,0),B(1,0),C(0,3),D(1,1),E(2,

3),F(15,12),G(–3,–5)underarotationwiththeindicatedcenterandtheindicatedangle:a)(0,0),90°b)(0,0),180°c)(0,0),–90°d)(1,0),90°e)(1,1),90°f)(2,3),180°

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g)(–1,–3),–90°h)(1,–2),45°

17.FindimagesforthepointslistedinExercise12.16undereachtranslationwithvectora)P(0,0)andQ(5,0)b)P(0,0)andQ(2,–3)c)P(2,5)andQ(1,7)d)P(3,–2)andQ(3,–5)18.Provethattheproductofarotationandatranslationisarotation.

19.ThecenteroftherotationβαofTheorem14.15isattheintersectionofthelinesmandn.Locatethecenteroftherotationαβ.Whenisαβatranslation?

SECTION15 HALFTURNS

15.1TheoremAnisometrypreservesangles.FromTheorem13.16,areflectionpreservesangles;thatis,areflectionmaps

anangleintoacongruentangle.ThetheoremthenfollowsfromTheorem13.13.

15.2 It follows thatan isometry isacongruence transformation,and thateverycongruence transformation is an isometry. Thus isometries are the verytransformationsweseekforEuclideansuperposition.

15.3DefinitionAn involutoric isometryisanynonidentityisometrythesquareofwhichistheidentitymap.

15.4 Notice that it is customary not to call ι involutoric even though ι2 = ι.Exercise11.7helpstoclarifythisidea.

15.5DefinitionAhalfturnaboutpointP(orareflectioninpointP)isarotationaboutPthroughanangleof180°.ItisdenotedbyσP(SeeFig.11.6.)15.6Theterm “reflection in point P” is quite indicative of how this isometrymaps thepointsof theplane. In thischapter,however,weshall strictly reserve the term“reflection” for a reflection in a line. Of course, the term “halfturn” alsoindicates quite well the idea of this isometry. Although a halfturn is just a(special) rotation, it has been given its own name (halfturn) because of itsimportanceand itsuniqueproperties. It is theonly involutoric rotationand, infact,theonlyinvolutoricdirectisometry.Theorem15.7establishesthathalfturnsandreflectionsaretheonlyinvolutoricisometries.

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15.7TheoremEachinvolutoricisometryiseitherareflectionorahalfturn.Letσ be the involutoric isometry and letσ(A) =B withA ≠B. SinceA =

σ2(A), thenσ(B)=σ(σ(A))=σ2(A)=A.ChoosePequidistantfromAandBsothatPABisanisoscelestriangle.Letσ(P)=P′sothatσ(∆PAB)=∆P′BA.NowP′hasexactlytwopossiblelocations:sothatP=P′orelsePAP′Bisarhombus.Inthe first case,σ is a reflection in the perpendicular bisector ofAB, and in thesecondcaseσisahalfturnaboutthemidpointofABbyTheorem13.8.

15.8 Theorem If σp is a halfturn about point P, and a and b are any twoperpendicularlinesthroughP,thenσP=σbσa=σaαb.

15.9CorollaryReflectionsintwoperpendicularlinescommute.

15.10 Only involutoric isometries will be denoted by σ (lower-case Greeksigma); that is, only reflections and halfturns. A proof that a halfturn isinvolutoric is left for the exercises.Which involutoric isometry is intended ismadeclearby thesubscripton thesigma;upper-case italic lettersshallalwaysrefertopoints,andlower-caseitalicletterstolines.Henceσaisareflection,andσAisahalfturn.

15.11TheoremTheproductσBσAisatranslationthroughvector2 .LetσA(P)=P′andσB(P′)=P″(seeFig.11.7b).IntrianglePP″P″,segment

ABjoinsthemidpointsoftwosides,soABisparallelandcongruenttohalfthethirdsidePP″.Hence =2 ,andthetheoremfollows.

15.12 Theorem The product σc σB σA is a halfturn about pointD, the fourthvertexofparallelogramABCD.

ThetheoremcanbeprovedeasilyasacorollarytoTheorem15.11.ItisalsoquiteinstructivetoproveTheorems15.11and15.12bywritingeach

given halfturn as a product of two reflections in perpendicular lines, one ofwhichischosenthrough(orparallelto)ABineachcase.

ExerciseSet15

1.ProveTheorem15.1.

2.Explainhowahalfturncanbeconsideredtobea“reflectioninapoint.”

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3.Provethatareflectionisinvolutoric.

4.Provethatahalfturnisinvolutoric.

5.ProveTheorem15.8.

6.ProveTheorem15.12asacorollarytoTheorem15.11.

7.ProveTheorem15.12bywritingeachhalfturnasaproductofreflectionsassuggestedinthetext.

8.FindimagesforeachofthepointsA(0,0),B(1,0),C(0,3),D(1,1),E(2,3),F(15,12),G(–3,–5)underaglide-reflectionwithvectorPQandwiththegivenequationformirror.a)P(0,0),Q(1,0),and=0b)P(0,0),Q(1,0),andy=5c)P(2,5),Q(2,–1),andx=–4d)P(0,0),Q(l,1),andy=x

9.RepeatExercise15.8forahalfturnaboutthegivenpoint:a)(0,0)b)(5,0)c)(2,3)d)(–3,7)10.ProveTheorem15.11byfactoringσBandσAintoproductsofreflectionshavingacommonmirrorm=AB.

11.UseTheorem15.12andaconversetoTheorem15.11toprovethattwotranslationscommute(Theorem14.6).

SECTION16 PRODUCTSOFREFLECTIONS

16.1TheoremIflinesa,b,cformapencil, thenthereisalined inthatpencilsuchthatσcσbσa=σd.Conversely,ifσcσbσa=σd,thenlinesa,b,c,dbelongtoapencil.

Giventhatlinesa,b,cbelongtoapencil,thenfindlinedsothatσbσa=σcσdbyoneofTheorems14.4and14.10(seeFigs.11.8aand11.8b).Thenσcσbσa=σd.Theargumentreversestoestablishtheconverse.

16.2CorollaryIflinesa,b,cformapencil,thenσcσbσa=σaσbσc

16.3TheoremIfσgσbσaisaglide-reflectionwithbothaandbperpendiculartog,then,lettingB=g∩bandA=g∩ahave

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16.4 Theorem Each product σc σb σa is a glide-reflection. (A reflection is aspecial glide-reflection.) The theorem is trivial if a, b, c form a pencil. Soassumethat(forexample)onlyaandbintersectatP(seeFig.16.4a).RotateaandbaboutPintolinesa′andb′sothatσb′σa′=σbσaandb′isperpendiculartocatapointQ(Fig.16.4a).Similarlyrotateb′andcaboutQintolinesb″andc′sothatσc′σb″=σcσb′isperpendiculartob″(Fig.16.4c).Nowwehave

aglide-reflection,sincea′andc′arebothperpendiculartob″.

Figure16.4a

Figure16.4b

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Figure16.4c

16.5TheoremEachisometrycanbewrittenasaproductoftheformσbσaoroftheformσBσa.

16.6TheoremAllreflectionsandproductsofreflections,thatis,allisometries,formasubgroupofthetransformationgroupofTheorem12.17.

16.7TheoremAllrotationsandtranslationsformasubgroupofthegroupofTheorem16.6.

16.8TheoremAllhalfturnsandtranslationsformasubgroupofthegroupofTheorem16.7.

16.9TheoremAlltranslationsformasubgroupofthegroupofTheorem16.8.

16.10 Theorem All translations along a fixed line form a subgroup of thegroupofTheorem16.9.

16.11TheoremAnisometrywithnoinvariantpointiseitheratranslationoraglide-reflectionaccordingtowhetheritisdirectoropposite.

Suchan isometrycouldnotbea reflectionora rotationsinceeachof theseisometrieshasat leastone invariantpoint.Thecenterofa rotation is invariantandeachpointonthemirrorofareflectionisfixed.

16.12TheoremAnisometrywithatleastoneinvariantpointiseitherarotationorareflectionaccordingtowhetheritisdirectoropposite.

16.13 Theorem An isometrywithmore than one invariant point is either theidentityorareflectionaccordingtowhetheritisdirectoropposite.

16.14TheoremThechartinFig.16.14summarizestherelationsbetweeneachisometryanditsrepresentationintermsofreflections.

16.15 The theorems of this section tie together the algebraic properties ofreflectionsandproductsofreflectionswiththegeometricpropertiesoflinesandpoints.Thisinterplaybetweenalgebraandgeometryissufficientlyimportanttowarrant one more section (Section 17) to collect, classify, and emphasizetheoremsthatshowthisrelationship.

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Figure16.14

ExerciseSet16

1.FillinthedetailsoftheproofofTheorem16.1.

2.ProveTheorem16.3.

3.ProveTheorem16.5.

4.ProveTheorem16.6.

5.ProveTheorems16.7through16.10.

6.ProveTheorem16.12.

7.ProveTheorem16.13.

8.JustifythestatementsaboutfixedpointsandfixedlinesinTheorem16.14.

9.Showthatarotationthatisnotahalfturnisnotinvolutoric.10.Showthatneitheratranslationnoraglide-reflectionwithanonzerovector

isinvolutoric.11.LetlinesmandnmeetatP.Showthatwhenbothmandnarefixedlines

underanisometry,thenPisafixedpoint.12.GivenpointAandlinem,findpointBandlinensothatσAσm=σnσB.

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13.Decidewhethereachsetoftransformationsformsasubgroupofthegroupofallisometries.a)AllreflectionsinmirrorsthroughagivenpointPandallrotationsaboutP(rememberthattheidentitycanbeconsideredarotation).

b)Allrotations.c)AllrotationsaboutafixedpointP.d)Allreflectionsinmirrorsparalleltoagivendirectionandalltranslationshavingvectorsperpendiculartothemirrors.

e)Allhalfturns.

SECTION17 PROPERTIESOFISOMETRIES;ASUMMARY

17.1 The theorems of this section summarize the algebraic and geometricproperties of isometries and especially the relations between the algebra andgeometry of isometries. Remember that lower case letters represent lines anduppercaselettersrepresentpoints.Althoughallthetheoremsofthissectionareof interest, perhaps special emphasis shouldbe laidupon theorems17.3, 17.4,17.8,17.13,and17.15.Theseresultswillprovequiteusefulintheapplicationsthat follow in the next three sections. Drawing a picture to illustrate eachtheorem and relating each part of the theorem to the figure should make thetheorem easier to remember. Be sure to draw such an illustration for eachtheorem.

17.2TheoremThefollowingconditionsareequivalenttooneanother:1)A∈b,2)σAσb=σbσA,3)A=σb(A),4)b=σA(b),5)σbσA(orσAσb)isinvolutoric,and6)σbσAisareflectioninthemirrorthroughAperpendiculartob.

17.3TheoremThefollowingconditionsareequivalent:1)a=boraandbareperpendicular,2)σbσa=σaσb,3)a=ab(a),4)(σbσa)2=θ,and5)σbσaiseithertheidentityorahalfturn.

17.4TheoremThe following conditions are equivalent: 1)a,b,c,d belong to apencil and the directed angle or directed distance froma tob is equal to thatfromctod,2)σbσaσdσc,3)σdσbσaisinvolutoric(andhenceitisthereflectionσc),4)σdσbσaisareflection(namelyσc),5)σcσdσbσa=ι,and6)σdσbσa=σdσbσa=σc

17.5Theorem Ifa≠c, then the followingconditionsareequivalent:1)b lies

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midwaybetweenaandcorbbisectstheanglebetweenaandc,2)σcσb=σbσa,3)c=σb(a),and4)σc=σbσaσb.

17.6TheoremIfA≠C,thenthefollowingconditionsareequivalent:1)bistheperpendicularbisectorofAC,2)σcσb=σbσA,3)C=σb(A)and4)σC=σbσAσb

17.7 Theorem If a ≠ c, then the following conditions are equivalent: 1) a isparalleltocandBliesmidwaybetweenthem,2)σcσB=σBσa3)c=σB(a),and4)σc=σBσaσB.

17.8 Theorem The following conditions are equivalent : 1) ABCD is aparallelogram,2)σBσA=σCσD3)σC=σCσBσA,and4)σDσCσBσA=ι.

17.9TheoremIfA≠C,thenthefollowingconditionsareequivalent:1)BisthemidpointofAC,2)σCσB=σBσA,3)C=σB(A),and4)σC=σBσAσB

17.10TheoremThefollowingconditionsareequivalent:1)A=B,2)A=σB(A),3)σBσA=σAσB,and4)σBσA=ι

17.11TheoremIfσCσbσAisinvolutoric,thenitisareflection.

17.12TheoremIfσCσBσaisinvolutoric,thenitisahalfturn.

17.13TheoremσCσBσAisalwaysinvolutoric,soσCσBσA=σAσBσC.

17.14TheoremσaandσAarealwaysinvolutoric.

17.15Theorem(σcσbσa)2isatranslation.

The theorem states that the square of each glide-reflection is a translation.Theorem 16.3 implies that the glide and the reflection of a glide-reflectioncommute.The squareof theglide-reflection, then, reduces to the squareof itstranslationsincethereflectionscanbearrangedsoastocancel.

17.16Toconcludethissectionweshowaninterestingrelation,aproductof22reflectionsthatisequaltotheidentity!Theimportanceofthisresultliesnotin

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its being a vital theorem that every student must be able to recall instantly.Neither is it likely that Thomsen’s relation would be an appropriate topic ofconversationatadinnerparty.Itisofinteresttousfortheuttersimplicityofitsproof.Beforereadingtheproofgiveninthetext,trytoimaginehowyouwouldprovethistheorem.

17.17TheoremThomsen’srelation.Foranythreelinesa,b,c,

Since(σaσbσc)2and(σbσcσa)2aretranslationsbyTheorem17.15,thentheycommute.Furthermore,

Nowthedesiredrelationissimplytheidentity

ExerciseSet17

1.Theorems17.3to17.12canbepairedsothatthesecondtheoremsaysforlinesandpointsapproximatelywhatthefirsttheoremofthepairsaysforpointsandlines.Thetwotheoremsofsuchapairarecalleddualsofeachother.Forexample,Theorems17.6and17.7areduals.a)ArrangeTheorems17.3to17.12indualpairs.b)FinddualsforTheorems17.2and17.2.c)WhatcanbesaidaboutadualforTheorem17.13?

2through14.ProveTheorems17.2to17.14.15.Foranythreelinesa,b,c,provethat

16.Showthat,foranyfourparallellinesa,b,c,d,

17.ShowthatThomsen’srelationandtheformulasofExercises17.15and

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17.16arestilltrueifeachreflectionσxisreplacedbyahalfturnσx.Explainhowtheproofsmustbechanged.

SECTION18 APPLICATIONSOFISOMETRIESTOELEMENTARYGEOMETRY

18.1WearenowreadytoapplythetheoryofisometriesdevelopedintheearliersectionsofthischaptertoproblemsinEuclideangeometry.Webeginwithverybasicpropertiesandprogress towardmoreadvancedideas throughSections19and 20.Many of the theorems and proofs presented here can be used in highschoolclasses,ifisometriesarepresentedsufficientlyearly.Indeed,mostofthetheorems in Sections 18 and 19 appear in high school geometry texts (withdifferent proofs). The high school student who understands the workings ofisometries should find the methods of transformations quite logical andunderstandable.

Theprocedureweareusingmightbestbecalledtransform-solve-transform,forwhenwearegivenaproblem,we transform it intoanewproblemthroughtheuseofisometries,solvethenewproblem,thentransformthesolutionbacktothe original problem. In Theorem 18.4, for example, the given problem istransformedintooneinwhichthetwotrianglesshareacommonside, thentheproblemissolved(thetheoremisproved)forthiscase,andfinallythissolutionisappliedtothegivenproblemofprovinganytwotrianglesthatsatisfytheSSSconditioncongruent.

Remember that we have assumed the SAS condition for congruence as apostulate,sothatothercongruenceconditionsmustbeproved.

18.2TheoremVerticalanglesarecongruent.

Figure18.1

LetABandCDmeetatPasinFig.18.1.ThenσP( APC)= BPDandσP(APD)= BPC,sincelinesthroughParefixedunderthemapσP.

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18.3 Theorem If a point is equidistant from two points, then it lies on theperpendicularbisectorofthesegmentbetweenthetwopoints.

LetAbeequidistantfromPandQas inFig.18.3.Letmbe thebisectorofanglePAQ.ThenσmmapslineAPintolineAQ,andsinceAP≅AQ,σm(P)=Q.HencemistheperpendicularbisectorofPQ.

Figure18.3

18.4TheoremTwo triangles are congruent if they satisfy theSSScondition ;that is, if the three sides of the first triangle are congruent respectively to thecorrespondingsidesofthesecondtriangle.

Let the sides of triangle ABC be congruent to the corresponding sides oftriangleA′B′C′.UseanisometrytomaptriangleA′B′C′totriangleABC″withC and C″ on opposite sides of line AB (Fig. 18.4). Since A and B are bothequidistant fromC and fromC″, then lineAB is the perpendicular bisector ofCC″.Hence the reflection in lineAB carries triangleABC″ into triangleABC.Now, since an isometry is a congruence transformation, the theorem follows;thatis,trianglesABCandA′B′C′arecongruent.

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%

Figure18.4

18.5TheoremTwo triangles are congruent if they satisfy theASAcondition;that is, if two angles and the included side of the first triangle are congruentrespectivelytothecorrespondingpartsofthesecondtriangle.

18.6TheoremTwo triangles are congruent if they satisfy theAAScondition;that is, if two angles and a side opposite one of them in the first triangle arecongruentrespectivelytothecorrespondingpartsinthesecondtriangle.

18.7TheoremParallellinescutbyatransversalformcongruentcorrespondingandalternateangles.

LettransversalpcuttheparallellinesmandnatpointsAandB(Fig.18.7).A translation through vectorAB carries the angles atA into those atB (sincelinesmandnareparallel),establishingthetheorem,

Figure18.7

18.8TheoremThebaseanglesofanisoscelestrianglearecongruent.

18.9TheoremThediagonalsofaparallelogrambisecteachother.

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LetNbethemidpointofdiagonalACofparallelogramABCD.ThenσAσN=σNσC.WemustshowthatNisthemidpointofBD;thatis,thatσDσN=σNσB

18.10 Corollary A parallelogram is symmetric with respect to the point ofintersectionofitsdiagonals.

18.11 Corollary A parallelogram and either diagonal form two congruenttriangles.

18.12TheoremTheperpendicularbisectorof the (major)baseofan isoscelestrapezoidistheperpendicularbisectoralsoofthesummit(minorbase).

ReflecttheisoscelestrapezoidABCDinitsbaseABintotheimagetrapezoidABC′D′(Fig.18.12).LetCC′andDD′cutABatEandF.TheanglesatEandFarerightanglesandDF≅CE.NowAD≅BCisgiven,andAD≅AD′,BC≅BC′andCC≅DD′.SoisoscelestrianglesADD′andBCC′arecongruent.ThentheircorrespondingaltitudesAFandBEarecongruent.Hencetheperpendicularbisector of AB is the perpendicular bisector of EF, hence also of CD, sinceCDFEisarectangle.

Figure18.12

18.13Corollary

a)Thebaseanglesofanisoscelestrapezoidarecongruent.b)Thesummitanglesofanisoscelestrapezoidarecongruent.c)Thediagonalsofanisoscelestrapezoidarecongruent.d)Anisoscelestrapezoidissymmetricwithrespecttotheperpendicularbisectorofitsbases.

18.14 Theorem If the base angles (or the summit angles) of a trapezoid are

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congruent,thenthetrapezoidisisosceles.

18.15TheoremInFig.18.15,ifAB≅BC≅DE,AD≅BE,andBD≅CE,thenA,B,andCarecollinear.

TrianglesABDandBCEarecongruentbySSSandBCEDisaparallelogram,so BD is parallel to CE. Similarly, AD is parallel to BE. Thus there is atranslation mapping triangle ADB into triangle BEC. Since a translationpreservesdirectionsoflines,itfollowsthatA,B,andCarecollinear.

Figure18.15

ExerciseSet18

1.ProveTheorem18.5.

2.ProveTheorem18.6.

3.ProvethattworighttrianglesarecongruentiftheysatisfytheHLcondition.

4.ProveTheorem18.8asacorollarytoTheorem18.3.

5.ProveTheorem18.8bysyntheticgeometryusingasanauxiliarylinefromtheapex(thevertexbetweenthetwocongruentsides)oftheisoscelestriangle,a)themedian,b)thealtitude,c)theanglebisector.

6.a)ProveTheorem18.8byreflectinginthebisectorofthevertexangle.b)ProveTheorem18.8byshowingtrianglesABCandACBcongruentbytheSASpostulate(wheresidesABandACaregivencongruent).

7.ProveTheorem18.9.

8.ProveCorollary18.10.

9.ProveCorollary18.11.10.ProveCorollary18.13.11.ProveTheorem18.14.

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12.Provethat(a)themedians,(b)thealtitudes,and(c)theanglebisectorsdrawnfromthebaseanglesofanisoscelestrianglearecongruent.

13.Provethatnomedianofascalenetriangleisperpendiculartothesideitbisects.

14.Provethatthediagonalsofarhombusareperpendicular.15.Provethatifthediagonalsofaparallelogramareperpendicular,thenthe

parallelogramisarhombus.16.ProveTheorem18.12forarectangle(assumedtrueintheproofofthat

theorem).

SECTION19 FURTHERELEMENTARYAPPLICATIONS

19.1TheoremAcircleissymmetricinanydiameter.

19.2 Theorem A diameter of a circle perpendicular to a chord of that circlebisectsthechord.

19.3TheoremTwochordsofacirclearecongruentifftheyareequidistantfromthecenter.

Given that the chords are equidistant from the center, then reflect in thatdiameter that bisects the angle between the radii perpendicular to the chords.Establishtheconversebyarotation.

19.4 Corollary Congruent chords of a circle intercept congruent arcs of thecircle,andconversely.

19.5 Theorem The angles of intersection of two intersecting circles arecongruent.EachofthecentersAandBofthetwocirclesisequidistantfromPand Q, the points of intersection of the circles, so A and B lie on theperpendicularbisectorofPQ.ReflectthefigureinlineAB.ThecirclesmapintothemselvesbyTheorem19.1andPmapsintoQ.Thetheoremfollows.

19.6TheoremTheareaofaparallelogramisequaltothatofarectanglewiththesamebaseandaltitude;thatis,itistheproductofitsbaseanditsaltitudetothatbase.

In parallelogramABCD (Fig. 19.6), there is a translation a that carriesADintoBC.Assuming,without lossofgenerality, thatangleA is less thana rightangle,dropperpendicularDEtosideAB.Letα(∆ADE)=∆BCF.ThenE,B,and

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F are collinear, so DEFC is a rectangle, and its area is equal to that ofparallelogramABCD.

Figure19.6

19.7TheoremTheareaofatrapezoidisequaltotheproductofthealtitudeandhalfthesumofthebases.

PerformahalfturnaboutthemidpointMofanonparallelsideADtoformaparallelogram B′C′ BC as shown in Fig. 19.7. Then apply Theorem 19.6 toestablishthetheorem.

Figure19.7

19.8TheoremTheareaofatriangleishalftheproductofanysideasbaseandthealtitudetothatside.

19.9 Theorem The midpoints of the sides of a quadrilateral form aparallelogram.

LetABCDbeaquadrilateralhavingM,N,O,Pasmidpointsof itssidesasshowninFig.19.9.NowσPσOσNσMisatranslation,andatranslationthroughanonzerovectorhasnofixedpoints.But

sowemusthaveσPσOσNσM=ι.ThusMNOPisaparallelogram.

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Figure19.9

19.10TheoremIfABQPandBCRQareparallelograms,thensoalsoisACRPaparallelogram.

Wehave(seeFig.19.10)

soACRPisaparallelogram.

Figure19.10

19.11TheoremIfABPQandBCQRareparallelograms,thensoalsoisACPRaparallelogram(Fig.19.11).

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Figure19.11

19.12TheoremIfA,B,C,D,E,Paresixpointslyingonacircle,andlinesa,b,careconcurrent,whereaistheperpendicularbisectorofABandofDE,bisthe perpendicular bisector ofBC andEF, and c is that forCD, then c is theperpendicularbisectoralsoofFA(Fig.19.12).

Figure19.12

19.13TheoremIfPisthemidpointofABandofDE,QisthemidpointofBCandofEF,andRisthemidpointofCD,thenRisthemidpointalsoofFA.

SeeFig.19.13.Since(σRσQσP)2=ι,thenwehave

thenRisthemidpointofAF.

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Figure19.13

ExerciseSet19

1.ProveTheorem19.1.

2.ProveTheorem19.2.

3.ProveTheorem19.3.

4.ProveCorollary19.4.

5.ProveTheorem19.8.

6.ProveTheorem19.11.

7.ProveTheorem19.12.

8.Whatfiguredothefourinternalanglebisectorsofaquadrilateralform?

9.Provethatthemediansofatriangle,asvectors,formatriangle.10.Provethatthemidpointofthehypotenuseofarighttriangleisequidistant

fromallthreevertices.11.Provethatthetwotangentsfromapointtoacirclearecongruent.12.ReviewtheproofofTheorem8.8(thattheEuclideanandmodern

compassesareequivalent).13.Letlinesa,b,cformatrianglewhoseorthictriangleisDEF.Provethatσc

σbσaistheglide-reflectionwhosemirroristhelineDF,andwhosevectorisdirectedfromDtowardFandhaslengthequaltotheperimeteroftheorthictriangle.

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14.Thecoingame.Twoplayersalternatelyplacecoinsoneatatimeonarectangulartabletop,withoutlettingthecoinsoverlap.Thewinneristhelastplayerwhocanfindaspaceforacoin.Assumingthatyouplayfirst,whatisawinningstrategy?Forwhatshapesoftablesdoesyourstrategyapply?(Forpracticalpurposesonemightplaythegamewithquartersornickelsona3-by-5card.)15.Thepolygongame.Acoinisplacedoneachoftheverticesofaregularpolygon.Eachplayerinturnremoveseithertwocoinsfromadjacentverticesoronecoinfromanyvertex.Thewinneristheonewhotakesthelastcoin.Assumingthatyouplaysecond,whatisawinningstrategy?Forwhatotherconfigurationsofcoinsdoesyourstrategyapply?

SECTION20 ADVANCEDAPPLICATIONS

20.1TheoremTheperpendicularbisectorsofthesidesofatriangleconcur.

20.2Firstproof.Lettheperpendicularbisectorsofsidesa,b,coftriangleABCbedenotedbyd,e,f.(SeeFig.20.2.)Then

soσfσeσdisanoppositeisometrywithafixedpoint.ByTheorem16.14itisareflectioninaline;thatis,d,e,fconcurbyTheorem16.1.

Figure20.2

20.3Secondproof.UsingFig.20.2again,letdandfmeetatOanddenoteline

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BObyg.ThentheisometryσdσgσfisareflectioninalinemthroughO.Since

itfollowsthatmistheperpendicularbisectorofAC.

20.4TheoremThebisectorsoftheanglesofatriangleconcur.DenotethebisectorsofanglesBandCintriangleABCbyeandf.Leteandf

meetatI (seeFig.20.4).Thenσfσgσe,whereg is theperpendicular fromI tosidea,isareflectioninalinemthroughIbyTheorem16.1.Since(σfσgσe)(c)=b,thenmisthebisectorofangleAbyTheorem17.5.

Figure20.4

20.5NotethatanytwoofthebisectorsinTheorem20.4canbetakenasexternalandthethirdoneinternal.

20.6TheoremIfσcσAσB=σv,σAσBσc=αv,andσBσCσA=αW,then

WehavebothσUσCandσCσVequalto

Theotherequationsfollowinasimilarmanner.

20.7 Theorems 20.6 and 20.9 are purely algebraic in statement and in proof,based on only the group properties of halfturns. Geometric interpretations ofthese algebraic theorems are given in the corollaries that follow the theorems.

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Results such as these illustrate the close connections between algebra andgeometry.

20.8CorollaryIfCABU,ABCV,andBCAWareparallelograms,thenABCisthemedial triangle for triangleUVW. Furthermore, the altitudes of triangle ABCconcurbyTheorem20.1sincetheyaretheperpendicularbisectorsofthesidesoftriangleUVW(Fig.20.8).

Figure20.8

20.9TheoremIfσC(U)=V,σA(V)=W,andσW(U),thenσU=σBσAσC,σV=σAσBσCandσW=σBσCσA.

SincethehalfturnσBσAσCleavespointUfixed,thenitisthehalfturnaboutpointU,etc.

20.10 Corollary The triangle (ABC) formed by joining the midpoints of thesides of a triangle (UVW) has sides parallel to and congruent to halves of thesidesofthegiventriangle(Fig.20.8).

20.11 Theorem The medians of a triangle meet at a trisection point of eachmedian.

LetGbetwo-thirdsofthewayfromAtothemidpointA′ofthejDppositesideoftriangleABC(Fig.20.11).Sincevector istwicevector thenσAσG=σGσA′σGσA′WemustshowthatσCσG=σGσC′σGσC′,whereC′isthemidpointofsideAB.ByTheorem17.9,sinceA′andC′aremidpointsoftheirrespectivesides,thenσB=σA′σCσA′andσB=σC′σAσC′=ιsoσC′=σBσC′σA=σA′σC′σA′σC′σA.

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Figure20.11

Nowwehave

becauseGistwo-thirdsofthewayfromAtoA′.Similarly,Gistwo-thirdsofthewayfromBtoB′andthetheoremfollows.

20.12Theorem IfCeviansa′b′c′ in triangleABC concur, then their isogonalconjugatesa″,b″c″concuralso.[Isogonalconjugatelinesaretwolinesthroughavertexofanangleandsymmetricwithrespecttothebisectoroftheangle.]

Figure20.12

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Wearegiven(Fig.20.12)σcσa′=σa″σb,σbσc′=σc″σa,σaσb′=σb″σcand(σa′σb′σb′)2=ι.Wemustprovethat(σa″σb″σb″)2=ι.Tothatend,

20.13TheoremLetPbeanypointnoton triangleABCand letD,E,Fbe thereflectionsofPinsidesa,b,c.ThenthecircumcenterOoftriangleDEFistheisogonal conjugatepointofpointP in triangleABC. [Twopoints are isogonalconjugatepointswithrespecttoatriangleiffthethreeceviansdeterminingonepointareisogonalconjugatelinesoftheceviansdeterminingtheotherpoint.]

Figure20.13

ByExercise20.1appliedtotrianglePEF,theperpendicularbisectorofEFistheisogonalconjugateoflineAPinangleCAB(Fig.20.13).

20.14TheoremTheorthocenterandthecircumcenterofatriangleareisogonalconjugatepoints.

InTheorem20.13 let thegiven trianglebeABC and its circumcenterbeP.ThentrianglesPEFandABChavecommonmidpointsMandNforthesidesPEandAC,andsidesPPandAB (seeFig.20.13). It follows thatEFisparalleland

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congruenttoBC.SimilarlyforABandDEandforCAandFD.HencetrianglesABC andDEF are congruent.Now the circumcenterP of triangleABC is theorthocenter of triangle DEF. Hence the circumcenter of triangle DEF is theorthocenteroftriangleABC.ThetheoremfollowsfromTheorem20.13.

20.15TheoremIntriangleABC,Sisthecentroidiff

SupposefirstthatwearegivenσSσCσSσBσSσA=ι.SinceA′isthemidpointofsideBC,thenσB=σA′σcσA′,byTheorem17.9.Now

whichstatesthatSistwo-thirdsofthewayfromAtoA′.HenceSisthecentroidG.(SeeFig.20.15.)

Figure20.15

Conversely,ifSisthecentroid,then(σSσC)(σSσB)(σSσA)=ιsincethethreemedians,asvectors,formatriangle,byTheorem6.7.SeealsoExercise19.9andAnswer19.9.

20.16Corollary The centroid of a triangle is at a trisection point of each

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median.

20.17 Theorem Fagnano’s Problem. In a given acute triangle inscribe atriangleRSThavingminimumperimeter.

ChooseanythreepointsR,S,TonthethreesidesofthegiventriangleABC,asshowninFig.20.17.Letσb(R)=R′andσc(R)=R″.ThentheperimeterpoftriangleRSTisgivenby

so,forfixedR,pisleastwhenR′STR″isastraightline.

Figure20.17

SinceAR″≅AR≅AR′and R″AR′≅2 AbyconstructionofR′andR″,thentriangleAR′R″isdeterminedwhenRisfixedandthelengthofR′R″variesdirectly with that of AR. Thus R′ R″ is a minimum when AR is as short aspossible;thatis,whenARisperpendiculartoBC.SimilarlySandTmustalsobefeetofaltitudesoftriangleABC.FromTheorem7.6,ifRSTistheorthictriangle,thenR′STR″isastraightline.Hencetheorthictriangleistheuniquesolution.

ExerciseSet20

1.AsacorollarytoTheorem20.1,showthatthedirectedanglefromdtogiscongruenttothatfrometof.

2.ProveTheorem20.4foroneinternalandtwoexternalbisectors.

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3.ProveCorollary20.8.

4.CompletetheproofofTheorem20.9.

5.ProveCorollary20.10.

6.IntheproofofTheorem20.11showthatGis2/3ofthewayalongmedianBB′

7.ProveTheorem20.14asacorollarytoTheorem6.20.

8.IntheproofofTheorem20.14showthattriangleDEFistheimageoftriangleABCinahalfturnabouttheninepointcenter.

9.ProveCorollary20.16.10.Provethataquadrilateralwiththreerightangleshasfourrightangles.11.Provethattheanglebisectorsofatriangleconcur,bytakingX,Y,Zasthe

pointsofcontactoftheincirclewiththesidesa,b,cofthetriangle,andd,e,fasthebisectorsofanglesA,B,C.Thenshowthat(σfσeσd)(Y)=Y.Completetheproof.

12.LetABCDbeaquadrilateralwithrightanglesatAandC.LetQ=σA(D)andR=σC(D).ProvethatH,themidpointofQR,istheorthocenteroftriangleABC.

13.Hjelmslev’sTheorem.Ifαisanisometrysuchthatα(m)=n,thenforeachpointPinm,thereisapointQinn,suchthatα(P)=Q.ProvethatthemidpointsofsuchsegmentsPQeitherallcoincideorarealldistinct.

14.Eves’problem.Afarmerhashishouseandbarnonthesamesideofastraightriver,atdistanceshandbfromtheriver,withthebarndunitsdownstreamfromthehouse.Towardwhichpointontheriver’sedgeshouldheheadfromthehouseinordertofillapailwithwaterandcarryittothebarnwhiletravelingtheshortesttotaldistance?

15.Provethattheperpendicularsdroppedfromtwoverticesofatriangleuponthemedianfromthethirdvertexarecongruent.

16.ThroughagivenpointPdrawalineequidistantfromgivenpointsAandB.17.Provethatatrapezoidinscribedinacircleisisosceles.

18.TrianglesABCandDEFhaveequalareasandcongruentbasesBCandEFlyingonthesamestraightline.VerticesAandDareonoppositesidesofthatline.ShowthatADisbisectedbytheline.

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19.MediansBB′andCC′oftriangleABCareextendedtheirownlengthstopointsB″andC″.ShowthatAliesonlineB″C″.

20.Fromapointinsideacircle,morethantwocongruentsegmentscanbedrawntothecircle.Provethatthepointisthecenterofthecircle.

21.AtangentintersectsatAandBtwoparalleltangentstoacirclewithcenterO.ProvethatthecircleonABasdiameterpassesthroughO.

22.[From“DynamicProofsofEuclideanTheorems”byR.L.Finney,MathematicsMagazine43(1970)pages177–185.]a)IfABMandCDMaresimilarly-orientedisoscelesrighttriangleswithrightanglesatM,provethatACandBDarecongruentandperpendicular.

b)IfXandYarethecentersofsquareserectedexternallyonsidesABandBCoftriangleABC,provethatXB′andYB′arecongruentandperpendicularatB′themidpointofsideAC.

c)LetW,X,Y,Zbethecentersofsquares(inorder)erectedexternallyonthesidesofaquadrilateral.ProvethatWYandXZarecongruentandperpendicularsegments.

d)Provethatthesegmentjoiningthecentersoftwosquareserectedexternallyontwosidesofatriangleiscongruentandperpendiculartothesegmentjoiningthecenterofthethirdsuchsquaretotheoppositevertex.

e)Showthattheword“externally”canbereplacedby“internally”inparts(b),(c),and(d).

SECTION21 ANALYTICREPRESENTATIONSOFDIRECTISOMETRIES

21.1Itwouldseemlogicaltowritefirsttherepresentationforareflection,thenfind a method of multiplying transformations since all other isometries areproductsof reflections. It ismuchsimplerhowever, to find representations fortranslationsandrotationsdirectlyfromtheirgeometricproperties.Infact,inthenextsectionweshallutilizetheresultsfoundhereinordertocalculatetheformofthegeneralreflection!

Incontrasttothemethodsandformulasusuallydevelopedinacalculusandanalyticgeometrycourse,weshallconsider translating, rotating,andreflectingallthepointsintheplaneinsteadofmovingthecoordinateaxes.Inthatrespectourmappingsarejusttheinversesofthemappingsconsideredinthecalculus.

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21.2SupposethateachpointP(x,y)istobetranslatedthroughvectorv= topointP′(x′y′)whereA(h,k)(Fig.21.2).ThenPismovedhunitstotherightandk units up, so that x′ = x + h and y′ = y + k. These equations define thetranslation,sowehaveprovedthefollowingtheorem.

Figure21.2

21.3TheoremLetv= whereO(0,0)andA(h,k).ThenthetranslationthatmapseachpointP(x,y)throughvectorvtopointP′(x′,y′)isgivenby

21.4DefinitionThevectorvofTheorem21.3issaidtohavecomponentshandkandwewritev=(h,k)forthisvectororforanyvectorv= whereC(a,b)andD(a+h,b+k).

21.5ConsiderrotatingeachpointP(x,y)aboutpointO(0,0)throughangleθtopointP′(x′ ,y′).Letthedirectedanglefromthepositivex-axistorayOPbeϕ,andletm(OP)=r.Then(seeFig.21.5)

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Figure21.5

Theanglefromthepositivex-axistorayOP′isthenθ+ϕ,andwehave

and

These equations determine the rotation, so we have proved the followingtheorem.

21.6 Theorem The rotation about the origin through angle θ, mapping eachpointP(x,y)intoP′(x′,y′),isgivenby

21.7Letαandβbetwotransformationsoftheplane,andletα(x,y)=(x′,y′)and

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β(x′,y′)=(x″,y″).Then,bydefinitionoftransformationmultiplication,wehave(βα)(x,y)=(x″,y″).

21.8WemayaccomplisharotationaboutpointC(h,k)throughangleθbyfirsttranslating throughvectorv (–h,—k)so that thepointC(h,k) ismoved to theorigin, second rotating through angle θ about the origin, and third translatingbackthroughvector–v=(h,k).Lettingα,β,andα–1denotetheseisometries,wehave

and

Calculatingα–1βαbyeliminatingx′,y′,x″,andy″fromtheequationsabove,wehavethefollowingtheorem.

21.9TheoremTherotationaboutpointC(h,k)throughangleθisaccomplishedby

21.10TheequationsofTheorem21.9aresufficientlycomplicatedthatthereaderisurgedtorememberthemethod(translatetotheorigin,rotateabouttheorigin,thentranslatebackagain)ratherthantomemorizetheformulas.

21.11 Thus translations and rotations about the origin have relatively simple

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formulations, so their analytic forms may be used whenever convenient.Rotatingaboutotherpointsprobablyshouldbeavoidedwheneverpossiblewhenusinganalyticrepresentations.Ofcourse,linesandothercurveswhoseequationsareknownmayberotatedandtranslatedbymakinguseoftheequationsgiveninTheorems 21.3, 21.6, and 21.9.Muchmore convenient are the implicit formsfoundbysolvingtheseequationsforxandyintermsofx′andy′.ForarotationthisisaccomplishedmosteasilybynotingthatP(x,y)isfoundfromP′(x′,y′)bythe inverseof thegivenrotation.The implicitequations for the translationandtherotationabouttheoriginare

Theequationsfortherotationmaybewritteninthesimplerform

although it may be more confusing to try to remember this simpler form inadditiontotheoriginalrotationform.Thatis,hereagainitisprobablybettertoremembertheideathattheimplicitformcomesfromarotationthroughangle–θinmappingP′toPratherthantorememberthespecificequations.

21.12ExampleRotate the line 2x + 3y = 5 through90° about the origin.Wehave

and

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Makingthesesubstitutionsinthegivenequationfortheline,weobtain2y′–3x′= 5. It is left to the reader to graph both equations to show that the result iscorrect.

Weconcludethissectionbystatingtheequationsforahalfturn.

21.13TheoremThehalfturnaboutthepointC(h,k)isgivenby

ExerciseSet21

1.Forthepoints0,A,C,DgiveninTheorem21.3andDefinition21.4,calculateanalyticallytheproductσCσDσAσOtoshowthatOADCisaparallelogram.

2.GraphthelinesofExample21.12.

3.ProveTheorem21.13fromageometricpicture.

4.ProveTheorem21.13asaspecialcaseofTheorem21.9.

5.Showanalyticallythattheproductoftwotranslationsisatranslation.

6.Showthattheproductoftworotationsabouttheoriginthroughanglesθandϕisarotationabouttheoriginthroughangleθ+ϕ.

7.ShowthattheproductofthetworotationsabouttheoriginwithangleθandaboutthepointC(h,k)withangleϕisatranslationiffθ+ϕisanintegralmultipleof360°.

8.FindananalyticrepresentationfortheinverseofthegeneraltranslationasgiveninTheorem21.3.

9.FindananalyticrepresentationfortheinverseoftherotationabouttheorigingiveninTheorem21.6.

10.FindananalyticrepresentationfortheinverseofthegeneralrotationasgiveninTheorem21.9.

11.FindananalyticrepresentationfortheinverseofthehalfturnasgiveninTheorem21.13.

12.ShowthattherotationaboutC(h,k)throughangleθcanbewrittenintheformbelow,andfindrandsintermsofh,k,andθ:

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13.ShowthattheequationsofExercise21.12showthateveryrotationcanbewrittenasarotationabouttheoriginfollowedbyatranslation.

14.Giventhata2+b2=1,showthateachtransformationoftheform

representstheproductofarotationandatranslation,andfindtheangleoftherotation.

15.ShowthattheisometryofExercise21.14isatranslationwhena=1;henceshowthattheseequationsrepresentalldirectisometries.

16.RotateA(1,0)through120°andthrough240°abouttheorigintolocatetheverticesofanequilateraltriangle.Showthatx=– isoneofitssides,androtateitthroughthesamerotationstofindequationsfortheothertwosides.

SECTION22 ANALYTICREPRESENTATIONSOFOPPOSITEISOMETRIES

22.1 A reflection in a coordinate axis has a very simple form. The followingtheoremisobvious.

22.2TheoremThereflectionσxinthex-axisandthereflectionσyinthey-axisaregiven,respectively,by

22.3 Reflections in other lines through the origin are handled without greatdifficulty.LetmbealinepassingthroughtheoriginwithangleofinclinationθasshowninFig.22.3.Letσxdenotethereflectioninthex-axisandletαdenotearotationabouttheorigininangle2θ.Sinceσmσx=α,wehave

thatis,thereflectioninlinemistheproductofareflectioninthex-axisfollowedby a rotation about the origin with angle 2θ. Since σx and α have therepresentations

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thenσm=ασxhastherepresentation

By setting θ = 0 and θ = 90° and comparing the results with those given inTheorem22.2,onecanseethattheequationsforthereflectionσmasgivenaboveare indeed correct for reflections in the x- and y-axes. The results we haveobtainedarestatedasTheorem22.4.

Figure22.3

22.4 Theorem A reflection σm in the linem passing through the origin withangleofinclinationθhastherepresentation

22.5 When its mirror does not pass through the origin, we may handle thereflectionjustaswedidforarotationnotabouttheorigin.Thus,supposelinempasses throughpointA(h,k)with inclinationθ.First, translateA to theorigin,then reflect in line n through the origin and parallel to line m, and finally

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translatebackagain.Theresultsarestatedinthenexttheorem.

22.6TheoremTheequationsforthereflectionσminlinempassingthroughAQi,k) with angle of inclination θ are 22.7 This general form of Theorem 22.6 isagain too complicated to be worth memorizing, but it is easily reconstructedwhenneeded.

22.8Lastly, letusconsideraglide-reflectionαwhosemirrormpasses throughtheoriginwithinclinationθandwhosetranslationβisrunitsalongmmeasuredpositiveupwardor,forahorizontalline,positivetotheright.Thetranslationandthereflectionarethengivenby

Nowtheglide-reflectionα=σmβ=βσmisgiveninTheorem22.9.

22.9TheoremAglide-reflectionwhosemirrormpassesthroughtheoriginwithangle of inclinationθ andwhose translation is along linem through r units, rmeasured positive from the origin into the first two quadrants or along thepositivex-axis,andnegativeotherwise,isgivenby

22.10When themirrorof theglide-reflectionpasses through thepointA(h,k)insteadoftheorigin,theequationsforitsrepresentationarereadilydevelopedinthesamemanneraswehavedoneearlierforthereflectionandfortherotation.Thisformulationforthegeneralglide-reflectionisleftasanexercise.

ExerciseSet22

1.ProveTheorem22.2.

2.ProveTheorem22.6.

3.Showanalyticallythattheproductofreflectionsinlinesmandnthroughthe

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originwithinclinationsθandϕisarotationabouttheoriginwithangle2(ϕ–θ).

4.Showanalyticallythattheproductofreflectionsinparallellinesmandnofinclinationθwithmpassingthroughtheorigin,andnpassingthroughA(h,k),isatranslation.Finditsvector.

5.ShowthattheequationsofTheorem22.9foraglide-reflectionarearrivedatwhetheroneanalyticallyusesα=σmβorα=βσm,whereβandσmarethetranslationandreflectiongivenin22.8.

6.FindananalyticrepresentationfortheinverseofthereflectionofTheorem22.4.

7.FindananalyticrepresentationfortheinverseofthereflectionofTheorem22.6.

8.Findananalyticrepresentationfortheinverseoftheglide-reflectionofTheorem22.9.

9.Formulatetheequationsforthegeneralglide-reflection.10.Findananalyticrepresentationfortheinverseofthegeneralglide-reflection

foundinExercise22.9.11.Showthateachreflectioncanbewrittenintheform

12.Showthateachglide-reflectionalsohastheformofExercise22.11.Henceeveryoppositeisometryhasthisform.

13.ShowthattheformofExercise22.11representsareflectioninalinethroughtheorigin,followedbyatranslationinthevector(c,d)notnecessarilyparalleltothemirrorofreflection.Henceallsuchequationsrepresentoppositeisometries.

14.Showthateachisometryhastheanalyticform

andthatitisdirectiftheplussignholdsandoppositeiftheminussign

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applies.Conversely,showthateverytransformationgivenbytheequationsaboveisanisometry.Hencetheseequationscharacterizeisometriesanalytically.

15.ShowthattheproductoftwoisometriesoftheformgiveninExercise22.14isanotherisometryofthatform.

16.Showthateachisometrycanbewrittenastheproductofeitherarotationabouttheoriginorareflectioninalinethroughtheorigin,followedbyatranslation.

17.Findimplicitformsfor:a)ThereflectionsofTheorem22.2b)ThereflectionofTheorem22.4c)ThereflectionofTheorem22.6d)Theglide-reflectionofTheorem22.9e)Thegeneralglide-reflectionfoundinExercise22.9

18.FromtheimplicitformsforareflectionfoundinExercise22.17,findanequationfortheimageofacirclex2+y2–4x–2y=0inareflectionintheliney=x.Findanyfixedpointsonthecircleandsketchagraph.

19.Showthat,intheformgivenforaglide-reflection,ifweletthetranslationhavevector ,takeralwayspositive,andtakeθasthedirectedangle(lessthan360°)fromthepositivex-axistotherayOA,thentheformulasgivenarestillcorrect.

20.Showanalyticallythataproductofthreereflectionsinmirrorsthroughtheoriginisareflectioninamirrorthroughtheorigin.

21.Showanalyticallythataproductofthreereflectionsinparallelmirrorsisareflectioninamirrorparalleltothegivenmirrors.

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3 SIMILARITIESINTHEPLANE

SECTION23 THEREBIRTHOFMATHEMATICALTHINKING

23.1 During the Dark Ages, beginning with the fall of Alexandria in 641,mathematical thinking sank to its nadir. Althoughminor discoveries appearedfromtimetotime,thenextthousandyearswereessentiallybarren.Furthermore,absolutelynoprogressinthemathematicalmethodoccurreduntilthenineteenthcentury. Thus, from the time of its first zenith about the time ofArchimedes,morethan2000yearspassedbeforeasecondzenithapproached.23.2TheArabs,bycartingawayandtranslatingintoArabicsomeofthelibraryatAlexandria,preservedmuchoftheGreekknowledge,latertransmittingitbackintoEurope.TheyalsoadoptedtheHindunumeralsystemandpasseditalongtoWesternmathematicians in lateryears.Thus theprimaryfunctionof theArabswasthatofcustodianofGreeklearning.23.3 At the end of the Dark Ages some knowledge began to filter back intoEurope,andthetwelfthcenturybecameatimeoftranslators,theancientGreeklearningbeing translated fromArabic intoLatin, the languageof theemergingscholars. In the thirteenth century universities began to spring up all over theContinent as people began again to delve into the mysteries known to theGreeks.23.4The fourteenth centurywasmarkedby theBlackDeath and theHundredYears’ War, to the detriment of mathematics. The fifteenth and sixteenthcenturiesbegantoshowpromiseofabrightfuturewithadvancesinarithmetic,algebra, and trigonometry. Printing was invented, so knowledge could bedispersedmuchmorerapidlyandwidelythaneverbefore.Thestagewassetforthegreatsceneswhichbegantobeplayedintheseventeenthcentury.23.5In1647WilliamOughtred(1574–1660)usedπ/δasthesymbolfortheratioπ=3.14159….Hegavea totalofmore than150symbols,mostly inalgebra.The straight logarithmic slide rule was his invention. Also he and one of hisstudents each independently invented the circular slide rule, it is said that hiswifewas so frugal that shewould not permit him the use of a candle for hisnighttime studies.Heworked late into thenight, sometimes skipping sleep fortwoorthreenightsinarow.Andheissupposedtohavedied“inatransportofjoy”whenheheardof therestorationofCharlesII to thethrone.“Itshouldbeadded,bywayofexcuse,”DeMorgancommented,“thathewaseighty-sixyearsold.”23.6 Johannes Kepler (1571–1630), a close friend of Galileo Galilei (1564–1642), spent a full 22 years analyzing against observed data the theories he

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assumedforthemotionsoftheplanets.Hefinallyconcludedthatalltheplanetstravelinellipticalorbitswiththesunatonefocus,thattheradiusvectorfromthesuntoagivenplanetsweepsoutequalareasinequaltimes,andthatthesquareoftheperiodofrevolutionofaplanetisproportionaltothecubeofthelengthofits orbit’s major axis. By dividing a circle into infinitely many congruentisoscelestriangles,heanticipatedthecalculus.Kepler’spersonal lifewas filledwith tragedy.Smallpoxat theageof four lefthim with very poor eyesight. His youth was unhappy and his marriage wasmiserable.His favoritechilddiedfromsmallpoxandhiswifewent insaneanddied.Agluttonforpunishment,hecarefullyweighedthegoodandbadpointsofeleven girls before choosing the worst one to be his newwife, resulting in asecondmarriageevenmoreunhappythanhisfirstone.Hismotherwasjailedasawitch,andhenearlysufferedthesamefatewhilefightingtofreeher.Shediedwithina fewmonthsofher release.Hewas fired fromone lectureshipandhispaywasalwayslate.Finally,hediedofafeverwhileonthewaytocollectsomeofhisbackwages.23.7PierredeFermat(1601–1665)ledaquietpersonallife,ratherunusualforamathematician.Althoughhismainworkwasinnumbertheory,hisDemaximusetminimisanticipatedthecalculus.23.8 Itwas commonlyheld at this time that the infinite couldnot possibly becomprehensibletoman,whohadafinitemind.Againequaltothechallenge,DeMorgan stated that by that reasoning, “who drives fat oxen should himself befat.”Nonetheless, itwas to be 200 years before themathematical infinitewasunderstood.23.9 René Descartes (1596–1650), a man with a violent temper, was thediscoverer of analytic geometry in 1637. Although others before him(Apollonius, Vieta, Oresme, Cavalieri, Roberval, and Fermat) had appliedalgebra and coordinates to specific curves, Descartes was the first one tointroduce a coordinate system to be applied to all geometric curves. Heintroducedexponents(asinx3forxxx)andtheruleofsignsthatbearshisnameand is used for determining the numbers of positive and negative roots of apolynomialequation.23.10FermatoncecriticizedDescartes,causinghisinfamoustempertoexplode.Descartes attacked the former’s “method of tangents”with a vengeance.Eventhough Descartes was dead wrong, he continued the controversy long after itshouldhavebeenforgotten.HediedbecauseofarequestfromQueenChristinaofSweden:Alwaysfrail,hehadbeenpermittedasachildtoariseeachdaywhenhewas ready to do so, generally toward noon. In later years he cherished hismornings in bed. Young Christina requested him to come to Sweden as her

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personal tutor, so great was her thirst formathematics. After refusing severaltimes,hefinallyagreedtoleavehiswarmHolland.Tohishorror,shedemandedherlessonsat5a.m.inacold,draftylibrary.Unabletostandtheclimateandtheearlyhours,hebecameillanddiedjustfourmonthslater.

It issaid that thereal reasonforhisdeathwas this :Christina,havingbeenraisedtobequitemasculine,wasquiteanequestrian,ridingdailysometimeafterherlessonfromDescartes(pronouncedday-CART).Hadshereversedtheorderofthesetwodailyevents,poorRenéwouldundoubtedlyhavelivedmuchlonger.Foritiswellknownthatoneshouldneverput“Descartesbeforethehorse.”23.11 In 1639GirardDesargues (1593–1662) introduced projective geometry.Hiswork,comingjusttwoyearsafterDescartes’analyticgeometryandwritteninaveryeccentricstyle,attractedlittleattentionandwassoonlostforovertwohundred years. He introduced the concept of a pencil of lines or planes, andprovedhisfamoustwo-triangletheorem.23.12 Blaise Pascal (1623–1662), not permitted to studymathematics until hehad thoroughly mastered Latin and Greek, set down his own axioms anddefinitionsforelementarygeometry,andthenprovedthatthesumoftheanglesofatriangleisequaltotworightangles.Hisfather,wellversedinmathematics,caughthimstudyingthistheoremandwassoamazedthatheweptforjoy.Pascalwrotehistreatiseonconicswhenhewasonly16yearsofage.Fromhismystichexagramtheoremhehimselfdeducedmorethan400corollaries!23.13 John Wallis (1616–1703) was the first to treat the conic sectionsanalytically as second-degree polynomials. He introduced the symbol ∞ forinfinityanddiscoveredtheinterestinginfiniteproduct

23.14Thediscoveryofthecalculusin1665byIsaacNewton(1642–1727),andindependently in 1675 byGottfriedWilhelmLeibniz (1646–1716) highlightedthisveryproductivecenturyinwhichmathematicalthinkingdevelopedagaintoaboutthestandardtheGreekshadreached.WeshallnotdwellhereontheworksofNewtonandLeibniz.Theirdiscoveryofthecalculuswasanaturaloutgrowthofanalyticgeometryandthe“methodofindivisibles”whichhadbeenappliedtomanyindividualsituationsbyearlierworkers.23.15 Now mathematicians, drunk with the power of the calculus, blindlyapplied its methods to all sorts of problems, completely ignoring the shakyfoundationsonwhichithadbeenconstructed.Manycontradictionsarose,toberesolved in the nineteenth century with a theory of limits. But now with theapplication of the methods of algebra and analysis to geometry, modern

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geometrywasborn.Andtheproductiveseventeenthcenturycametoanend.ExerciseSet231.ProvesyntheticallytheDesarguestwo-triangletheorem(Theorem4.7)forthecasewherethetwotriangleslieindistinctintersectingplanes.

2.Pascal’smystichexagramtheoremstatesthatifanot-necessarily-convexhexagonisinscribedinaconicsection,thenthethreepointsofintersectionofpairsofoppositesidesarecollinear.Provethistheoremforthecasewheretheconicsectionisacircle.

3.Givenfivepointsthatlieonaconicsection,usePascal’smystichexagramtheorem(Exercise23.2)tolocateotherpointsontheconic.

4.Useadeskcalculator(orcomputer)tocalculatethevalueoftheproductofthefirstten(or100)factorsofWallis’expressionforπ/2(see23.13).Howrapidlydoesthisproductconverge?

5.In1806,whilestillastudent,C.J.Brianchon(1785–1864)provedthatifahexagoniscircumscribedaboutaconicsection,thenthethreediagonalsthatjoinoppositeverticesareconcurrent.ShowthatBrianchon’stheoremisthedual(see4.9)ofPascal’stheorem(Exercise23.2).

SECTION24 INTRODUCTIONTOSIMILARITIES

24.1Inthischapterweshalldoforsimilarities—mappingsthatcarryfiguresintosimilar figures—what we did for isometries in Chapter 2. We begin with ahomothety, a simple stretch ormagnification of the planewhichmultiplies alldistances from a fixed point called the center by the same ratio k ≠ 0. Thismapping is adirect similarityandcarrieseach line intoaparallel lineandhasjustitscenterforaninvariantpoint.Conversely,ifanytwosimilarfigureshavecorrespondingsidesparallel,thenthereisahomothetythatmapsoneofthemtothe other and its center is the point of concurrence of the lines joiningcorrespondingvertices(Fig.24.1).

Figure24.124.2Twocircleshavetwocentersofhomothetyifthecirclesarenotconcentric.

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ThesecentersarelocatedonthelineofcentersofthecirclesandmaybefoundbydrawingparalleldiametersAB andA′B′ in the twocircles (Fig.24.2).Thenthe linesAA′ andBB′meet at one center of homothety, the linesAB′ andA′Bmeetattheother.Thesecentersofhomothety,alsocalledcentersofsimilitude,divideharmonicallythelineofcentersofthetwocircles.

Figure24.224.3Twohomothetiescommutewhenandonlywhentheyhavethesamecenteror at least one of the ratios is +1. The product of two homotheties is either ahomothetyoratranslation,thelatteroccurringwhentheirratiosarereciprocalsofoneanother(seeTheorem25.9).24.4 A similarity is the composition of an isometry and a homothety. Anysimilarity that is not an isometry is either a rotation followed by a homothetywiththesamecenteroritisareflectionfollowedbyahomothetywhosecenterlies on the mirror, the first being direct, the second product opposite (seeTheorems26.9and26.10).24.5Any twogiven segmentsAB andA′B′ are related by just two similarities(withA′theimageofAandB′theimageofB),onedirectandtheotheropposite.Any two similar triangles are related by just one similarity. In this sense, ifsimilarity a maps triangleABC to triangleA′B′C′, we agree that α–l mappingtriangleA′B′C′totriangleABCdoesnotconstituteadifferentsimilaritybetweenthesetriangles.24.6 The theory described above will be developed in the next two sections.With that foundationwe shall then have a sufficient basis to proceed to threesectionsofapplicationstoelementarygeometry.Afinalsectionisdevotedtotheanalyticrepresentationsofsimilarities,enablingustousethesetransformationsin the Cartesian plane. Thus this chapter parallels closely, but more briefly,Chapter2.ExerciseSet241.Showthatasimilaritypreservesangles.2.Wherearethetwocentersofhomothetyforthebaseandsummitofanisoscelestrapezoid?

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3.Showthattheproductoftwohomothetieswiththesamecenterisahomothety,andfinditsratioandcenter.

4.Provethatiftwosimilartriangleshavetheircorrespondingsidesparallel,thenthelinesjoiningcorrespondingverticesconcur,sotheyarerelatedbyahomothety.

5.Giventwodirectlysimilartriangles,showhowtofindarotationandahomothety(notnecessarilywiththesamecenter)whoseproductmapsonetriangletotheother.

6.Giventwooppositelysimilartriangles,showhowtofindareflectionandahomothety(whosecenterdoesnotnecessarilylieonthemirror)whoseproductmapsonetriangletotheother.

7.Findthecentersofhomothetyoftwononintersectingcircleswithradii3and5,andfindtheratiosinwhichthesecentersdividethelineofcentersofthecircles.

8.Provethatthecentersofhomothetyoftwocirclesdividethelineofcentersofthecirclesharmonically.

9.Provethattheconstructionforthecentersofhomothetyoftwocircles,givenin24.2,iscorrect.

10.Findahomothetyotherthantheidentitymap(homothetywithratio=+1)thatcarriesagivenrectangleintoitself(notnecessarilypointbypoint).

11.DenotethehomothetywithcenterAandratiokbyH(A,k).GiventhatO(0,0),P(l,0),andQ(0,2)arepointsintheCartesianplane,findtheimageoftriangleOPQundereachhomothety.a)H(O,1)b)H(O,2)c)H(O,–1)d)H(O, )e)H(A,2),whereA(2,0)f)H(A,–

g)H(B,2),whereB(4,0)h)H(B,– )12.Findallhomothetiesthatareinvolutoric.

13.Findtheinverseofthehomothetywithagivencenterandratio.14.Showthat,foragivenhomothetyα,thesmallestpositivenforwhichαn=

a)is1iffα= ,b)is2ifftheratioofthehomothetyis–1,andc)doesnotexistotherwise.

SECTION25 HOMOTHETY

25.1DefinitionAhomothetyH(O,k),whereOisafixedpointintheplaneandkisanonzerorealnumber, is that transformationthatmapspointO toitselfandmapsanyotherpointPtoapointP′suchthatO,P,P′arecollinearandOP′=k·OP.PointOiscalledthecenterandktheratioofthehomothety.

Thehomothetyisabuildingblockforsimilaritymappingsinaboutthesameway as a reflection is a building block for isometries. One cannot obtain all

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similaritymappingsfromproductsofhomothetiesalone,buttheyarenecessaryandbasictosimilarities.Inthissectionweshallstudytheelementarypropertiesof homotheties and their products. In the next section we shall show therelationshipbetweenhomothetiesandsimilarities.25.2TheoremThehomothetyH(O,k)mapsa linesegmentAB intoaparallelsegmentA′B′withA′B′=|k|·AB.

LetH(O,k)mapAtoA′,BtoB′,andanyotherpointPonsegmentABtoP′(Fig.25.2).Then BOP= B′OP′and

HencetrianglesBOPandB′OP′aresimilar,soB′P′/BP=|k|andB′P′isparalleltoBP.SimilarlyP′A′/PA= |k| andP′A′ isparallel toPA.Now, sinceBPA is astraightline,thensoalsoisB′P′A′astraightline,andB′A′/BA=|k|.

Figure25.225.3CorollaryAhomothetymapsanygiventriangleintoasimilartriangle.Ingeneral,itmapseachpolygonintoasimilarpolygon.25.4CorollaryAhomothetypreservesanglesbetweenlines.25.5TheoremAhomothetyH(O,k)mapsacircleofradiusrintoanothercirclewhoseradiusr′is|k|·r,andwhosecenterC′isthehomotheticimageofthecenterCofthegivencircle.

LetH(O,k)mapthecenterCofthegivencircletoC′,andanypointPonthecircle topointP′, as inFig.25.5.SinceCP=r, the radiusof thegivencircle,thenC′P′=|k|·rbyTheorem25.2.Thatis,thelocusoftheimagesofpointPonthe given circle is another circle of radius |k|·r and centerC′ the homotheticimageofC.

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Figure25.525.6TheoremForanytwogivencirclesthereisahomothetythatmapsonetotheother.If thecirclesareunequalandnotconcentric, thentherearetwosuchhomotheties.

Iftheunequal,nonconcentriccirclesofradiirandr′donotintersect,thenthecenters are at the meeting of their common external tangents (with ratio ofhomothety r′/r) and themeeting of their common internal tangents (with ratio–r′/r).Inallcaseswemaylocatethesecenters(Fig.25.6)bytakingthecentersof the given circles as C and C′, then erecting perpendiculars to the line ofcentersCC′ atC andC′.Let theseperpendiculars cut circleC atP andQ andcircleC′atP′andQ′.Thecentersofhomothetyareat the intersectionsofPP′andQQ′andofPQ′andP′Q.Wedenote thesecentersby IandE, Ibeing theinternal center of similitude and lying between C and C′ and E being theexternalcenterofsimilitudeandlyingoutsidesegmentcc′.

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Figure25.625.7Theorem If two similar noncongruent triangles are so oriented that eachside of one is parallel to the corresponding side of the other, then there is ahomothetythatmapsonetriangletotheother.

Let the triangles be ABC and A′B′C′, as in Fig. 25.7. Since pairs ofcorrespondingsidesareparallel,thenthetwolinesofeachpairmeetonthelineat infinity; that is, the two triangles are coaxial. By Desargues′ two-triangletheorem(Theorem4.7),thesetrianglesarealsocopolaratapointO.ItfollowsthatpointOservesasthecenterofhomothetyandthattheratioofhomothetyisOA′/OA=±A′B′/AB.

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Figure25.725.8 Theorem A homothety is determined by two distinct points and theirimages.

Let thehomothetymappointsA andB toA′ andB′, respectively.Then thecenterofhomothetyisthepointofintersectionofthelinesAA′andBB′.Itsratiois°A′B′/AB,theminussignoccurringifthecenterliesbetweenAandA′theplussignotherwise.25.9TheoremIfjk≠1,thentheproductofthetwohomothetiesH(O,k)and

H(Q,j)isthehomothetyH(P,jk)wherePiscollinearwithOandQ.Ifjk=1,thenPbecomesidealandtheproductofthetwohomothetiesisatranslation.

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Figure25.9Since each homothety maps a triangle into a similar triangle with sides

parallel to thegiven triangle, it follows that theirproductdoes the same.Thuseitherproductisahomothetyofratiojk≠1,oratranslationwhenjk=1.

LetH(O,k)mapΔABCtoΔA′B′C′andletH(Q,j)mapΔA′B′C′toΔA″B″C″(seeFig.25.9).SinceAB,andA″B″,areallparallel,theymeetatanidealpointI.ThustrianglesAA′A″andBB′B″arecopolaratI.BytheDesarguestheorem,theyarecoaxial;thatis,pointsO,P,Qarecollinear.25.10 Theorem The product of two homotheties having the same center iscommutative, and is ahomothetywith the samecenterandwith ratioequal totheproductoftheratiosofthegivenhomotheties.25.11TheoremThecenterofahomothetyof rationotequal to+1 is theonlyfixedpointofthehomothety.Linesthroughthecenteraretheonlyfixedlines.25.12Theconceptofdirectoroppositeapplies tosimilar figuresaswellas tocongruentfigures.Appropriatedefinitionsareleftfortheexercises.25.13TheoremAhomothetyisadirecttransformation.25.14TheoremAhomothetyofratio±1istheidentitymap.

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25.15Theorem The inverse of the homothetyH(O, k) is the homothetyH(O,1/k).25.16TheoremAllhomothetiesandtranslationsformatransformationgroup.25.17TheoremAllhomothetieshavingthesamecenterformatransformation

group.ExerciseSet251.ProveCorollary25.3.2.ProveCorollary25.4.3.Locatethecenterofhomothetyandfinditsratiofortwoequalnonconcentriccircles.

4.Provethatthecentersofsimilitudeoftwocircleslieonthelineofcentersofthecircles.

5.Provethat,asstatedintheproofofTheorem25.6,theintersectionsofthecommontangentsoftwononintersectingcirclesaretheircentersofhomothety.

6.ProvethattheconstructionintheproofofTheorem25.6thatapplies“inallcases”doesindeedgivethecentersofsimilitude.

7.IntheproofofTheorem25.8,showthatthecenterliesbetweenAandA′iffitliesbetweenBandB′.

8.InTheorem25.9,showthatPdividesOQintheratio(j–1)/(j(k–1)).9.UsetheresultofExercise25.8toshowthatwhenj≠1andk≠1,thenthehomothetiesofTheorem25.9donotcommute.Illustratethisnoncommutativitywithafigure.

10.ProveTheorem25.10.11.ProveTheorem25.11.12.TheproductH(B,1/k)·H(A,k)isatranslation.Whattranslation?13.ProveTheorem25.13.14.ProveTheorem25.14.15.ProveTheorem25.15.16.ProveTheorem25.16.17.ProveTheorem25.17.18.Definedirectandoppositesimilarfigures.19.Showthataproductofthreehomothetiesisahomothetyoratranslation.20.GeneralizeExercise25.19toaproductoffourormorehomotheties.SECTION26 SIMILARITY

26.1DefinitionAsimilarityisamapoftheplanethatcarrieseachpointpairA,BintoapointpairA′,B′,suchthatforsomefixedpositiverealnumberk,A′B′=k·AB.Thenumberk iscalled theratioof thesimilarity.The termsimilitude is

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alsousedforthismapping.Thissectioncompletesourstudyofsimilaritiesand theirproperties,so that

wemayproceedrapidly to theapplications inSections27 to29.Although thetheorywedevelophereisnotasextensiveasthatforisometriesinChapter2,itisquitesufficientforourpurposes.26.2TheoremAsimilarityofratio1isanisometry.26.3TheoremAsimilaritymapssegmentsintosegments.

LetasimilarityofratiokmapAtoA′andBtoB′.TakeanypointPbetweenAandBandlettheimageofPbeP′.Then

soP′liesonsegmentA′B′.26.4TheoremAsimilarityisatransformationoftheplane.26.5TheoremAsimilaritypreservesangles.

MarkpointsBandConthetwosidesofangleAtoformatriangleABC.LetthesimilaritymaptriangleABCtotriangleA′B′C′byTheorem26.3.Then

ThustrianglesABCandA′B′C′aresimilarbySSS.Henceandthetheoremfollows.26.6 Corollary A similarity maps a triangle into a similar triangle. Moregenerally,itmapsapolygonintoasimilarpolygon.26.7TheoremAsimilarity isdeterminedbyany threenoncollinearpointsandtheirimages.26.8TheoremThereareexactly twosimilarities,onedirect andoneopposite,thatmapanypointpairA,BintoanyotherpointpairA′,B′(understandingthatA≠B,A′≠B,A′istheimageofA,andB′istheimageofB).26.9 Theorem Each direct similarity of ratio k that is not an isometry is theproduct of a rotation and a homothety of ratio k having the same center.Furthermore,suchaproductiscommutative.

Clearlysuchaproductisadirectsimilarity.ByTheorem26.8,thissimilarityisdeterminedbyasegmentABanditsimageA′B′.LetAA′andBB′meetatQ,anddrawthecirclesthroughA,B,Q,andthroughA′,B′,QtomeetagainatO,asshowninFig.26.9.Thenwehave

and

bythepropertiesofanglesinscribedinacircle.HencetrianglesABOandA′B′O

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aresimilar.Now,ifABisrotatedaboutOthroughangleAOA′toA1B1thenA1B1isparalleltoA′B′andtrianglesOA1B1andO,A′B′aresimilar,soB1liesonlineOB′. Thus the homothety H(O, k) maps A1Bl to A′B′. This rotation andhomothetysatisfythetheorem.Now,ifthefigureformedbyO,A1,B1,A′,B′isrotated aboutO through angleA′OA (the inverse of AOA′), thenA1B1 willcoincidewithAB.Itfollowsthatthehomothetyandrotationcommute.

Figure26.926.10TheoremAnyoppositesimilarityofratiokthatisnotanisometryistheproduct of a reflection, and a homothety whose center lies on the mirror.Furthermore,suchaproductiscommutative.

LetthesimilaritymapsegmentABtoCD,andletthelinesABandCDmeetatQ(Fig.26.10).OfthetwobisectorsoftheanglesatQ,callthatonensothatσnmapsABtoEFwhereEFandCDhavethesamesense.Drawanyotherlinen1parallelto(anddistinctfrom)n.ReflectABinlinen1toA1B1.LetO1bethecenter of the homothety that maps A1B1 to CD. Draw linep through O1perpendicular to n. Reflect AB in line p to A″B″. Now the center O of thehomothetythatcarriesA″B″toCDliesonp.DrawlinemthroughOandparallelton,andletσm(AB)=A′B′.Thenmisthemirrorofreflection,andOisthecenterofhomothetymappingABtoCD.

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Figure26.10Proofs of the validity of the construction, and of the commutativity of the

productofthesetransformationsareleftforthereadertosupply.To verify the construction, first show that all such points A1 and A′,

reflectionsofAinmirrorsparallelton,lieonalinethroughAandperpendicularton.ThenshowthatallthecentersO1lieonanotherlinepperpendicularton.Show that the reflection ofAB toA″B″ in p makesA″B″ parallel toCD andoppositeinsense,sothatthecenterOofthehomothetythatcarriesA″B″toCDliesonp.ShowthatH(O,–k)·σpmapsAB(toA″B″andthen)toCD.ShowthatthismapisequaltoH(O,k)·σ0·σp=H(O,k)·σm.

Toestablishthecommutativity,reflectthefigureformedbyO,A′B′,andCDinlinem,andconsideritsimage.26.11TheoremAllsimilaritiesformatransformationgroup.26.12TheoremAlldirectsimilaritiesformatransformationgroup.

ExerciseSet261.ProveTheorem26.2.2.ProveTheorem26.4.3.Provethatasimilaritymapscirclesintocircles.4.ProveCorollary26.6.5.ProveTheorem26.7asacorollarytoTheorem13.6.

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6.ProveTheorem26.8.7.a)DiscussthecasewhenthecirclesintheproofofTheorem26.9are

tangent.b)DiscussthecasewhenoneorbothofthecirclesintheproofofTheorem9.26reducestoastraightline;thatis,whenA,B,Qarecollinear,orwhenA′,B′,Qarecollinear.

8.ProvethecommutativityofTheorem9.26assuggestedinthetext.9.OnacleansheetofpapertracethetwosegmentsofFig.26.9labeledABandA′B′.RelabelthefirstoneBA;thatis,interchangethelabelsonitsendpoints.PerformtheconstructionofTheorem9.26forthesenewsegments.

10.RepeatExercise26.9forthesegmentsABandCDofFig.26.10,andusingtheconstructionofTheorem26.10.

11.ProvetheconstructionofTheorem26.10.12.ProvethecommutativityofTheorem26.10.13.ProveTheorem26.11.14.ProveTheorem26.12.15.Ingeneraltherearetwohomothetiesthatmaponecircletoanother,anda

homothetyisadirectsimilarity.Theorem26.8statesthereisjustonedirectsimilaritythatmapsonesegmenttoanother.a)Explainthisapparentparadox.b)Findanoppositesimilaritymappingonecircletoanother.c)Aretheremorethantwodirectsimilaritiesthatmaponecircletoanother?Defendyouranswer.

d)Howmanyoppositesimilaritiesmaponecircletoanother?e)Inthissamesense,howmanydirectandoppositesimilaritiesmaponeequilateraltriangletoanother?HerewearepermittingvertexAoftriangleABCtomapintoanyoneofthethreeverticesoftheimagetriangle.

f)Repeatpart(e)forasquare.g)Repeatpart(e)forageneralrectangle.

SECTION27 APPLICATIONSOFSIMILARITIESTOELEMENTARYGEOMETRY

27.1Inthisandthenextsectionarepresentedapplicationsofsimilaritiestohighschoolgeometry.InSection29,moreadvancedapplications,applicationsinthedomainof collegegeometry, arepresented.Certainlymanyof these theorems,evenintoSection29,arereadilyusableinhighschoolgeometryclassrooms.27.2 Theorem Three parallel lines cut off proportional segments from twotransversals.

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If the transversals are parallel, then the theorem follows immediately sinceopposite sides of the resulting parallelograms are congruent. So supposetransversalsABC andA′B′C′ meet atP (Fig. 27.2). ThenH(P,PB/PA) mapstrianglePAA′toPBB′andH(P,PC/PB)mapstrianglePBB′toPCC′.Now

followsfromsimilartriangles.

Figure27.227.3 Corollary A line parallel to one side of a triangle cuts off proportionalsegmentsfromtheothertwosides.27.4 Theorem In trapezoid ABCD with parallel sides AB and CD, let thediagonalsmeetatE.ThentrianglesABEandCDEaresimilar.

Under the homothetyH(E, –AE/EC) pointC maps toA, andD maps to apoint onEB (Fig. 27.4). Since the lineCDmaps to the parallel throughA, itfollowsthatDmapstoB.HencetheimageoftriangleCDEistriangleABE,soΔABE~ΔCDE.

Figure27.427.5 Theorem The common internal tangents of two nonintersecting circlesmeetonthecircles’lineofcenters.

LetthetangentsTT′andUU′meetatCasinFig.27.5.ThehomothetyH(C,–CT′/CT)mapsonecircletotheother.ByTheorem25.5,itmapsonecentertothe other. But a point and its image are collinear with the center of the

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homothety.Thetheoremfollows.

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Figure27.527.6 Theorem The figure formed by joining the midpoints of the sides of asquareisasquarehavinghalftheareaofthegivensquare.

LetObethecenterofthesquareABCD(Fig.27.6).ThenOisthemeetingofthemediansA′C′ andB′D′. Since the diagonals and also themediansmeet atright angles, and are bisected by the center, it follows that the similaritycomposedofa45°rotationandahomothetyofratioOA/OA′,bothwithcenterO, maps square ABCD into quadrilateral A′B′C′D′. SinceOA/OA′ = , thetheoremfollows.

Figure27.627.7ProblemLetPbeafixedpointonacircle.FindthelocusofthemidpointsofallchordsPA.

AllsuchmidpointsM lieonthehomotheticimageofthegivencircleunderthehomothetyH(P,1/2)(Fig.27.7).Hencetheirlocusisacirclewithradiushalfthatofthegivencircle,andtangenttoitinternallyatP.

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Figure27.727.8TheoremIfPTisatangent,andPABasecantfromanexternalpointPtoacircle,thenPA·PB=(PT)2.

Reflect trianglePAT in the internalbisectorof angleP, and thenapply thehomothetyH(P,PB/PT)totheresult.ThenPTmapsintoPB,andAmapsintoapointonPT(Fig.27.8).Since PBTand PTAarebothmeasuredbyhalfofarcAT, these angles are congruent.Thus the homothety and reflectionmapPTAinto PBT,anditfollowsthattrianglePATmapsintotrianglePTB.NowwehavePA/PT=PT/PB,andthetheoremfollows.

Figure27.827.9 Theorem The altitude to the hypotenuse of a right triangle is the meanproportionalbetweenthesegmentsintowhichitdividesthehypotenuse.

LetCFbe thealtitude to thehypotenuseof right triangleABC as shown inFig.27.9.InrighttriangleACF, ACF=90° A.Hence ACF= B.Thusthere is a similarity (a 90° rotation and homothety with center F) mappingtriangleACFtotriangleCBF.HenceAF/FC=FC/FB,establishingthetheorem.

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Figure27.927.10TheoremThemedianofa trapezoidhas lengthequal tohalf thesumofthebasesofthetrapezoid.

LetMandNbethemidpointsofthenonparallelsidesofthetrapezoidABCD,sothatMNisthemedian.LetthesidesADandBCmeetatP(Fig.27.10).ThenH(P,PM/PD)mapssummitDCtomedianMN,andH(P,PM/PA)mapsbaseABontomedianMN.Hence

from which, by writing the first equation slightly differently, and thenmultiplyingthetwoequationsabovesideforside,weobtain

Figure27.10SolvingeachoftheseequationsforDM/PD,obtain

andfinally,

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ExerciseSet271.ProveCorollary27.3.2.ProveTheorem27.6byreflectingtrianglesAA′D,BB′A′,CC′B′,andDD′C′intheirhypotenuses.

3.FromagivenpointA,straightlinesaredrawntopointsPonagivensegmentBC.FindthelocusofallpointsMthatdividethesesegmentsinagivenratior.

4.FindtheratiooftheareasofthetwotrianglesABEandCDEinthetrapezoidofFig.27.4.

5.Provethatthediagonalsofatrapezoidintersectatapointthatdividesthemintoproportionalsegments,andfindtheratioofproportionality.

6.Provethatthesegmentjoiningthemidpointsofthediagonalsofatrapezoidhaslength(b–s)/2,whenbandsarethelengthsofthebaseandsummit.

7.ChordsABandCDofagivencirclemeetatpointE.FindthesimilaritythatmapstriangleACEtotriangleDBE.

8.FindthelocusofthemidpointsofallsegmentsdrawntoacirclefromafixedpointPintheplaneofthecircle.

9.Provethatthecommonexternaltangentstotwocirclesmeetontheirlineofcenters.

10.Twocirclesaretangentinternallyandhaveratio2:1fortheirradii.Provethatachordofthelarger,fromtheirpointoftangency,isbisectedbythesmaller.

11.Twocirclesofradii3and7aretangentexternally.TheircommonexternaltangentsmeetatP.FindthelengthofthetangentfromPtothelargercircle.

12.GiventhatPABandPCDareanytwosecantsfromexternalpointPtoacircle,findthesimilaritythatmapstrianglePAContotrianglePDB.

13.ExtendsideBAofrhombusABCDtoapointE.ThroughEdrawaparalleltoBC,andthroughAaparalleltothediagonalBD,andlettheseparallelsmeetatF.ProvethattriangleEAFisisosceles.

14.FrompointPonlegACofrighttriangleABCdropperpendicularPQtothehypotenuseAB.ProvethattrianglesAPQandABCaresimilar.StatewhatsimilaritymapstriangleAPQtoABC.

15.DropperpendicularsfromgivenpointsDandFonlegsBCandCAofrighttriangleABCtopointsEandGonthehypotenuseAB.ProvethattrianglesAFGandDBEaresimilar.

16.MediansAA′andBB′oftriangleABCmeetatG.ProvethattrianglesABGandA′B′Garesimilar,andfindtheirsimilaritymap.

SECTION28 FURTHERELEMENTARYAPPLICATIONS

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28.1TheoremLetAbethemidpointofarcCDofagivencircle.LetchordABcut chordCD atM. ThenAM*AB is independent of the location ofB on thecircle.

InFig.28.1anglesACDandCBAarecongruent,sincetheyaremeasuredbyhalves of the congruent arcs AD and CA. Thus triangles AMC and ACB aresimilar,sothereisasimilaritymappingonetriangleontotheother.ThenAM/AC=AC/AB,soAM·AB=(AC)2,aconstant.

Figure28.128.2 Theorem The triangles ACD and BCE formed from a triangle ABC byaltitudesADandBEaresimilar.

SincetheserighttrianglesshareanacuteangleatC(Fig.28.2),itfollowsthatthereisasimilarity(areflectioninthebisectorofangleC,andahomothetywithcenterC)thatmapsthesetrianglesonetotheother.

Figure28.228.3ProblemLetPbeavariablepointonthesemicircleofdiameterAOB.LettheperpendicularfrompointBtothetangentatPmeetlineAPatpointX.FindthelocusofX.

SinceOPisperpendiculartothetangentatP,thenOPandBXareparallelasshown inFig.28.3.HenceH(A, 2)mapsOP toBX.Then the locusofXis the

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homotheticimageofthesemicircleonwhichPlies,asemicircleofcenterBandradiusBA.

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Figure28.328.4 Theorem Let perpendiculars erected at arbitrary points on the sides oftriangleABCmeetinpairsatpointsP,Q,R.ThentrianglePQRissimilartothegiventriangle.

SeeFig.28.4.A90°rotationoftriangleABC intotriangleA′B′C′ leavesthesidesof triangleA′B′C′parallel to thecorrespondingsidesof trianglePQR.ByTheorem 25.7 an appropriate homothetywillmap triangleA′B′C′ into trianglePQR.Thetheoremfollows.

Figure28.428.5TheoremLetAB,CD,andEFeachbeperpendiculartoBDsothatADandBCmeetatE,andpointsAandClieonthesamesideofBD.ThenEFishalftheharmonicmeanofABandCD.

Recallthattheharmonicmeanofaandbis2ab/(a+b).LetAB=a,CD=b,EF=x,BF=w,andFD=v(Fig.28.5).ThenH(B,u/(u/(u+v))mapstriangleBCDtoBEF,sox/b=u/(u+v).AlsoH(D,v/(u+v))mapstriangleDABtoDEF,sox/a=v/(u+v).Thenwehave

soEFishalftheharmonicmeanofABandCD.

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Figure28.528.6TheoremThebisectorofaninternalangleofatriangledividestheoppositesideintosegmentsproportionaltotheadjacentsides.

LetAUbisectangleAoftriangleABC(seeFig.28.6).LetH(B,BC/BU)maptriangle BAU to triangle BPC. Then . It follows also thatBU/UC=BA/AP.Furthermore,sinceAUisparalleltoPC,then

sotriangleAPCisisoscelesandAP AC.HenceBU/UC=BA/AC.

Figure28.628.7TheoremGiven that theperpendicularbisectorsof thesidesconcur, thenthealtitudesofatriangleconcur.

LetH(G, –2) map triangle ABC into triangle A1 B1 C1 (Fig. 28.7). ThenverticesA,B,CbecomethemidpointsofthesidesoftriangleA1B1C1andsincethe corresponding sides are parallel, the altitudes AD, BE, CF become theperpendicularbisectorsof thesidesof triangleA1B1C1.The theoremfollows,since we assumed that the perpendicular bisectors of the sides of a triangleconcur.

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Figure28.728.8ComparetheproofofTheorem28.7withCorollary20.8.28.9 The method of homothety is quite useful for constructions. The basicprinciple is illustrated in the next two items. In each case it is not possible tosatisfy all the required conditions immediately. So we solve the problempartiallybyomittingoneof theconditions.Thenahomothetymaps thepartialsolutionintoacompletesolution.Oneisremindedofagambitinchess,whereaplayergivesupapiecetohisopponentinordertogainawinningposition.28.10ProblemLocatepointsDandEonsidesABandACofagiventriangleABCsothatBD DE EC.

ArbitrarilychoosepointsD′andE′onsidesABandAC(Fig.28.10)sothatBD′ E′C. [Herewegiveup thecondition thatD′E′ shallbecongruent to theothertwolengths.]LetthecircleD′(B)cuttheparalleltoBCthroughE′withinthe triangle at E″. Draw through E″ a parallel A′C to AC, and observe thattrianglesABCandA′BC′aresimilarandthat theproblemissolvedfor triangleA′BC′.[WehaveregainednowtheconditionBD′ D′E″ E C′attheexpenseofthesizeofthetriangle.]ThehomothetyH(B,BC/BC′)mapsthissolutionontotriangleABC.ThismapmaybeaccomplishedbydrawingBE″ tocutACatE,oneofthedesiredpoints.

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Figure28.1028.11ProblemInscribeasquareinagivensemicircle.

Inthesolvedproblem,clearlythecenterOofthesemicircleisthemidpointofone sideof the square.Letuskeep thisconditionand temporarilyoverlooktheconditionthattwoverticesmustlieonthesemicircle.OnthediameterAOBofthesemicircle,constructanyconvenientsquareP′Q′R’s′withOthemidpointofthesideP′Q′lyingonAB(Fig.28.11).Now,withOascenter,projectverticesR′andS′ontothesemicircletothedesiredpointsRandS.TheothertwoverticesPandQaremerelythefeetoftheperpendicularstoABfromSandR.

Figure28.1128.12ComparetheconstructionofProblem28.11withthatofProblem8.13.ExerciseSet281.ChordABiscongruenttotheradiusofagivencirclewithcenterO,OMisperpendiculartoABatM,andDisthefootoftheperpendiculardroppedfromMtoOA.FindtheareaoftriangleMDA.

2.InacirclechordsABandACarecongruent.ChordADcutssegmentBCatE.ProvetrianglesABDandAEBaresimilar.

3.ThreecircleswithcentersA,B,CpassthroughO.DiametersOAA′,OBB′,OCC′aredrawn.ProvethatthesidesoftriangleA′B′C′passthroughtheotherpointsofintersectionofthecircles.

4.Thebaseandsummitofatrapezoidhavelengthsbands.FindtheratioinwhichalinesegmentMNdividesthealtitudeofthetrapezoidwhenMNis

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paralleltothebase,terminatesonthesides,andhaslengthm.5.TakeanypointFonsideABofparallelogramABCD,andletDFmeetdiagonalACatEandsideCB(extended)atG.Provethat(DE)2=EF·EG.

6.Provethattheanglebetweenthebaseofanisoscelestriangleandthealtitudetooneofthecongruentsidesiscongruenttohalfthevertexangle.

7.InisoscelestriangleABCwithAB AC,letthealtitudefromBmeettheperpendiculartoBCatCinpointD.TakepointEonBDsothatED EC.ProvethattrianglesABCandECDaresimilar.

8.TakepointsXandYonsidesABandACoftriangleABCsothatXYisparalleltoBC.FromthemidpointA′ofsideBCdrawlineA′XtomeetCAatM,andlineA′YtomeetBAatN.ProvethatMNisparalleltoBC.

9.Provethatthebisectorofanexternalangleofatriangledividestheoppositesideexternallyinsegmentsproportionalto′theadjacentsides.

10.TakepointsD,E,FonsidesBC,CA,ABofequilateraltriangleABCsothatBD CE AF.ProvethattriangleDEFisequilateral.

11.IntriangleABC, andAUisthebisectorofangleA.Provethat(AB)2=BU·BC.

12.ConstructatriangleABCgiventhemeasuresof A,a+c,anda+b.13.ConstructasquaresothatonesideliesalongthebaselineBCofagiven

triangleABCanditsothertwoverticeslieonraysBAandCAoutsidethetriangle.

14.Inagiventriangleinscribeatrianglehomothetictoanothergiventriangle.15.Inagiventriangleinscribeaparallelogramhomothetictoagiven

parallelogram.16.Inscribeasquareinagivencircularsector.17.Inscribeasquareinagiventriangle.18.Inscribearectanglesimilartoagivenrectangle:a)inagivencircleb)ina

givensemicirclec)inagiventriangled)inagivensquare,sothatonevertexoftherectangleliesoneachsideofthesquare.

SECTION29 ADVANCEDAPPLICATIONS

29.1TheoremThemediansofatriangleconcur.Since the medial triangle A′B′C′ has sides parallel to those of the given

triangleABC,thereisahomothetymappingonetriangletotheother.Thecenterofthehomothetyisthepointofconcurrenceofthelinesjoiningcorrespondingvertices,thatis,themediansoftriangleABC.29.2TheoremTheorthocenterH, thecircumcenterO,andthecentroidGofatriangle,alllieonalinecalledtheEulerlineofthetriangle,andHG=2GO.

Theorem28.7showedthatH(G,– )mapstriangleABCanditsorthocenterH

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intotriangleA′B′C′anditsorthocenterwhichisthecircumcenterOfortriangleABC.Hencethetheorem.29.3CorollaryThedistancefromthecircumcentertoasideofatriangleishalfthedistancefromtheorthocentertotheoppositevertex.29.4 Theorem The ninepoinl circle theorem. The midpointsA′,B′,C′ of thesides, the feetD, E, F of the altitudes, and the midpointsNa, Nb, Nc of thesegments joining theorthocenterH to theverticesof triangleABC,all lieonacirclewhosecenterNisthemidpointoftheEulersegmentHO.

We show that the homothetyH(H, 2)maps the nine listed points onto thecircumcircle,fromwhichthetheoremallfollows(Fig.29.4).

Figure29.4LetdiameterAOcutthecircleagainatP.Then ABP= ACP=90°,since

eachangleisinscribedinasemicircle.NowPCandBEarebothperpendicularto AC, and BP and EC are both perpendicular to AB. Thus BPCH is aparallelogram,soitsdiagonalsbisectoneanother;thatis,HPpassesthroughA′andHP=2HA′.HencethehomothetyH(H,2)mapsA′ toPonthesemicircle.SimilarlyB′andC′arealsomappedontothecircumcirclebyH(H,2).

LetaltitudeADcutthecircumcircleagainatA1.Then

sincethefirsttwoanglesareeachinscribedinarcBA1andthelastequalitiesareseenfromrighttrianglesBADandBCF.SinceCDisperpendiculartoHA1and,triangleCHA1isisosceles,soitsaltitudeCDbisectsitsbaseHA1.HenceH(H,

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2)mapsDtoA1.SimilarlytheotherfeetEandFofthealtitudestotriangleABCaremappedtothecircumcirclebyH(H,2).

ClearlyH(H,2)mapsNa,Nb,NctoA,B,Conthecircumcircle.ThereforetheninepointsA′,B′,C′,D,E,F,Na,Nb,Ncalllieonacirclehalfthesizeofthecircumcircle,andwithcenterofhomothetyH.ThusH(H,2)alsomapsthecenterN of this ninepoint circle to the centerO of the circumcircle. That is,N liesmidwaybetweenHandO.29.5 Theorem The altitude to the hypotenuse of a right triangle divides thehypotenuseintosegmentsproportionaltothesquaresoftheadjacentsides.

LetCFbethealtitudetothehypotenuseofrighttriangleABC.SincetrianglesACFandCBFaresimilar(showninFig.27.9),then

Bymultiplyingtheseequationssideforside,obtain

29.6ProblemConstructarighttrianglegivenitsperimeterandtheratioofthesquaresofitslegs.

Letpdenotetheperimeterandm/nthegivenratioasshowninFig.29.6a.OnabaselinemarksegmentsA′F′=mandF′B′=n,anddrawasemicirclewhosecentershallbeOonA′B′asdiameter(Fig.29.6b).LettheperpendiculartoA′B′atF′cut

Figure29.6athe semicircle atC′. Then triangleA′B′C′ is similar to the desired triangle, byTheorem 29.5. On the tangent atB′ to the semicircle,markB′P′ equal to theperimeteroftriangleA′B′C′,andalsoB′Q=p.LettheperpendiculartoB′QatQmeetOP′atP.TheparalleltoB′P′throughPcutslineA′B′atB.PointsAandCarelocatedhomothetictoA′andC′incenterOandlyingonthecircleO(B).Theproofoftheconstructionisleftasanexercise.

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Figure29.6b

29.7ProblemConstructatrianglegivenitsanglesanditsaream2.Givenitsangles,onecanconstructasimilar triangleA′B′C′,anditsaltitude

A′D′,asshowninFig.29.7a,alongwiththegivenlengthm.Now BC·AD=m2,

%

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Figure29.7a

Figure29.7b

so m is the mean proportional between BC and A′D′. Construct the meanproportionalm′between B′C′andA′D′onalinesegmentP′Q′R′,whereP′Q′ B′C′andQ′R′ A′D′(Fig.29.7b).DrawthesemicirclehavingcenterOonP′R′as diameter. Let the semicircle cut the perpendicular toP′R′ atQ′ in pointS′.ThenQ′s′isthemeanproportionalm′between B′C′andA′D′.LocateSonOS′so thatQS =m andQS is perpendicular toP′R′ atQ. ThenQS andQ′S′ arehomotheticincenterO.ThecirclewithcenterOandradiusOScutsthelineP′R′atpointsPandRsothatPQ BCandQR AD.ThedesiredtriangleiseasilyconstructedbymarkingAonA′D′sothatAD′ QR.ThenABandACaredrawnparalleltoA′B′andA′C′.

29.8 Problem Through a given point P draw a line to pass through theinaccessiblepointofintersectionoftwogivenlinesmandn.

Figure29.8

DrawtwolinesPQandPRthroughPsothatQliesonm,andRonn,asinFig.29.8.NowdrawQ′R′frommtonandparalleltoQR.Drawlinesparallelto

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PQandPRthroughQ′andR′tomeetatP′.ThentrianglesPQRandP′Q′R′arehomothetic,solinePP′passesthroughthepointofintersectionofQQ′andRR′.

ExerciseSet29

1.IntriangleABC,showthatGandHdivideNOinthesameratiosinternallyandexternally,ratios and–

2.IntriangleABC,showthatH(G,–2)alsomapstheninepointcircletothecircumcircle,henceHandGarethetwocentersofhomothetyforthesecircles.

3.ProvethatthefourEulerlinesofthefourtrianglesofanorthocentricquadrangleABCHareconcurrent.Findtheirpointofconcurrence.

4.ProveCorollary29.3.5.ProvetheconstructionofProblem29.6.6.ProvetheconstructionofProblem29.7.7.Throughoneofthetwopointsofintersectionoftwocircles,drawalineonwhichthetwocirclesinterceptcongruentchords(butnotthecommonchordofthetwocircles).

8.Throughoneofthepointsofintersectionoftwocircles,drawalineonwhichthetwocirclesinterceptchordshavingagivenratior.

9.Giventhreeconcentriccircles,drawasecantsothatthesegmentbetweenthefirstandsecondcirclesiscongruenttothatbetweenthesecondandthirdcircles.

10.Constructatriangle,givenanangle,thelengthofthebisectorofthisangle,andtheratioofthetwosegmentsintowhichthisbisectordividestheoppositeside.

11.Constructatrapezoidgiventhenonparallelsides,theanglebetweenthem,andtheratiooftheparallelsides.

12.Constructasquaregiventhesumofitssideanddiagonal.13.Constructarighttriangle,giventheperimeterandtheratioofitslegs.14.Provethatthefourcentroidsofanorthocentricquadrangleforman

orthocentricquadranglehomothetictothegivenone.15.Provethattheninepointcircleofanorthocentricquadrangleisconcentric

withthatoftheorthocentricquadrangleformedfromthefourcentroids.

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16.Provethatthefourpointsofanorthocentricquadranglearethecentroidsofanotherhomotheticorthocentricquadrangle.[ThisistheconverseofExercise29.14.]

17.Thecircumcentersandthecentroidsofanorthocentricquadrangleformtwoorthocentricquadrangleshavingthesameninepointcenter,thecenterofsimilitudeofthesequadrangles.Provethistheoremandfindtheratioofsimilitude.

18.ProvethattheorthocenterofthetriangleformedbythecentroidsoftrianglesHBC,HCA,HAB(HistheorthocenteroftriangleABC)isthecentroidoftriangleABC.

19.Provethatthelocusofthethirdvertexofalltrianglesdirectlysimilartoagiventriangleandhavingthefirstvertexatafixedpointandthesecondvertexalongastraightline,isastraightline.

20.IfthesecondvertexofthetriangleofExercise29.19istolieonacircleinsteadofastraightline,provethatthelocusofthethirdvertexisahomotheticcircle.

21.Constructatrianglesimilartoagiventriangle,havingonevertexatafixedpointandtheothertwoverticeslyingontwofixedlines.

22.Constructatrianglesimilartoagiventriangleandhavingitsthreeverticeslyingonthreegivenlines.

23.ProvethatthemedianAA′andthesegmentB′C′joiningthemidpointsofsidesCAandABofatriangleABCbisecteachother.

24.Findthelocusofthecentroidofatrianglehavingonesideandthecircumcirclefixed.

25.Provethattheproductoftwosidesofatriangleisequaltotheproductofthecircum-diameterandthealtitudetothethirdside.[ThisisTheorem6.20.]

SECTION30 ANALYTICREPRESENTATIONSOFSIMILARITIES

30.1 Theorem The homothety H(O, k) centered at the origin O(0, 0) isdeterminedbytheequations

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Clearlytheindicatedequationsmultiplyeverydistancefromtheoriginbythefactor|k|.

30.2 By translating from point P(a, b) to the origin, applying the homothetyH(O, k), then translating back to P again, we obtain the equations for ahomothetywithratiokandcenterP:

whichreducetothoseindicatedinTheorem30.3.

30.3 TheoremA homothetyH(P, k) centered at pointP(a,b) has the analyticrepresentation

whichisahomothetycenteredattheoriginfollowedbyatranslation.

30.4Weseethateachhomothetycanbewrittenasahomothetycenteredattheoriginfollowedbyatranslation.Observealsothat,whenkisfactoredfromeachright-hand side, the homothety can be written as a translation followed by ahomothetycenteredattheorigin:30.5TheoremAhomothetyH(P,k)centeredatpointP(a,b)hastheanalyticrepresentation

whichisatranslationfollowedbyahomothetycenteredattheorigin.ByExercise22.14,everyisometryhasanequationoftheform

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direct if the plus sign holds, and opposite if the minus sign holds. Eachsimilarity, then, being the product of a homothety and an isometry, can bewritteninthefollowingform.

30.6TheoremEachsimilaritycanbewrittenanalyticallyintheform

where (a2 +b2)l/2 = k, the ratio of the homothety. It is direct if the plus signholds,andoppositeiftheminussignholds.

30.7 Theorem Every set of equations of the form given in Theorem 30.6representsasimilarity,provideda2+b2≠0.

30.8ExampleFindequationsforthetwosimilaritiesthatmapthesegmentOAto the segmentBCwhereO(0,0),A(1,0),B(2,3), andC(5,7).For thedirectisometryaswritteninTheorem30.6,obtaintheequations

whichyield

FortheoppositeisometryofTheorem30.6,wehave

whichyield

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ExerciseSet30

1.ProveTheorem30.1.2.ShowalgebraicallytheequivalenceoftheformsofTheorems30.3and30.5.3.ShowgeometricallytheequivalenceoftheformsofTheorems30.3and30.5.

4.ProveTheorem30.6.5.ProveTheorem30.7.6.FindthevalueoftheratiokinthesimilaritiesofExample30.8.7.FindtheangleofrotationforeachofthesimilaritiesofExample30.8.8.WritethesimilaritiesofTheorem30.6analyticallyasproductsof(1)ahomothetycenteredattheorigin,(2)arotationabouttheorigin,(3)areflectionifnecessary,and(4)atranslation.

9.ShowthattheproductoftwosimilaritiesoftheformsgiveninTheorem30.6isanothersimilarityofthatsameform.

10.FindthedirectsimilaritythatmapssegmentABtosegmentA′B′wherethesefourpointshavethecoordinates:a)A(1,0),B(2,0),A′(1,0),B′(2,0)b)A(1,0),B(2,0),A′(2,0),B′(1,0)c)A(1,0),B(2,3),A′(–1,2),B′3,–3)d)A(1,0),B(2,3),A′(–3,–3),B′(–1,2)11.FindtheoppositesimilaritythatmapseachsegmentABtoA′B′forthesetsofpointsgiveninExercise30.10.

12.FindequationsforthesimilaritythatcarriestriangleABC,whereA(0,0),B(1,0),andC(0,2),intotriangleA′B′C′,wherea)A′(3,0),B′(3,2),C′(7,0)b)A′( ,0),B′(4,4),C′( ,5)c)A′(0,0),B′(3,0),C′(0,6)d)A′(–3,–2),B′(–4,–2),C′(–3,–4)e)A′(–5,5),B′(–6, ),C′(–2,3)

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4 VECTORSANDCOMPLEXNUMBERSINGEOMETRY

SECTION31 THESEARCHFORTHEMEANINGOFCOMPLEXNUMBERS

31.1Throughoutthehistoryofmathematicsitseemsthattherulehasbeentouseeachnewentityforafewhundredyearsbeforedecidingjustwhatitisyouareusing. The early thinking bymathematicians about complex numberswas notunlike the muddy ideas expressed in other areas of mathematics, butoccasionallyitwasdottedwithabitofinsight.31.2 Inanattempttojustifyusingnegative, irrational,andimaginarynumbers,the English mathematician George Peacock (1791–1858) stated his absurd“principle of permanence of forms,” that “equal general expressions ofarithmetic are to remain equal when the letters no longer denote simplequantities,andhencealsowhentheoperationischanged”!Hewastryingtostatethat the same rules applied to negatives, etc., as apply to positive rationalnumbers.Forexample,sincea+b=b+awheneveraandbarepositiverationalnumbers,thena+b=b+ashouldholdforallnumbers.

Such fuzzy thinking never really contributed to the advancement ofmathematics. And especially not when done as recently as the nineteenthcentury.31.3Theearliestglimmeringsof theexistenceof imaginarynumbersappearedsome 1100 years ago, when the Hindu Mahavira made the very intelligentstatement that“anegativenumberhasnosquareroot.”Furthermore,hewiselymade no attempt to work with these nonexistent square roots of negativenumbers.31.4LucaPacioli (1445–1509)stated the rule that theproductof twonegativenumbersisapositivenumber,butusedthatfactonlyrarely.Andtothestudentwithlittleunderstandingofnegativenumbers,thisruleisfarfromobvious.Heislikelytofeel thataproductsuchas(–2)×(–3)means(not2)×(not3),andifyoumultiplysomethingthatisnot2bysomethingnot3,thentheresultcertainlyisnot6.Thusheconcludesthat(–2)×(–3)=–6.31.5RafaelBombelli (1526–1573) improvedalgebraicnotation,wroterootsofcubic equations as sums of imaginary numbers, and formulated rules forhandlingimaginaries.31.6RenéDescartes(1596–1650)statedthatapolynomialequationofdegreenhas nomore thann roots, and itwas hewho applied those unfortunatewords“real”and“imaginary”tonumbers.Atthistimeimaginarieswereconsideredtobe“uninterpretableandevenself-contradictory.”Thattheywereusedwithever-increasingfaithisoneofthegreatmysteriesofhumannature.

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31.7RogerCotes(1682–1716),bysimpleformalism,gavetheforerunnertothefamous theorem of DeMoivre. The first real bit of mathematics of complexnumbers thus was born. After Cotes’ death Newton lamented, “If Cotes hadlivedweshouldhavelearntsomething.”31.8ThisbringsustoAbrahamDeMoivre(1667–1754),whosefamoustheoremstates

forn a positive integer, introducing complex numbers into trigonometry. It issaid thathediedofanarithmeticprogression.Suddenly,at the ripeoldageof87,hefoundthathewasrequiringsome15minutesmoresleepeachnightthanthe night before. When the terms of this progression reached 24 hours, heexpiredinhissleep.31.9A firm believer in themanipulation of formulas, Leonhard Euler (1707–1783) was the first to state DeMoivre’s theorem in its present form, and toextend it to all values ofn. It was in 1777 that he suggested i for andcalled a2 + b2 the norm of a + bi. Euler was a rather ordinary person. For amathematician.When inBerlin, he answered theQueen’s questioning only inmonosyllables.Whensheaskedwhy,herepliedthatitwasbecausehehad“justcomefromacountrywhereeverypersonwhospeaksishanged.”31.10Inordertogivesomesortofmeaningtocomplexnumbers,CasparWessel(1745–1818), a Norwegian who spent much of his life as a surveyor for theDanishAcademy of Sciences, formulated their geometric interpretation in theplane.Heinterpretedadditionandmultiplicationastranslationandrotation(seeSections33and34).Hisverycompletepaperappearedin1799,twoyearsafterhis discovery. Unfortunately it lay unnoticed by the mathematical world fornearly100years.31.11 As so often occurs in the history of science, another person, thebookkeeper Jean Robert Argand (1768–1822) of Geneva, arrived at the sameinterpretationin1806.Hispaper,althoughnotasclearasWessel’s,waswidelyread.HencetheplaneofcomplexnumbersiscalledanArganddiagram.31.12 Now, simply replacing coordinates (a, b) of points in the plane bycomplexnumbersa+biprovidesanicepicture,aids the imagination,andhasworthwhile applications in geometry, trigonometry, and electrical engineering,but innowaydoes itprove thatcomplexnumbersexist inalgebra.Therewasstillnojustificationforusinganimaginarynumberasarootofanequation.Butthenineteenthcenturymarkedthebeginningofeffortstosolvetheproblemsinthe foundations ofmathematics. So the timewas ripe for someone to find thetrue meaning of complex numbers in mathematics, and again two men,

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independentlyofoneanother,providedidenticalanswers.31.13In1825CarlFriedrichGauss(1777–1855),thegreatestmathematicianofmodern times, stated that “the truemetaphysics of is illusive.”Thus herecognized the problem at hand, a giant step forward! Since his doctoraldissertation was a proof of the fundamental theorem of algebra–that eachpolynomial with complex coefficients has at least one complex zero–he wasquitewellversedinimaginarynumbers.31.14In1831hefoundthe“truemetaphysicsof .”Fromabasisintherealnumber systemhedefinedanalgebraoforderedpairs (a,b)wherein equality,addition,andmultiplicationaregivenby

and

In the resultingalgebraoneprovesquite readily that theseorderedpairs (a,b)behavelikecomplexnumbersa+bi.Thustheseorderedpairsofrealnumbersare complex numbers. The mathematician is now satisfied of their existence.Thisisthe“truemetaphysicsof .”31.15 Gauss did not publish his work, not unusual for him, and quiteindependently the IrishmanWilliamRowanHamilton (1805–1865) performedthesame feat in1835andpublishedhiswork.LaterGaussclaimed theearlierdiscovery,sothecomplexplaneissometimescalledtheGaussplane.31.16Itwouldseemthattheorderedpairnotation(a,b)isnothingmorethanathin disguise for a + bi, but the algebraic theory of the former is rigorous,eliminating thatmysteriousentity i,which stands for the (nonexistent?) .Howmanyhighschoolgraduateseventodaybelievethat“istandsforthesquarerootof–1,anumberyoucannottakethesquarerootof”?Evenaslateas1873,theLaroussedictionarystatedthatimaginarynumbersare“impossible”andthatalgebrauptothenhadfound“onlytwoimpossibleentities:thenegativeandtheimaginary.”It isacuriositythat“theimaginary”cametobeunderstoodbefore“thenegative.”ExerciseSet311.a)Fromthetwoequations

Bombelliobtained

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Showthis.b)Usepart(a)tofindp2+qwhen .Thensolveforqandsubstitutethatvalueintotheequation

,obtaining4p3–15p=2,whichhasarootp=2.Nowshowthat

2.ProveDeMoivre’sTheorem(see31.8).3.AssumingDeMoivrenormallyrequired8hoursofsleep,howmanydayswasitbeforeheexpiredafterhisprogressionstarted(see31.8)?

4.UsingGauss’definitionsforequality,addition,andmultiplicationasgivenin31.14,showthata)(a,b)+(0,0)=(a,b)b)(a,b)(1,0)=(a,b)c)(a,b)+(–a,–b)=(0,0)d)(a,b)(a/(a2+b2),–b/(a2+b2))=(1,0)when(a,b)≠(0,0)e)(0,1)(0,1)=(–1,0)5.Interprettheorderedpairs(a,b)ofExercise31.4ascomplexnumbersa+bi,andshowthateachequationinthatexerciseistrueforcomplexnumbers.Especially,whatistheinterpretationforpart(e)?

SECTION32 INTRODUCTIONTOCOMPLEXNUMBERS

32.1Many educatedpeople retain a subtle distrust of complexnumbers.Theyhavenotovercomethefeelingthat“iisasymbolforthesquarerootofanumber(–1)thatyoucannottakethesquarerootof.”Thuscomplexnumbersarethoughttobemysteriousquantities(ornonquantities)thatarenotunderstoodatall.Theyarethoughttobeentitiesthatarepurelyimaginary(inthenontechnicalsenseofthatword),andthathavenoconceivableuse.32.2 Nothing could be further from the truth! The analysis of electric andelectroniccircuitryreliesmostheavilyonthecomplexnumbersystem.Similaranalysiscanbemadeinmechanics,too.Itiswithgreatreluctancethatwedonottake thespacehere toshowsuchapplications. Instead,weshall investigate thevery elegant applications of complex numbers to plane geometry andtrigonometry.32.3 It ishopedthat,afterreadingthematerial in thischapter,youwillhaveamuchfinerappreciationoftherealnessofcomplexnumbers, thatyouwillfeelcomfortable in their presence, and that you will be able to apply them toappropriate problems in geometry and, perhaps to a lesser extent, intrigonometry. In particular, you should understand just what is meant by thatmysteriousequationi2=–1,andbytheplussigninthecomplexnumber2+3i.Does itmeanaddition? If so, thenwhat is the sumof2and3i?Doyouget5somethings?Again,since2+3i≠3+2i,whichisthegreater?Is2+3i>3+

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2i,oris3+2i>2+3i?Justwhatismeantbysayingthatthecomplexnumberscannotbeordered?32.4Inshort,thischapterproposestoshowthereaderthatthecomplexnumbersarejustasrealtoworkwithastherealnumbers.Tothisend,weshallapproachthegeometryofcomplexnumbersthroughadiscussionofvectorsintheplane.32.5Acomplexnumbermaybethoughtofasapolynomiala+biinthesymboli,withrealcoefficientsaandb,subjecttotheconditionthat

Equality,addition,andmultiplicationforcomplexnumbersareexactlythesameasforpolynomials,rememberingthati2isalwaysreplacedby–1:

and

This interpretation of complex numbers as polynomials is quite satisfactory tothemathematician,butitdoesnotprovidethestudentwithanysortofconcretepicture ofwhat complex numbersmean.He certainly does not expect ever tomeet one on the street corner. The polynomial idea is convenientmathematically,easytotalkabout,butcompletelysterileforyoungstudents.32.6Realnumbersareeasilypicturedaspointsonanumber lineorrealaxis.Furthermore, this interpretation is easily modified to allow real numbers asvectorsalongtherealaxis.Thus5–3=2means“travel5unitsinthepositivedirection and then 3 units in the negative direction. The result is the same astraveling2unitsinthepositivedirection.”(SeeFig.32.6.)Thus5,–3,and2canbethoughtofasvectors(directedlengths)of5,3,and2unitsalongtherealaxisintheappropriatedirections.

Figure32.632.7 In just the same way, we associate points in the plane with complexnumbers.ThepointP(a,b)intheCartesianplaneisassociatedwiththecomplex

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numbera+bi,andhencewiththevector fromtheoriginOtothepointP.Thenthex-axisiscalledtherealaxis(Re)andthey-axistheimaginaryaxis(Im)(seeFig.32.7a).WhenweconsiderthetwopointsPandQasrepresentingthecomplexnumbersa+biandc+di, thesum(a+c)+(b+d)imaybeshownvectoriallyquitenicely.WhenweletSbethepointrepresentingthatsum,thenPOQSisaparallelogram;thatis, isthevectorsumof and (see33.6through33.9).Figure32.7bshowsthisaddition.

Figure32.7a

%

Figure32.7b32.8 Vectors in the plane certainly are real entities; a distance in a givendirection isnotsimplyafigmentof the imagination.Henceweshallbeginourstudy of complex numbers by the slightly round-about route of vectors andvector addition, which will lead naturally into complex numbers and theiraddition.32.9 To obtain an added bonus, we define multiplication of vectors from anaturalgeometricdesiretorepresentrotationsandhomothetiesonvectors.This

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multiplicationturnsouttobetheusualmultiplicationofcomplexnumbers(see33.10 through33.12 and34.8 through34.13).Thus from thegeometric vectorinterpretation, we obtain the complex numbers. This geometric pictureintroduces a “reality” to the complex number system that few high schoolstudentsareevergiventheopportunitytoappreciate.32.10Althoughourprimarypurposeinstudyingvectorshereistoleadintothecomplexnumbersystem,thegreatimportanceandusefulnessofvectorsassuchshouldnotbeoverlooked.Manytheoremscanbeprovedbeautifullybymeansofvectors. Some examples are given in 33.15 to 33.21, 34.17, 34.24, and theexercisesinSections33and34.Again,vectormethodsdonotprovidetheonlytoolinmathematics,buttheydoprovideatoolthatisbothusefulandimportant.32.11 Section 35 leads you from vectors into complex numbers, and theremainder of this chapter develops and illustrates the geometry of complexnumbers.Theteacherwhounderstandsthismaterial ismuchbetterpreparedtoimparttohisstudentsagenuinefeelingforcomplexnumbers.ExerciseSet321.Order(>)amongtherealnumbershasthefollowingthreeproperties:1)Trichotomy.Everyrealnumberxsatisfiesexactlyoneofthethreestatements,x>0,x=0,or–x>0.2)Closureofthepositivenumbersunderaddition.Ifa>0andb>0,thena+b>0.

3)Closureofthepositivenumbersundermultiplication.Ifa>0andb>0,thenab>0.Thefollowingchainoftheoremsdemonstrateswhythecomplex

numberscannotbeorderedinthesamewayasaretherealnumbers.Proveeachtheoremtocompletethisdemonstration.a)Itcannotbetruethat–1>0.b)1>0.c)Assumethatthecomplexnumbersdosatisfythethreepropertieslistedabove.Showthatitisnottruethati>0.

d)Itisnottruethat–i>0.e)i≠0.f)Thethreeorderpropertiesarenotsatisfiedbyi.g)Thesetofallcomplexnumberscannotsatisfythethreeorderproperties.h)Itmakesnosensetodiscusswhether3+2i>2+3ior2+3i>3+2iistrue(see32.3).

2.Calculatethefollowing,a)(2+3i)+(3–2i)b)(2+5i)+(2–5i)c)(2+5i)–(7+6i)d)(2+3i)(3+2i)e)(2+3i)(3–2i)f)(2+3i)(2–3i)g)(2+i)/(1

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–i)h)(3+2i)/(4–3i)i)(1+i)12j)(5+10i)/(2+i)k)Istheanswertopart(b)acomplexnumber?Explain.

3.Solveforrealnumbersxandy.a)2x+3i=5–yib)7x+8yi=1+ic)5+x+y=2y+(2x–y)id)x2–y2+2xyi=–1e)x2–y2+2xyi=if)(x+yi)2=ig)(x+yi)2=–ih)(x+yi)2=7–24ii)(x+yi)2=–7+24i

4.Showthat:a)i2=–1b)i3=–ic)i4=1d)i4k+m=imforintegralke)1/i–i–1=–if)i–m=i4k–mforintegralk

5.Evaluate:a)i5b)i6c)i7d)i8e)i42f)i7183g)i–2h)i–3i)i–4j)i–5k)i–6i)i7m)i–8n)i–35o)i–2161p)5i9–8i5+7i4–3i–3+7

6.Whatistheresultoftraveling5ydeastandthen10ydnorth?Whatistherelationbetweensuchtravelsandthecomplexnumber5+10i?

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7.Interpret2–3iintermsoftravelingasinExercise32.6.8.Writeoutyourownexplanationofjustwhat“i2=–1”meanstoyou.Saveyouressayandreaditaftercompletingthischapter.Similarly,giveaninterpretationtotheplussigninacomplexnumbersuchas2+3i.

9.Isianumber?Explain.10.Oneisoftentold“youcannotdothus-and-so”bymathematicsteachers

whenthestatementiscorrectonlyinanappropriatecontext.Someexamplesofsuchthingsthat“cannotbedone”aregivenbelow.Explainthecircumstancesunderwhichyoucannot,andunderwhichyoucan,doeachlistedtask,andthenperformthetask.a)Takethesquarerootof–5.b)Factorx2–7.c)Factorx2+2.d)Factorx2+4x+5.e)Canceltheb’sinthequotientab/bcoftwotwo-digitnumberstogeta/c.Example: .

f)Findastatementthatcannotbetrueandcannotbefalse.g)Findtwononparallellinesthathavenopointsincommon.h)Findnumbersaandbsothata2+ab+b2isasquarenumber.i)Findthreedistinctlinesm,n,psothatnandpareeachperpendiculartom,butnandparenotparallel.

SECTION33 VECTORS

33.1DefinitionAvectorintheplaneisanorderedpair(a,b)ofrealnumbersaandb.Vectorsaredenotedbyboldfacelower-caseRomanlettersu,v,w,…orbypointpairswithanarrowoverbar .33.2DefinitionVectorsu=(a,b)andv=(c,d)arecalledequaliffa=candb=d.33.3Moreinformally,avectorisadirecteddistance;thevector(5,–3)means“5units in the x-direction and –3 units in the y-direction.” We do distinguishbetweenthedirectedsegment goingspecificallyfromthegivenpointytothegivenpointBandthevector determinedbythatdirectedsegment.ThusreferstothespecificdirectedsegmentfromA toB,whereas simplyisthatdirectionanddistancewithno reference to specific initial and terminalpoints.Wewrite = onlywhenA=CandB=D.Wewrite = wheneverABDCisa(perhapsdegenerate)parallelogram.Weneverwrite = .33.4DefinitionGiventhepointsP1(x1,y1)andP2(x2,y2),wewrite

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33.5Thatis, isthevectorrepresentingthedirecteddistancenecessarytotravelfromP1toP2(Fig.33.5).Itfollowsimmediatelythatthevector =(x,y)whenOandPhavecoordinatesO(0,0)andP(x,y).33.6DefinitionLetA,B,Cbepointssuchthat =vand =wforgivenvectorsvandw.Wedefinevectoradditionby

33.7Informally,wethinkofaddingthevectorsvandwbyplacingtheheadofvectorvonthetailofvectorw,thentakingastheirsumthevectorfromthetailofv

Figure33.5to the head of w as in Fig. 33.7. That vector addition is associative andcommutativeisstatednext,thecommutativityillustratingthatthesumv+wisgivenbytheparallelogramlaw;thatis,intheparallelogramhavingvandwassides,v+wisthatdiagonalthatemanatesfromthevertexwherethetwovectortailsmeet.

Figure33.733.8TheoremVectoradditioniscommutativeandassociative.

TheproofofthistheoremisindicatedinFigs.33.7and33.8,thedetailsbeing

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leftforthereadertosupply.

%

Figure33.833.9TheoremWhenu=(a,b)andv=(c,d),then

33.10DefinitionWhenkisarealnumberandv=(a,b)isavector,wedefinescalarmultiplicationasfollows.Wesaythatthescalarproductofkandvis

33.11 Thus scalarmultiplication of a vector by a real number (or scalar) k issimplyastretchorhomothetyofratiok.Thedirectionofthevectorisunchangedwhenk<0andisreversedwhenk<0.Themagnitudeorlengthofthevectorismultipliedbythefactor|k|.33.12TheoremWhencandkarerealnumbersanduandvarevectors,then1)c(kv)=(ck)v=(kc)v=k(cv),2)k(u+v)=ku+ku,3)(c+k)v=cv+kv.33.13DefinitionWhenuandvarevectors,thenwewrite

Thisoperationonvectorsiscalledvectorsubtraction.33.14 The properties developed so far are quite adequate to prove manytheorems concerning points and lengths in polygonal figures. Some examplesfollow.33.15TheoremThediagonalsofaparallelogrambisecteachother.LettheparallelogrambeABCDwith and .

33.16Firstproof.LetthediagonalsmeetatEasinFig.33.16.Now

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Figure33.16SinceEispartwayalongACandpartwayalongDB,therearepositiveonstantsmandn,eachlessthen1,suchthat

Nowweconsider foranotherpointofview:

Equatingthesetwoformsfor ,wehave

so

This last equationcanbe trueonly if thecoefficientsofbothvectorsarezero,sinceneithervectorisamultipleoftheother.Thus

fromwhichweobtainm=n= .Hence

establishingthetheorem.33.17Secondproof.LetE be themidpoint ofAC, andF themidpoint ofBD.Then and

. Since the vectorsand areequal,itfollowsthatpointsEandFcoincide;thatis,thetwo

diagonalsmeetattheirmidpoints.33.18TheoremThemediansofatriangleconcuratatrisectionpointofeach

median.Letusshowthat thepoints2/3of thewayalongeachof the threemedians

coincide.Tothatend(seeFig.33.18),letu= andv= .Then =v–u.Now,formediansAA″,BB″,CC″,inturn,wehave

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Figure33.18Sincetheselastthreevectorsareallequalandhavethesameinitialpoints,theirterminalpointscoincide;thatis,themediansconcuratatrisectionpointofeachmedian.33.19TheoremThealtitudesofatriangleconcur.LetusthistimedefinethetriangleABCintermsofthevectors =u,

=v,and =wfromitscircumcentertoitsverticesasinFig.33.19.Thenthevectorsu, v, w all have the same length, the circumradius. LettingA1 be thefourth vertex of the parallelogram BOCA1, then BOCA1 is a rhombus whosediagonals are perpendicular and bisect each other at pointA′, themidpoint ofsideBC.Furthermore,vectorOA1=v+w.LetpointHbedefinedbymakingAOA1Haparallelogram.Hence

Since OA1 is perpendicular to BC, then so also is AH perpendicular to BC.Furthermore,bythesymmetryoftherepresentationforOH,itfollowsthatBHisperpendiculartoAC,andCHisperpendiculartoAB.HenceHisthemeetingofthethreealtitudes.

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Figure33.1933.20Wecallspecialattentiontotheequation

provedinTheorem33.19.Itstates that thevectorfromthecircumcenter totheorthocenterofatriangleisequaltothesumofthevectorsfromthecircumcentertothethreevertices.33.21TheoremTheorthocenter,circumcenter,andcentroidofatriangleare

collinearand .Referring to thenotation in theproofofTheorem33.19andFig.33.19,we

have

Now maybecalculatedfromtheequationsofTheorems33.18and33.19,since .Thus

fromwhichthetheoremfollows.ExerciseSet331.ProveTheorem33.8.

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2.Showthatthesetofallvectorsisclosedundervectoraddition;thatis,showthatthesumofanytwovectorsisalwaysanothervector.

3.ProveTheorem33.9.4.ProveTheorem33.12.5.Showthatthesetofallvectorsintheplaneisclosedundervectorsubtraction,whichisneithercommutativenorassociative.

6.Showthatifu+v=w,thenu=w–v.7.Showthatvectoradditioniswelldefined,thatis,ifA,B,C,D,E,Faresixpointssuchthat and ,then .

8.Showthatthevector0=(0,0)istheadditiveidentity,andthateachvectorv=(a,b)hastheadditiveinverse–v(–1)v=(–a,–b).

9.Provethatthemidpointsofthesidesofaquadrilateralformaparallelogram.10.Provethatthesegmentjoiningthemidpointsoftwosidesofatriangleis

paralleltoandequaltohalfthethirdside.11.AssumingthatOisanypointintheplaneoflineAB,andthatMisthe

midpointofsegmentAB,showthat (ComparewithTheorem2.12.)12.GiventhatOisanypointintheplaneoflineAB,andthatPisapointonlineABthatdividessegmentABintheratior/s,showthat .

13.WhenOisanypointintheplaneofatriangleABC,showthat

whereA′,B′,C′arethemidpointsofthesidesofthetriangle.14.Provethatthelinesjoiningagivenvertexofaparallelogramtothe

midpointsofthetwooppositesidestrisectthatdiagonalnotthroughthegivenvertex.

15.Provethatinanytrianglethethreemedians,asvectors,formatriangle.16.Inatriangle,dotheanglebisectors,asvectors,formatriangle?17.Showthatthesegmentjoiningthemidpointsofthenonparallelsidesofa

trapezoidisparalleltothebasesandcongruenttohalftheirsum.SECTION34 VECTORMULTIPLICATION

34.1BythePythagoreantheorem,thelength(ormagnitude)ofthevectorv=(a,b)isgivenby(a2+b2)1/2.Weusethefamiliarabsolutevaluebarstodesignatethislength.34.2DefinitionIfv=(a,b),thenwedefinetheabsolutevalueormagnitude|v|ofthevectorvby

34.3Theorem34.4statesthatabsolutevalueasappliedtovectorshastheusual

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propertiesofabsolutevalue.34.4TheoremIfuandvarevectorsandcisascalar,then1)|v|≥0,2)|v|=0iffv=(0,0),denotedby0,3)|cv|=|c||v|,4)|u+v|≤|u|+|v|,5)|u–v|≥|u|–|v|.34.5DefinitionIf|u|=1,thenuiscalledaunitvector.34.6 Theorem Each unit vector u has the form (cos θ, sin θ) where θ is thetrigonometric angle in standard position made by the vector u when it isconsideredtoemanatefromtheorigin(seeFig.34.6).

Figure34.634.7Theorem Each vectorv can bewritten in the formcuwhereu is a unitvectorandcisanonnegativescalar.Ifv≠0,thisrepresentationisunique.

Simplytakec=|v|andu=(1/c)V.34.8 It is convenient to define vector multiplication in two stages. First, weconsideraunitvectorasa rotationabout theorigin throughangleθwhere theunitvectoru=(cosθ,sinθ).Theproductofthetwounitvectorsu=(cosθ,sinθ)andv=(cosϕ,sinϕ)becomesthatunitvectorrepresentingtheproductofthetworotations.Wehave

byapplyingtheformulasfromtrigonometryforcos(θ+ϕ)andsin(θ+ϕ).(SeealsoExercise21.6.).Theformaldefinitionfollows.SeeFig.34.8,where,letting

, and P(1, 0), we have triangles OPQ andOP′Q′congruentbySAS.

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Figure34.834.9DefinitionIfu=(a,b)andv=(c,d)areunitvectors,thenwedefinetheirproduct*uvby

34.10 Theorem Unit vector multiplication is commutative, associative, anddistributiveovervectoraddition,hastheidentity(1,0),denotedby1,andeachunitvectoru=(a,b)hasamultiplicativeinversegivenbyu–1=(a,–b).

Clearly the vector1 = (1, 0) is equal to (cosO, sinO), so this vector is themultiplicative identity. This result is important enough for us to restate thenotationinthenextdefinition.Since(a,b)(a,–b)=(a2+b2,–ab+ab)=(1,0)whenever(a,b)isaunitvector,itfollowsthatu–1=(a,–b)isthemultiplicativeinverseofu=(a,b).(Notethatu–1isthereflectionofuinthex-axisasmirror(seeFig.34.10).Therestof

Figure34.10thisproof,thecommutativity,associativity,anddistributivity,followeitherfromthe algebraic equations or from the geometric interpretation of unit vector

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multiplicationastheproductofrotations.

34.11DefinitionLet1=(1,0),calledthevectorone.

RecallingDefinition34.5,we see that1 is aunitvector,but thatnot everyunitvectoris1.

34.12Nowweturnourattentiontomultiplicationofanytwovectorsv1andv2.ByTheorem34.7, letu1 andu2 beunitvectorsandc1 andc2 be scalars,bothnonnegative,suchthatv1=c1u1andv2=c2U2.Itseemsnaturaltodefinetheirproductby

Geometrically,v1v2isarotationandahomothety,whichcanbedisplayedasinFig. 34.12. Let , and . ThentrianglesOPQ andOP′Q′ are similar.That is,v1v2 is the resultof rotatingv1throughtheangleθ2of

Figure34.12

v2 and magnifying it by the ratio |v2|. Curiously, this definition of vectormultiplication gives rise to the same algebraic formulation as for unit vectors.Thisfactwenowstateasourformaldefinitionofvectormultiplication.

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34.13DefinitionIfu=(a,b)andv=(c,d)areanytwovectors,thenwedefinetheirproductby

34.14 Theorem Vector multiplication is commutative and associative,distributiveovervectoraddition,andhasidentity1=(1,0).Eachvectorv=(a,b)≠(0,0)hasthemultiplicativeinverse

Theexpressionfortheinversev–1ofvectorvfollowsfromtheequation

Thus

34.15Theorem|uv|=|u||v|.

34.16Weclosethissectionwithapairofapplications.

34.17ExampleFindavectorconditionfortwovectorsu=(a,b) 0andv=(c,d)≠0tobeorthogonal(perpendicular)(Fig.34.17).

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Figure34.17

Letusdenotethevector(0,1)=(cos90°,sin90°)byi.Theniisthevectorcorrespondingtoa90°rotation.Sinceuandvareorthogonal,thenarotationofeither±90°carriesuintothedirectionofv.Thuswemaywrite

astheconditionforperpendicularityofthevectorsuandv.Then

Thatis,

Eliminatingkbetweenthesetwoequations,weodtain

asanecessaryandsufficientconditionforthetwononzerovectorsu=(a,b)andv = (c, d) to be orthogonal. In terms of the dot product (see the footnote toDefinition34.9),thisequationstatesu·v=0.Intermsofvectormultiplication,ifwewrite =(a,–b),calledtheconjugateofu(itsmirrorimageinthex-axis),thisconditionfororthogonalitymaybewrittenas

fortheexpressionontheleftbecomes

Thislastvectoris0iffac+bd=0.

34.18Itshouldbestatedformallythattwonotationswereintroducedinthislast

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example.Wedoso.

34.19DefinitionLet =(a,–b)denotethevectorconjugateofthevectorv=(a,b).

34.20Theorem and .

34.21Theorem .

34.22Theorem .

34.23DefinitionLeti=(0,1).

34.24ExampleCeva’stheorem.ThreeceviansAD,BE,CFfortriangleABCareconcurrentiff

Weshallproveonly that theequation is truewhen theceviansconcur.Theconversecanbeestablishedbyvectormethodsassuggestedintheexercises.

Let =uand =v.Letmandnbescalarssuchthat =muand(seeFig.34.24).Then

Figure34.24

Thentherearescalarsrandssuchthat

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Hence

and

Sinceu andv havedistinctdirections, it follows that eachcoefficientmustbezero;thatis,

Thus

sowehave

Nowtherearealsorealnumberspandq,suchthat

Thenwehave

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Bycollectingterms,obtain

andagainthecoefficientsofuandvmustbothbezero,yielding

andhence

Finally

34.25Theexamplesgivenintheselasttwosectionsindicatethatvectormethods

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canbequiteusefulforcertainproblems,buttherearecaseswhenothermethodsareeasier.Vectorsareaconvenientandhelpful tool ingeometry,but theyarenot the universal tool. In the next few sectionswe shallmodify our approachslightlyinordertoincreasethisusefulness.

ExerciseSet34

1.ProveTheorem34.4.2.ProveTheorem34.6.3.Provethat|v|=|–v|.4.ProvethattheformulagiveninDefinition34.9agreeswiththeconceptofmultiplicationgiveninthediscussionof34.8.

5.CompletetheproofofTheorem34.10.6.CompletetheproofofTheorem34.14.7.ProveTheorem34.15.8.ProveTheorem34.20.9.ProveTheorem34.21.10.ProveTheorem34.22.11.ProvetheconversepartofCeva’stheorem(Example34.24)byvector

methods.12.LetvectorsuandvformthelegsCAandCBofarighttriangleABC.Show

thatthePythagoreanrelationisequivalenttotheconditiongiveninExample34.17.

13.Showthatthemidpointofthehypotenuseofarighttriangleisequidistantfromthethreevertices.

14.LetvectorsuandvformthesidesCAandCBoftriangleABC.ShowthatthebisectorofangleCsatisfiestheequation

15.Provethatthethreeanglebisectorsofatriangleconcur.16.AsinExercise34.14,findanexpressionforeachanglebisector,asavector,

intermsofthevectors fromanypointPintheplanetothe

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verticesofthetriangle.17.InparallelogramABCD,showthat

18.Provethatthediagonalsofarectanglehaveequallengths.19.Provethatthediagonalsofarhombusareperpendicular.20.Provethateverypointontheperpendicularbisectorofasegmentis

equidistantfromtheendsofthesegment.21.Provethatwhentwomediansofatrianglearecongruent,thenthetriangleis

isosceles.

SECTION35 VECTORSANDCOMPLEXNUMBERS

35.1Continuingourdevelopmentofthealgebraofplanevectors,letusexaminemorecloselythetwounitvectors1=(1,0)andi=(0,1).Thesevectorsformabasis forplanevectors; that is,eachvectorv canbe representeduniquelyasalinearcombinationof1andi,asasumoftheform

whereaandbarescalars.

35.2TheoremLetv2denotevv.Then12=1.For12=(1,0)(1,0)=(1°1–0°0,1°0+0°1)=(1,0)=1.

35.3Geometrically,1 isarotationthrough0°;thatis,1 is theidentityrotation.Thisideawillleadusquitenaturallyintocomplexnumbers.

35.4Theoremi2=–1.Fori2=(0,1)(0,1)=(0°0–1°1,0°1+1°0)=(–1,0)=–1.

35.5AnotherimportantgeometricresultisgiveninTheorem35.4.Itstatesthat,sinceiisa90°rotation,theni2istheproductoftwo90°rotations,hencea180°rotation.Sincea180°rotationisrepresentedvectoriallyby(cos180°,sin180°)= (–1, 0) = –1, we have i2 = –1. Again, this equation simply states that theproductoftwo90°rotationsisa180°rotation.

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35.6TheoremTherealnumbersxandthevectorsx1areringisomorphic;thatis, letting the real number x correspond to the vector x1, and denoting thiscorrespondencebyx~x1,wehave1)x=yiffx1=y1,2)(x+y)~(x1+y1),3)(xy)~(x1)(y1).

Thestatementx1=y1isequivalentto

whichistrueiffx=y.Similarlypart(2)isestablished,sincex1+y1=(x+y)1byTheorem33.12and(x+y)~(x+y)1.Also

35.7Theorem35.6statesthattherealnumbersxandthevectorsx1alongthex-axisbehavethesamealgebraically,sowemaytreatthemthesame.Whetherweperformarithmeticoperationsonrealnumbersoronthecorrespondingvectors,the results are the same, so itmakes no difference inwhich systemwework.Thisisclearwhenwerealizethatthegeometryofvectorsalongthereallineisan interpretation or a model of the algebra of real numbers. It means that,algebraically,wecouldsimplyomitwritingthevector1intheformx1withoutcausing confusion. The convenience of the ideas presented in these last threetheoremsisstatedinthenexttheorem.

35.8TheoremWhenvectorsu=(a,b)arewrittenaslinearcombinationsoftheunitvectors1andi,asu=a1+bi,andwhenthevector1isomitted,asu=a+bi, thenthesevectorscanbeaddedandmultipliedjustasrealpolynomials in iareaddedandmultiplied,providedi2,wheneveritappears,isreplacedby–1.

Lettingu=a+biandv=c+di,wehave

and

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Thustheresultsarethesamewhichevertechniqueisusedtoaddortomultiply.

35.9ExampleTomultiplythetwovectorsu=(2,–3)andv=(1,5),weneednolonger remember the complicated formula given in the definition of vectormultiplication(Definition34.13).Rather,weusethemethodofTheorem35.8toobtain

35.10 This multiplication and addition remind one of the correspondingoperations on complex numbers. Indeed, considering i as the imaginary unitinsteadofasavector,thealgebraicoperationsareidentical.Itfollowsthatoneinterpretation for the complex number system is the system of vectors in theplane,justasrealnumberscanbeinterpretedasvectorsalongaline.

35.11 We now turn our attention to the geometry of complex numbers, firstinterpreting in the complex number plane the vector concepts we havedeveloped.

35.12DefinitionToeachpointZoftheplaneweassociatethecomplexnumberz=a+biwhereZhascoordinates(a,b).TheplanethusinterpretediscalledtheGaussplane,pointZistheimageofthecomplexnumberz,andzistheaffixofZ(seeFig.35.12).

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Figure35.12

35.13Itfollowsthat

Also,ofcourse,thesumoftwocomplexnumberscorrespondstothevectorsumoftheirassociatedvectors,andtheproductoftwocomplexnumberscorrespondstothevectorproductoftheirvectors.

35.14 Now is an appropriate time for you to reinvestigate the questionsconcerning complex numbers posed in 32.3. You should be in a much betterpositionnowtofindlogicalanswers.

35.15LetusreexaminethevectorpropertiesestablishedinSections33and34asthey apply to complexnumbers.Twocomplexnumbersa +bi andc +di areequal,a+bi=c+di,iffa=candb=d(Definition33.2).Theirsumisgivenby(Theorem33.9)

Addition of complex numbers is commutative and associative (Theorem33.8)andhasidentity0=0+0i(Exercise33.8).Theadditiveinverseofz=a+biis–z=–a–bi(Exercise33.8).

Theproductofa+biandc+diisgivenby(Definition34.13),

Thismultiplication is commutative, associative, and distributive over addition

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(Theorem34.14).Ithasidentity1=1+0i,andeachnonzerocomplexnumberz=a+bihasthemultiplicativeinverse(Theorem34.14),

Theorem35.4statesthati2=–1.

35.16InTheorems34.6and34.7itwasseenthateachcomplexnumberz=a+bicanbewrittenintheform(seeFig.35.16)

where

andwhere

Figure35.16

Fromthediscussionin34.8,itisclearthatif

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then

Theabsolutevalueofz=a+biisgivenby

where =a –bi, thecomplex conjugate of z (Definitions34.2 and34.19andTheorem34.21).And for complexnumbers z andw,wehave (Theorems34.4and34.14)

and

Also and ,byTheorems34.20and34.21.

35.17 Definition If z = r(cos θ + isin θ) = a + bi then a + bi is called therectangular form of z,a is its realpart andb is its imaginary coefficient.Wewrite

The symbol i is called the imaginaryunit.Also r(cosθ + isinθ) is called thepolarortrigonometricformforz,risitsabsolutevalue(seeDefinition34.2)oritsamplitude,andθisitsangleorargument.Itisconvenienttodefinecisθ(read“sistheta”)oreiθby

Theexpressionreiθiscalledtheexponentialformofz.

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35.18DefinitionInthecomplexnumberz=x+yi,ifx=0,thenziscalledpureimaginary;ify=0,thenziscalledreal.

35.19Therectangularforma+bi isespeciallyusefulforpurposesofadditionand subtraction, not as convenient formultiplication and division, and almostuseless for powers and roots. On the other hand, with the trigonometric andexponential forms of complex numbers addition is next to impossible, butmultiplication, division, raising to a power, and taking roots are easilyperformed.HenceitisimportanttobefamiliarwithallthreeformspresentedinDefinition35.17.TheconvenienceofthetrigonometricandexponentialformsisillustratedinExercises35.6through35.11and35.18through35.23.

35.20Inacourseincomplexanalysis,itiscustomarytouseonlyradianmeasurefor anglesθ, especiallywhenusing the formeiθ. Sincewe are concernedhereonlywith the convenienceof such forms to geometry,we shallmakeno suchrestriction.Infact,weshallgenerallyusedegreemeasureforourangles.

ExerciseSet35

1.Multiplythevectors(2,3)and(4,5),thenmultiplythecorrespondingcomplexnumbersandshowthattheproductscorrespond.

2.RepeatExercise35.1usingadditioninsteadofmultiplication.3.Answerthequestionsposedin32.3.Compareyouranswerswiththosegivenearlier.Again,savetheseanswersuntilthischapterhasbeencompleted.

4.Showthatwhenz=a+bi≠0,thenz–1=(a–bi)/(a2+b2).5.Showthatwhenz≠0,thenz–1= /|z|2.6.Showthateiθeiϕ=ei(θ+ϕ)7.ProveDeMoivre’stheorem:Ifnisanaturalnumber,then(cisθ)n=cisnθ.

8.Showthat(eiθ)n=einθforallnaturalnumbersn.9.Provethat(cisθ)/(cisϕ)=cis(θ–ϕ).10.ShowthatbothDeMoivre’stheorem(Exercise35.7)andthestatementof

Exercise35.8holdtruefornanyinteger.11.RewritethestatementofExercise35.9inexponentialform.12.Evaluate(5cis30°)4.

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13.Evaluate .

14.Showthateiτ+1=0,aninterestingidentityrelatingfiveofthemostimportantconstantsinmathematics!

15.Showthatz= iffzisreal.16.Showthatz=– iffzispureimaginary.17.Showthatz+ , ,and(z– )/iareallreal.18.Showthatifw=r1/ncis((θ+360k°)/n)foranyintegersn>0andk,thenwn

=rcisθ.Hencewisannthrootofrcisθ.19.Findthethreecuberootsof8.20.Findthefivefifthrootsof–1.21.Findthetwosquarerootsofi.22.Findthetwosquarerootsof–i.23.Findthefourfourthrootsof–1.24.Showthat:a)|z|=| |

b)c)d)e) –z

25.Showthat .ComparewithExercise34.17.

26.Solve2z+ =5+iforz,makinguseofthefactthatwheneveragivenequationw=zistrue,thentheequation isalsotrue.

27.(5–2i)+ =–16iforz28.solve|z|–z=1+2iforz.29.solve|z|–z=1+2iforz.30.showthatifa +b=zhasnosolutionz,then|a|=1.

SECTION36 TRIANGLESINTHEGAUSSPLANE

36.1Inthisandtheremainingsectionsofthischapter,alllower-caseitaliclettersshall denote complex numbers; i, of course, is still reserved for the imaginaryunit.Unlessstatedotherwise,upper-caseitaliclettersdenotethepointswhoseaffixes

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arethecorrespondinglower-caseitalic letters.Thuswenowbegintoshowtheassociationbetweenplanegeometryandthealgebraofcomplexnumbers.

36.2TheoremRe(z)=(z+ )/2andIm(z)=(z– )/2i.

36.3TheoremTheimageofthepointwhoseaffixiszundera:1)halfturnabouttheoriginis–z,2)reflectioninthex-axisis ,3)reflectioninthey-axisis– ,4)rotationthroughangleθabouttheoriginiszeiθ=zcisθ,5)translationthroughvector isz+w,6)rotationthrough90·abouttheoriginisiz.

Theproofsofthesestatementsareleftfortheexercises.

36.4 Theorem 33.6 tells how the complex numbers z andw should be added.Geometrically, z +w is the affix of the fourth vertex S of the parallelogramhaving affixes z, 0, w for three consecutive vertices (see Fig. 36.4a). Figure36.4b shows the corresponding construction for the product zw: LetU be theimageof1.ConstructtriangleOWPsimilartotriangleOUZ.ThenPistheimageofzw.

Figure36.4a

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Figure36.4b

36.5ProblemToconstructw–z,takew–zastheaffixofthefourthvertexDoftheparallelogramOZWD(seeFig.36.5).

%

Figure36.5

36.6 Problem To construct z–1 when z ≠ 0, let 1 be the affix ofU andconstructRsothattrianglesORUandOUZaresimilar.ThenRistheimageofz–1(seeFig.36.6).

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Figure36.6

Proofsoftheselasttwoconstructionsareleftasexercises.

36.7TheoremAgiventriangleABCisequilateraliff

whereλequalseither

Since λ equals either e120°i or e240°i, then (b –a)λ rotates sideAB througheither120°or240°intosideBC,and(c–b)λrotatessideBCthroughthesameangleintosideCA(seeFig.36.7).IneithercasethetriangleABCisequilateral,counterclockwiseinthefirstcaseandclockwiseinthesecond.Conversely,eachequilateraltriangleisformedbysuchrotations.

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Figure36.7

36.8Beforecontinuingourdevelopment,letusdigressbrieflytorecallsomeofthebasicpropertiesofdeterminants.Personsorclasseswhohavenotyethadtheopportunitytostudythistopicmaywishtospendsomeextratimeherelearningthe basic concepts of determinant algebra. Although the development givenbelowissufficientforourmodestneeds,mosttextsonanalyticgeometryoroncollege algebra contain a fuller treatment of the subject. Put simply, adeterminantisanumberassociatedwithasquarearrayormatrixofnumbersinthemannerprescribedbythedefinitionsthatfollow.

36.9DefinitionAsecond-orderdeterminantisdefinedby

36.10DefinitionAthird-orderdeterminantisdefinedby

36.11Example

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Also

Althoughwedonotneed it asyet, it isconvenient todefinea fourth-orderdeterminantatthistime.

36.12DefinitionAfourth-orderdeterminantisdefinedby

36.13Fifth-andhigher-orderdeterminantsaredefinedin just thissamemannerin terms of determinants of the next-lower order. Of the many properties ofdeterminants,weselectonlyafewbasiconestopresenthere.

36.14TheoremThevalueofadeterminantisunchangedifamultipleofonerowisaddedtoadifferentrow,orifamultipleofonecolumnisaddedtoadifferentcolumn.

36.15TheoremIfalltheelementsofaroworofacolumnofadeterminantaremultipliedbyaconstant,thenthevalueofthedeterminantismultipliedbythatconstant.

36.16CorollaryIfalltheelementsofaroworofacolumnofadeterminantarezeros,thenthevalueofthedeterminantiszero.

36.17TheoremIf two rowsor twocolumnsofadeterminantareproportional,thenthevalueofthedeterminantiszero.

36.18Havingpresentedthesefewbasicpropertiesofdeterminants,wearenowabletoreturntothegeometryofcomplexnumbers.Somegeometricconditions

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lendthemselvestomostelegantdeterminantforms.

36.19Theorem Two (perhaps degenerate) trianglesABC andDEFare directlysimilariff

Let the two triangles be similar and suppose that the similarity that mapstriangleABCtoDEFiscomposedofarotationthroughangleθandahomothetyofratior,andletz=rcisθ.Thene–d=z(b–a)andf–d=z(c–a).(Notethatthecentersoftherotationandthehomothetyneednotbelocated.)Now

by subtracting multiples of the last column from the other two columns(Theorem36.14)andthenapplyingTheorem36.17.

Conversely,ifthedeterminantiszero,then

fromwhichitfollowsthatthereisacomplexnumberzsuchthat

ButthentrianglesABCandDEFaresimilar,

36.20CorollaryAtriangleABCisequilateraliff

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36.21CorollaryTwotrianglesABCandDEFareoppositelysimilariff

36.22CorollaryAtriangleABCisisosceleswithapexAiff

ExerciseSet36

1.ProveTheorem36.2.2.ProveTheorem36.3.3.ProvetheconstructionofProblem36.5.4.ProvetheconstructionofProblem36.6.5.Showthatwhenz=reiθ,thena)cosθ=(Re(z))/|z|=(z+ )/2(z )1/2

b)sinθ=(Im(z))/|z|=(z– )/2i(z )1/2c)tanθ=(Im(z))/(Re(z))=(z= )/i(z+ )6.ProveTheorem36.14.

7.ProveTheorem36.15.8.ProveCorollary36.16.9.ProveTheorem36.17.10.a)Provethatasecond-orderdeterminantiszeroiffitsrowsare

proportional;thatis,iffthereisaconstantksuchthateachelementofoneoftherowsisktimesthecorrespondingelementoftheotherrow.

b)Showthat“rows”canbereplacedby“columns”inpart(a).c)Showthatparts(a)and(b)arenottrueforathird-orderdeterminant.

11.ProveCorollary36.20.12.ProveCorollary36.21.13.ProveCorollary36.22.14.Showthatthetrianglewhoseverticeshaveaffixesa,b,andzb+(1–z)ais

directlysimilartothetrianglewithverticeswhoseaffixesare0,1,andz.

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15.Showthat

16.FindanexpressionsimilartothatinExercise36.15fora)sin BACb)tan BAC

17.Giventhatλ=cis60°inTheorem36.7,thenshowthatthefigurecannotbeatriangle.

18.ShowthattriangleABCisequilateraliff(c–b)2=(b–a)(a–c).19.ShowthattriangleABCisequilateraliff

20.WhatistrueoftriangleABCif

21.WhatistrueoftriangleABCif

22.Showthatwhentworowsortwocolumnsofadeterminantareinterchanged,thenthevalueofthedeterminantismultipliedby–1.

23.Showthat

24.GeneralizeExercise36.23.25.Investigatehowtoexpandathird-orhigher-orderdeterminantintermsof

rowsotherthanthefirstrow(asdefinedinDefinitions36.10and36.12)and

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alsointermsofcolumns.26.Defineafifth-orderdeterminant.Howmanytermsappearinthecomplete

expansionofasecond-,third-,fourth-,andfifth-orderdeterminant?27.ForanytwopointsAandBintheplane,letusdefineA*B=C,wherepoint

CistakensuchthattriangleABCisdirectlysimilartoagiventrianglewhoseaffixesare0,1,z.a)Showthat*isabinaryoperationonthepointsoftheplane.b)Showthat*isnotcommutative.c)Showthat*isnotassociative.d)ShowthatA*A=A;thatis,ifA*Aistobedefinedmeaningfully,thenwemusthaveA*A=A.

e)Showthat,givenanytwoofthethreepointsA,B,andC,wecansolvetheequationA*B=Cforthethirdpoint.

f)Provethemedialproperty:(A*B)*(C*D)=(A*C)*(B*D).g)Provethe“distributivelaw”:A*(B*C)=(A*B)*(A*C).h)Showthat(A*B)*A=A*(B*A).Whenthegiventriangleisequilateral,findthiscommonvalue.

28.Provethat|z+w|2=|z|2+|w|2+2Re(z )29.Provethat|z–w|2=|z|2+|w|2–2Re(z )30.Provethat|z+w|2+|z–w2=2(|z|2+|w|2).

SECTION37 LINESINTHEGAUSSPLANE

37.1TheoremPointsA,B,Carecollineariff

37.2TheoremAnequationforthelinepassingthroughpointsAandBis

Clearlytheequationissatisfiedwhenz=aandwhenz=b.ThetheoremthenfollowsbyTheorem37.1.

As an alternative method of proof, expand the determinant to obtain the

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equation

Writingx+iyforz,so =x–iy,wefindthattheequationreducesto

Sinceeachofthesecoefficientsispureimaginary,whenwemultiplythroughbyi,weobtainanequationoftheformαx+βy+γ=0forrealα,β,γ.Thisisanequationforastraightline.

Theorem37.2givesusaformforeverystraightline,andweobservethenextresultasadirectcorollarytotheproofgivenforthattheorem,yieldingasimpleequationforaline.

37.3CorollaryEachlineintheGaussplanehasanequationoftheform

37.4TheoremThelineaz+az+b=0,withbreal,issatisfiedby–b/2aandbyia–b/2a,andsoisperpendiculartothelineOA.

Thatthetwopointssatisfytheequationisstraightforwardalgebra,sincebisreal.Then the line joins these twopoints, so it is parallel to the line from theorigintotheimageofia,a90°rotationofthelineOA.

37.5TheoremTwolines ,and ,withb1andb2real,areparalleliffa1/a2isreal.

37.6TheoremThetwolinesofTheorem37.5areperpendiculariffa1/a2ispureimaginary.

TestingforparallelismorperpendicularitytwolineswhoseequationsareoftheformgiveninTheorem37.3isquitesimple,accordingtoTheorems37.5and37.6. We may test similarly any two lines whose equations are given inparametricform,asindicatedinTheorem37.7andCorollaries37.8and37.9.

37.7 Theorem The line through A and parallel to OB has the parametricequation,withrealparametert,

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37.8CorollaryThelinesz=a+tbandz=c+ud,withrealparameterstandu,areparalleliffb/disreal.

37.9 Corollary The lines of Corollary 37.8 are perpendicular iff b/d is pureimaginary.

37.10TheoremTheareaoftriangleABCisequalto

whicheverispositive.Leta–c=αandb–c=β.ThentheareaoftriangleABCisgivenby |α||β|

sinC,soletusfindanexpressionforsinC(seeFig.37.10).Nowβ=(|β|/|α|)eiCα,sowehaveeiC=β|α|/|β|α.Itfollowsthat

NowtheareaKisgivenby

We obtained the plus sign in this case because triangle ABC was oriented

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counterclockwise,asshowninFig.37.10.[Wheredidweusethisinformation?]Theminussignholdswhenthetriangleisclockwiseoriented.

Figure37.10

37.11TheformulaofTheorem37.10isconvenientforfindingareasoftriangleswhen three specific points are given. Its theoretical uses also abound; forexample, the areaof such a triangle is zero iff the threevertices are collinear.Hence a test for the collinearity of three points A, B, and C is that thedeterminantofTheorem37.10isequaltozero.(SeeTheorem37.1.)

ExerciseSet37

1.ProveTheorem37.1.2.ShowthatthelinethroughAandparalleltoOBhastheequation

3.through9.ProveTheoremsandCorollaries37.3through37.9.10.LetA(a1,a2),B(b1,b2),C(c1,c2)bethreepointsintheCartesianplane.

ShowthattheareaoftriangleABCisgivenbythefamiliarformulafromanalyticgeometry

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11.ShowthatwhenthetriangleABCisclockwiseorientedinTheorem37.10,thentheminussignholds.

12.Showthatthethreelines andbkreal,areconcurrentiff

13.Findthepointofintersectionofthetwolines

14.ShowthatthelinesOAandOBareperpendiculariffab+ab=0.15.ShowthatthetwolinesofTheorem37.5(andofExercise37.13)are

perpendiculariffa1a2+a1a2=0.16.ShowthatthelinethroughpointA,andperpendiculartolineOB,hasthe

equation

17.Findtheareaofthetrianglewhoseverticeshavetheaffixes:a)0,1,3–2ib)0,1,zc)3,2i,3+2id)5–3i,2+i,–1+2i

18.a)ShowthatthecentroidoftriangleABChasaffix(a+b+c)/3.b)OnthesidesofatriangleABCandexternaltoit,constructtrianglesABC′,BCA′,CAB′allsimilartoagiventrianglePQR.ProvethecentroidsoftrianglesABCandA′B′C′coincide.

c)IfthetriangleABCofpart(b)isequilateral,showthattriangleA′B′C′isalsoequilateral19.Showthatifrisreal,thenc=(a+rb)/(1+r)istheaffixofthepointthatdividessegmentABintheratior.

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20.LetA1B1C1andA2B2C2beanytwotriangles.LetA3B3C3beatrianglesuchthatA3,B3,C3divideA1,A2,B1B2,C1C2,respectively,inthesameratior.ShowthatthecentroidsGiofthethreetrianglesAiBiCiarecollinear,andfindtheratioinwhichG3dividesG1G2.

SECTION38 THECIRCLE

In this section we shall treat both circles and lines simultaneously, for theirequationsarise together.The treatmentwill followalong thesamepathas thatforlinesinSection37.

38.1TheoremAnequationforthecirclewithcenterCandradiusris

Thisequationiseasilyderivedbysquaringbothsidesoftheequation|z–c|=r.

38.2CorollaryAnycirclehasanequationoftheform

wherebisreal.Theaffixofitscenteris–a,anditsradiusisequalto(aa–b)1/2.

38.3TheoremAnequationoftheform

andwhere t isa realparameter, representsastraight linewhencd is real,oracircle inallothercases.Furthermore, each lineorcirclecanbewritten in thisparametricform.

Theproofof this theorem is straightforward, but quite involved.Henceweshallbreakuptheproofintothreesteps:(1)eliminatetheparametertandwritetheresultingequationinaconvenientform;thenweshallexaminethisequationwhen(2)cdisreal,and(3)inallothercases.Proof. 1) Taking the conjugate of each number in the given equation andrecallingthattisreal(sot=t),weobtain

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Nowsolvebothequationforttoobtain

Thuswehave

andfinally,

Inthislastequationthecoefficientofz ,andalsotheconstantterm,arereal.2)Ifcdisreal,thenitsconjugatecdisalsoreal,socd= .Thenequation(*)becomes

astraightlinebyTheorem37.3.3) In every other case, , so we may divide equation (*) by thatquantity,obtaining

anequationforacircle,since isreal.

38.4DefinitionIntheparametricequationofTheorem38.3,nofinitevalueoftyields z =a/c.Yet this point lies on the lineor circlewheneverc ≠ 0.Let usrectify this omission by defining t = ∞ as follows. Replacing t by 1/u and

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clearingfractions,weobtain

Inthislastform,setu=0togetz=a/c.Henceweagreethatt=∞meansu=0whereu=1/t,andwepermitooasavalueforarealparameter.Tothatend,itbecomesapparentthatwhent=–d/c,thenzdoesnotexist.Soweaddjustoneidealpoint∞ to theGaussplane, andwewritea/0=∞whena≠0.Now theequation of Theorem 38.3maps each real number including co to a complexnumberincluding∞.

38.5Theorem The parametric equation z = (at +b)/(ct +d) is determined towithinaconstantfactorinnumeratoranddenominatorbyanythreevaluesoftheparametertandthecorrespondingzvalues.

Geometrically,threepointsdeterminealineoracircle.Thealgebraisleftasanexercise.

38.6Itiscustomaryandconvenienttoconsiderthezvaluescorrespondingtot=0, 1, and∞, calling them z0, z1, and z∞. In the equation of Theorem 38.5 inparticular,notethat

Theseequationsyield

inwhicheithercordmaybechosenarbitrarily.Thentheotherthreeparametersareuniquelydetermined.

38.7 The parametric equation of Theorem 38.3 represents either no locus, asinglepoint,ortheentireplanewhenad–bc=0andwhenwrittenintheform

dependingonthenatureoftheconstantsa,b,c,d.Thedetailsarenotespecially

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importanttous.

38.8TheoremThelocusofallpointsZwhosedistancesfromtwofixedpointsAandBareintheratiok≡1,kreal,isacircle.

Setting|z–a|/|z–b|=kandsquaring,weobtain

whichmayberewrittenas

anequationforacircle.

38.9TheoremThecircleofTheorem38.8canbewrittenintheparametricform,withrealparametert,

Forthisequationholdsiff|z–a|\|z–b|=k.

38.10Theorem IfA ≡B, then the equation of Theorem 38.9 represents acirclewhenk≡1,kreal,andaline,theperpendicularbisectorofAB,whenk=1.

38.11 Theorem For fixed t and with real parameter k, the equation ofTheorem38.9representsacircleorlinethroughpointsAandB.

ExerciseSet38

1.ProveTheorem38.1.2.ProveCorollary38.2.3.ProveTheorem38.5.4.ProveTheorem38.9.

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5.ProveTheorem38.10.6.ProveTheorem38.11.7.FindthecenterandradiusofthecircleofTheorem38.3,part(3).8.ShowthateachlinecanbewrittenintheparametricformofTheorem38.3.9.ShowthateachcirclecanbewrittenintheparametricformofTheorem38.3.

10.Findparametricequationsforthelinesorcirclesgiventhatz0,z1,andz∞are:a)0,1,∞b)0,1,ic)0,1+i,3–id)2,1+i,3–ie)1+i,1+3i,2–if)a,b,c

11.Investigatethevariouscasesdiscussedin38.7.12.Findthevaluesofkforwhichz=aandforwhichz=binTheorem38.11.13.Showthatwhen|a|=1or|b|=1,then|(a–b)/(1–ab)|=1.14.Forgivenaandb,findthesmallestvalueof|(z–a)(z–b)|.15.Ifaandbaretheaffixesoftwoverticesofasquare,locatetheothertwo

verticesforeachpossiblepositionofthesquare.16.Thepolynomialequationz6–z5+z3–7z2+6z–6=0hastworootswhose

sumiszero.Findthem.17.LocatethecircumcenterandfindthecircumradiusoftriangleABC.18.Showthatifaz+bz=chasnosolution,then|a|=|b|.Istheconverse

statementtrue?19.SolvetheequationofExercise38.18.20.RereadSection31,especially31.14andExercises31.4and31.5.?

SECTION39 ISOMETRIESANDSIMILARITIESINTHEGAUSSPLANE

AsintheclosingsectionsofChapters2and3,thislastsectionofChapter4isdevotedtotheanalyticrepresentationsofisometriesandsimilarities.Wepresenthereequations for thesemaps in theGaussplane.Theorems39.5,39.7,39.11,andCorollary 39.13give themain results.Theorem39.8 states the interestingresult that taking conjugates of the points in the plane is “the basic” oppositeisometry.

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39.1 Theorem A translation through vector (mapping Z to Z′) has theequation

39.2TheoremArotationaboutpointAthroughangleθhastherepresentation

39.3TheoremAhomothetywithratiok(kreal)andcenterAhastheequation

39.4TheoremEachequationoftheform

represents1)atranslationifa=1,2)arotationifa≠1but|a|=1,3)ahomothetyifaisrealanda≠1,or4)theproductofarotationandahomothetyif|a|≠1.

This theoremisreadilyseen tobe truewhena iswritten in theexponentialformreiθ,sothat

Inthisformitisclearthatwehave1)arotationaboutO throughangleθ,2)ahomothetyincenterOandratior,andthen3)atranslationthroughvector .

39.5TheoremEachdirectsimilaritymaybewrittenintheform

39.6 Theorem Each direct similarity maps circles into circles and lines intolines.Thedirectsimilarityz′=αz+βmapstheequation

Thatis,

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Sincethislastequationhasthesamedenominatorastheoriginalonehad,itstillrepresents the same type of curve, circle or line, as the original equation,accordingtoTheorem38.3.Itislefttothereadertoshowthatsincead–bc≠0isgiven,thenthecorrespondinginequalityistrueinthenewequation.

39.7TheoremEachdirectsimilarityisdeterminedbytwogivenpointsAandBandtheirimagesA′andB′andhastheequation

ByTheorem36.19.

39.8TheoremAreflectioninthex-axishastheequation

39.9TheoremAreflection in the linemcrossing the realaxisatpointAwithangleofinclinationθhastheequation

Let x denote the real axis. Then σm = (σm σx)σx, a reflection in the x-axisfollowedbyarotationaboutAthroughangle2θ.ThetheoremnowfollowsfromTheorems39.8and39.2.

39.10TheoremA reflection in the linemparallel to the realaxisandpassingthroughpointBhastheequation

39.11TheoremEachoppositesimilarityhasanequationoftheform

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39.12TheoremAreflectioninthelineABhastheequation

For any given pointZ in the plane, trianglesABZ andABZ′ are oppositelysimilar,sowehave

byCorollary36.21;thatis,

whichmayberewrittenintheform

Nowfactorthisequationtoget

fromwhichthetheoremfollows.

39.13CorollaryAnoppositesimilarityisdeterminedbytwogivenpointsAandBandtheirimagesA′andB′,andhastheequation

39.14TheoremEachoppositesimilaritymapscirclesintocirclesandlinesintolines.

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ExerciseSet39

1.ProveTheorem39.1.2.ProveTheorem39.2.3.ProveTheorem39.3.4.Answeragainthequestionsposedin32.3.Compareyourcurrentanswerswiththosegivenearlier,5.ProveTheorem39.5.

6.CompletetheproofofTheorem39.6.7.VerifyTheorem39.9bywritingthedesiredreflectionastheproductofarotationaboutAthroughangle–θ,followedbyareflectionintherealaxis,andthenfollowedbyarotationaboutAthroughangleθ.

8.ProveTheorem39.8.9.UnderwhatconditionsonaandbisthedirectsimilarityofTheorem39.5involutoric?Hencededucetheconditionsunderwhichitrepresentsahalfturn.

10.ProveTheorem39.10.11.ProveTheorem39.11.12.UnderwhatconditionsonaandbistheoppositesimilarityofTheorem

39.11involutoric?Hencededucetheconditionsunderwhichitrepresentsareflection.

13.SolvetheequationofTheorem39.12forz′andwriteitintheformofTheorem39.11.

14.ProveTheorem39.14.15.Provethattheproductoftwotranslationsisatranslation.16.Findtheanalyticexpressionfortheproductoftworotations,andshow

whenthisproductrepresentsatranslation.17.Writeanequationforareflectionintheimaginaryaxis.18.Writeanequationforareflectionintheliney=x.19.Writeanequationforthehalfturnabouttheorigin.20.WriteanequationforahalfturnaboutpointA.

* Vectormultiplication,asdefinedhere,isneverdenotedbyaraiseddotasinu·vorbyacrossasinu·v.Thesenotationsarereservedfortwoothervectorproducts,thesocalledscalarordotproductu·v=ac+bd,andthecrossproduct,whichisdefinedinspaceofthreedimensions.

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5 INVERSION

SECTION40 MATCHLESSMODERNMATHEMATICS

40.1Duringtheeighteenthandnineteenthcenturiesmanyadvancesweremadeinanalyticgeometryandin thecalculus,aswellas insyntheticgeometry.Themethods of the calculus were applied to the study of various curves andproblems which defined curves, ushering in a new and fascinating aspect ofEuclideangeometry.Weshallnot,however,considersuchworkhere.Someoftheworkersthatareofinteresttousfollow.40.2 Giovanni Ceva (1648–1737), an Italian engineer, studied transversals ofgeometric figures,publishing the theoremwhichbearshisname (seeTheorem5.2)in1678inhisDelineisrectisseinvicemsecantibus. It iscuriousthat thistheorem should have escaped discovery for so long, since its dual,Menelaus’theorem,was“wellknown”inA.D.100,accordingtoMenelaus.40.3MatthewStewart (1717–1785), a professor ofmathematics at Edinburgh,extendedthetheoremsofCeva(seeExercise2.10).40.4 Solid analytic geometry was developed by Antoine Parent (1666–1716),andfurtheredbyAlexisClaudeClairaut(1713–1765),oneoftwentychildrenofateacherofmathematics.Heandabrotherwhodiedofsmallpoxatage16were“twoofthemostprecociousmathematiciansofall time,”accordingtoHowardEves.40.5In1706WilliamJones(1675–1749)wasthefirsttousetheGreekletterπtodenotetheratio3.14159…ofacircle’scircumferencetoitsdiameter.40.6 Projective geometry was rediscovered by Gaspard Monge (1746–1818),whose outstanding work spurred great interest among mathematicians in thisfield.Atapartyonceheoverheardascoundrelslanderingacertainwoman,andimmediatelyleapttoherdefense.Sometimelaterhewassingularlyattractedtoawoman,andwhen theywere introduced,he found that shewas theveryonewhosehonorhehaddefended.Later theyweremarried,andamore loyalwifewouldhavebeenverydifficultforanyonetofind.40.7 Constructions using compass alone were investigated by LorenzoMascheroni (1750–1800),whoproved that all ruler andcompassconstructionscouldbeaccomplishedwiththecompassalone.Thisassumes,ofcourse,thatalineisdeterminedwheneveranytwopointslyingonitarelocated.Hismethodofattackleanedheavilyonthereflectiontransformation.40.8 After reading Mascheroni’s work, Jean-Victor Poncelet (1788–1867)examined straightedge constructions, proving that all ruler and compassconstructions could be performedwith a straightedge alone, provided that one

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hadavailableatleastonecircleofanysizealongwithitscenter.40.9 JacobSteiner (1796–1863) investigated rulerconstructions further.Called“thegreatestgeometriciansincethetimeofEuclid,”heextendedPascal’smystichexagramtheorem,whichstates that the threepointsof intersectionofpairsofopposite sides of a hexagon inscribed in a conic section are collinear (seeExercise23.2),byshowingthatifthesixverticesaretakeninallpossibleorders,the resulting60“Pascal lines”pass threeby three through20“Steinerpoints.”TheseSteinerpoints liefourbyfouron15“Plticker lines.”OthersshowedthePascallinesalsoconcurthreebythreein60“Kirkmanpoints”whichliethreebythree upon 20 “Cayley lines” which pass four by four through 15 “Salmonpoints”!40.10Malfatti’s problem is to inscribe three circles in a triangle so that eachcircleistangenttotwosidesofthetriangleandtothetwoothercircles.Steinergaveasyntheticconstruction,notingthattheproblemhas32solutions.40.11 Joseph Diaz Gergonne (1771–1859), and Poncelet independently,discoveredtheprincipleofdualityinprojectivegeometry.GergonnealsosolvedtheproblemofApollonius: todrawwith rulerandcompassacircle tangent toeachofthreegivencircles.Thegeneralcasehaseightsolutions.40.12 His only claim to fame being the theorem that the ninepoint circle istangent to the four equicirclesof a triangle (seeTheorem44.6),KarlWilhelmFeuerbach (1800–1834) published his theorem when only 22. Having been apoliticalprisonerforatimeatage19,hegraduallywentmad,spendingthelastsixyearsofhislifelockedawayfromtheworld.40.13Duringthenineteenthcenturythethreefamousproblemsofantiquity, totrisectanangle,toconstructthesideofacubewhosevolumeistwicethatofagivencube, and to construct a square equal in area to agivencircle, all usingonly Euclidean tools, were proved impossible. This theorem, which leansheavilyonfieldtheory,provesthatnoonecaneverperformtheseconstructionsifheusestherulerandcompasscorrectly.Anyconstructionthatclaimstohavesolvedtheseproblemseithergivesonlyanapproximationtothecorrectresult,orusesthetoolsimproperly.Thisfacthasbeenmathematicallyproved—justasitiseasytoprove,fromtherulesofthegame,thatitisimpossibletohavethirteenkingsofonecolorinagameofcheckers.Also,πwasprovedirrationalin1770,andtranscendentalin1882.40.14 Other contributors to the growth of geometry include Arthur Cayley(1821–1895), Karl von Staudt (1798–1867), Michel Chasles (1793–1880),JohannPliicker (1801–1868),HenriBrocard (1845–1922),andEmileLemoine(1840–1912). In 1899 David Hilbert (1862–1943) published his famousGrundlagen derGeometrie (Foundations ofGeometry), a careful development

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ofEuclideangeometryfromasetofpostulates.40.15Themost significant advanceof thenineteenth centurywas a change inmathematical thinking. Euclidean geometry had been considered the only truegeometrythatcouldeverexist.Nodifferenttypeofgeometrywasconceivable.Butmathematicianshadbeenbothered formore than20 centuriesbyEuclid’sfifth postulate, the “parallel postulate.” His first four postulates are clear,concise, and reasonably obvious “truths” about the real world. His fifthpostulate,especiallyinitsoriginalform,isnoneofthese.Itisnotclearwithoutapicture,itisquitewordy,andnotatallobvious(seeExercise40.8).

Since theywerepositive that itwas true,mathematicians right fromGreektimesattemptedtoshowthatthefifthpostulatecouldbededucedfromtheotherfour. Ifso, then this troublesomeassumptioncouldbesimplydeletedfromthelistofpostulates.Fromtheirmanyattempts,severalstatementsequivalenttotheparallelpostulatewerediscovered,buteachattemptedproofwasshowntohavesomeflaw.40.16GirolamoSaccheri(1667–1733)in1733,JohannLambert(1728–1777)in1766,andAdrien-MarieLegendre(1752–1833)in1794to1823,eachattemptedto prove the parallel postulate, and in so doing, proved many basic andinterestingtheoremsinso-calledabsolutegeometry,Euclideangeometrywithoutthe parallel postulate. Had one of them recognized and admitted theimpossibility of his attempted program, hewould have been creditedwith thediscoveryofnon-Euclideangeometry.40.17Thenon-Euclideangeometryinwhichthroughapointnotonagivenlinethere is more than one line not intersecting (parallel to) the given line, wasdiscovered independently first by Carl F. Gauss (1777–1855), the greatestmathematician of his age, then by JohannBolyai (1802–1860), and finally byNicolaiLobachevsky(1793–1856).Lobachevsky,however,wasin1829thefirsttopublishhisfindings,sothisgeometrygenerallybearshisname.40.18Nowgeometrywas freeof its traditionalmold.Hereafter, ageometry issimply a collectionof assumptions about a set of things calledpoints, and thelogical results thereof. Many new geometries followed. In 1854 BernhardRiemann(1826—1866)developedageometry,realizedinthegeometryofgreatcirclesonasphere,inwhichtherearenoparallellineswhatever.40.19 Generalization now became the motto in mathematics. Henri Poincare(1854–1912), who showed Lobachevskian geometry to be as consistent asEuclidean geometry, developed the geometry of continuous functions, namelytopology.Maurice Fréchet (1878– ) in 1906 began the study of very abstractspaces. Today it is quite difficult to state just where geometry leaves off andotherbranchesofmathematicsbegin.

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40.20 In an attempt to obtain order out of the ever-increasing mass of newgeometries, Felix Klein (1849–1929) in 1872 defined a geometry to be “thestudyofthosepropertiesthatareinvariantwhentheelements(points)ofagivensetare subjected to the transformationsofagiven transformationgroup.”Thiscodification,whilequiteconvenientandusefuleventoday, isnowoutdatedbygeometriesthatdonotfitthisclassificationscheme.40.21Thusmathematicshaschangedfromthematerialaxiomatics,thestudyofthe real world, of the Greeks to the formal axiomatics of today. Now one iscompletely free to state a collection of postulates about any set of objects, solongasthepostulatesdonotcontradictoneanother(notalwaysaneasyquestiontoanswer),andtoinvestigatetheconsequencesthereof.Inthisway,hundredsofgeometriesandalgebrashavebeeninvestigated.Andtheendisnotinsight.Newmathematicsisbeingconstructeddaily.ExerciseSet401.Showhowtoconstruct,withEuclideancompassalone,eachofthepointsindicated.a)GivenpointsAandB,constructpointConlineABsothatd(AB)=d(BC).

b)GivenpointsA,B,X,constructthereflectionofXinlineAB.c)GivenpointsA,B,C,DwithCnotonlineAB,constructthepointsofintersectionoflineABwiththecircleC(D)(withcenterCandpassingthroughD).

d)LookupinEves’SurveyofGeometry,Vol.1,theMohr-Mascheroniconstructiontheorem,pages198–202.

2.GivenabisectedsegmentABandapointPnotonlineAB,constructalinethroughPparalleltoABusingruleralone.

3.GiventwoparallellinesandpointsAandBononeofthem,findthemidpointofsegmentABusingruleralone.

4.LookupinEves’SurveyofGeometry,Vol.1,thePoncelet-Steinerconstructiontheorem,pages204–209.

5.Showthat6points,nothreeofwhicharecollinear,aretheverticesof60differenthexagons.

6.a)TrisectarightangleusingEuclideantools;thatis,constructa30°angle.b)Explainhowitisstillimpossibleto“trisectanangle”inspiteoftheconstructionofpart(a).

7.LookupinEves’SurveyofGeometry,Vol.2,theproofoftheimpossibilityofthethreeconstructionproblemsofantiquity,pages30–38.Thismaterialisalsotobefoundinthesameauthor’sTheFoundationsandFundamental

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ConceptsofMathematics,rev.ed.,pages317–325.8.Euclid’sfifthpostulatereads,“Ifastraightlinefallingontwostraightlinesmakestheinterioranglesononesideofthestraightlinetogetherlessthantworightangles,thenthetwostraightlines,ifextendedsufficiently,willmeetonthatsideofthegivenlineonwhichthetwointerioranglesaretogetherlessthantworightangles.”Interpretthisstatementanddrawanexplanatoryfigure.Doyoufeelthatits“truth”is“obvious”?

9.ReferringtoKlein’sdefinitionofageometryasgivenin40.20,explainwhycongruenceisstudiedinChapter2(Isometries),andsimilarityofgeometricfiguresinChapter3(Similarities).WhatpropertiesshouldbestudiedinChapter5(Inversion)?

SECTION41 INVERSION

41.1 Although not generally considered a part of high school geometry,inversion should be known to the earnest student of geometry. Thistransformationdoesnotpreservecongruenceorevensimilarity,but it isusefulforthequantitiesitdoespreserve.Inversion,itwillbeseen,preservesthesetofallcirclesandlines;thatis,acirclemapsintoacircleoraline,andalineinvertsinto a circle or a line. It preserves cross ratios and angles between curves. Itseemshard to imagine that a transformationnotpreserving lines, distances, orshapes can be worth studying. Indeed, inversion is most worthwhile, as theapplicationsinSection44willdemonstrate.41.2 Inorder that inversionbea transformationof theplane, it isnecessary toappendtotheplaneexactlyoneidealpointinessentiallythesamemanneraswedidfortheGaussplaneinDefinition38.4.41.3DefinitionTheEuclideanplanewithoneidealpointorpointatinfinity(∞)appended is called the inversiveplane.This idealpoint is considered to lieoneverylineintheplane.41.4DefinitionLetCbeafixedpointintheplane,butnottheidealpoint,andlet r be a fixed positive number. For each pointP in the planewe define itsinverseP′inthecircleofcenterCandradiusrbytakingP′asthatpointonrayCPsuchthat

wheneverP≠C.TheinverseofCistheidealpoint,andtheinverseoftheidealpointisC(seeFig.41.4).Thismapofthepointsoftheplaneiscalledinversion(in a circle), or sometimes reflection in a circle, pointC is the center of theinversion,risitsradius,andr2isitspower.

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Figure41.441.5 Theorem Inversion is a transformation of the inversive plane and isinvolutoric.41.6 Theorem Points inside the circle of inversionmap to points outside thecircle,andpointsoutsidemap inside.Pointson thecircleare invariantandaretheonlyinvariantpoints.

IfPliesinsidethecircleofinversionwithcenterCandradiusr,thenCP<r.SinceCP·CP′=r2,thenCP′cannotbelessthanorequaltor,forthenCP·CP′<r2.HenceCP′>r,soP′liesoutsidethecircle.Therestoftheproofissimilar.41.7DefinitionAtransformationoftheplaneiscalledconformaliffitpreservesangles between curves. It isdirectly conformal if it preserves the sense of theangleaswell,andinverselyconformalifitreversesthesense.41.8Theorem Isometriesand similarities areconformal;direct similarities aredirectlyconformal,andoppositesimilaritiesareinverselyconformal.41.9 Theorem Let P and Q be two points not collinear with the center ofinversionC,andletP′andQ′betheirinverses.ThentrianglesCPQandCQ′P′areoppositelysimilar.

Lettingrbetheradiusofinversion,wehave,asseeninFig.41.9,

Itfollowsthat

Since also QCP = P′CQ′, then ΔCPQ ~ ΔCQ′P′ by SAS. Since thesimilaritycanberealizedbyreflectingtriangleCPQ in thebisectorofangleCandthenapplyingthehomothetyH(C,CQ′/CP),itfollowsthatthesimilarityisopposite.

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Figure41.941.10TheoremInversionisinverselyconformal.

Let the curves c and s intersect at pointA and suppose thatCPQ is a raydifferentfromCA,andintersectingcandsatPandQ,respectively,whereCisthecenterofinversion.LetA′,P′,Q′c′s′betheinversesofA,P,Q,c,s(seeFig.41.10).ThenP′∈c′andQ′∈s′.Now

byTheorem41.9.Bysubtractingcongruentanglesinthesesimilartriangles,weobtain

that is, angles PAQ and P′A′Q′ are oppositely congruent. Furthermore, theseangles are congruent no matter how small angle ACP (greater than zero)becomes.By taking the limit as ACP approaches0, anglesPAQ andP′A′Q′becomethe

Figure41.10anglesbetweenthecurvescands,andc′ands′.Itfollowsbythecontinuityofthe inversion transformation that these angles between the two curves andbetweentheirinversesareoppositelycongruent.

41.11TheoremAlinethroughthecenterofinversionisinvariant.

41.12TheoremAlinenotthroughthecenterCofinversionmapsintoacirclethroughC.

DropaperpendicularCPfromthecenterCtothegivenlinem.LetQbeanyotherpointonlinem(seeFig.41.12).LetP′andQ′betheinversesofPandQ.ByTheorem41.9, trianglesCPQ andCQ′P′ are similar.Hence CQ′P′ CPQ= 90°, so it follows thatQ′ lies on the circleof diameterCP′. Since the

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idealpointmapstoC,eachpointoflinemmapstosomepointonthatcircle.

Figure41.12

Furthermore,foreachpointQ′(otherthanC)onthatcircle,rayCQ′cutslinem inapointQ,andfromthefirstparagraphof thisargument,Q inverts toQ′.HencetheimageofthelinemistheentirecircleonCP′asdiameter.

41.13CorollaryA circle through the center of inversionmaps into a line notthroughthecenterofinversion.

41.14TheoremAcirclenot throughthecenterof inversionmaps intoanothersuchcircle.

Let that diameter of the given circle s that passes through the centerC ofinversioncuts inpointsAandB,and letA′andB′be their inversesas inFig.41.14.TakeanyotherPpointonsandletitsimagebeP′.ByTheorem41.9,

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Figure41.14

Fromthesesimilartrianglesobtain

sincePAC isanexteriorangleoftriangleAPB,whichtriangleisinscribedinasemicircle.HenceP′liesonthecircles′ofdiameterA′B′.Itreadilyfollowsthattheinverseofcirclesistheentirecircles′.

41.15 Example Find the inverse of a right triangleABC in vertexC, thevertexoftherightangle.

Draw the triangle as in Fig. 41.15a. Then assume a convenient radius ofinversion,saythelengthofthelongerlegCB.NowlinesCBandCAinvertintothemselves, so draw two perpendicular lines meeting atC, as in Fig. 41.15b.TakeB′ on one of them so thatCB CB′. TakeA′ on the other ray so thatCA·CA′=(CB)2.Since

Figure41.15a

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Figure41.15b

Figure41.15c

Figure41.15d

CA<CB,thenCA′>CB′.Finally,lineABinvertsintoacircles′throughA′andB′andthroughthecenterC.ThusweobtainFig.41.15b.Toseewhatarcsandsegments are the images of the sides AB, BC, CA, superimpose (perhapsmentally)triangleABContoFig.41.15btoobtainFig.41.15c.DrawtoanypointPonsegmentAB,forexample,rayCPtocutcircles′,theimageofsideAB,inpointP′.ThenP′ is the inverseofP, and since this is true, it follows that arcA′P′B′ is the inverse of segmentAB. Since the inverse ofC is the ideal point,

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thentheimagesofCAandCBarethe(infinite)segments,therays,fromA′andB′ to infinity in the directions away from C. These results are shown in Fig.41.15d.HereweseetheinverseA′B′C′oftriangleABC.

41.16Weseethatinversionpreserves,butreversesthesenseof,anglesbetweencurves, and that it maps circles and lines into circles and lines. The circle ofinversion is invariant, and circles concentric to it map into other concentriccircles.Astheradiusofsuchacircleshrinkstozerosothatthecircleshrinkstothe center of inversion, the radiusof its image circle increaseswithout bound.Thus,whenconsidering that thepointat infinity is the inverseof thecenterofinversion,oneisremindedof thefableof theriderwhowassoexcited thathe“jumpeduponhishorseandrodeoffmadlyinalldirections”.

ExerciseSet41

1.Findtheinverseofeachofthefollowingpointsinthecirclewithradius10andcenteredattheorigin.a)(6,8)b)(–6,8)c)(10,0)d)(0,–10)e)(1,0)f)(1,–1)g)(10,10)h)(3,5)i)(2,–7)j)(0,0)k)theidealpoint2.InverteachofthepointsinExercise41.1inthecirclecenteredat(5,0)withradius5.

3.ProveTheorem41.5.4.CompletetheproofofTheorem41.6.5.ProveTheorem41.8.6.ProveTheorem41.10forthecaseinwhichthereisnorayCPQ,differentfromCA,andcuttingcurvescandsinPandQ.

7.ProveTheorem41.11.8.ProveCorollary41.13.9.DrawthefigureforTheorem41.14assumingthecenterCliessomewhereinsidecircles.Howdoestheproofneedtobealteredforthiscase?

10.DrawtheinverseofarighttriangleABCwiththemidpointMofitshypotenuseascenterofinversion.

11.Drawtheinverseofasquareanditsdiagonals,withitscenterascenterofinversion.

12.Drawtheinverseofasquareanditsdiagonals,withavertexascenterofinversion.

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13.Invertanequilateraltriangleinitscentroid.14.Invertanequilateraltriangleinavertex.15.Provethatanycircleorthogonal(perpendicular)tothecircleofinversionis

invariant.16.Provethatanycirclethatisinvariantunderaninversionisorthogonalto,or

coincideswith,thecircleofinversion.17.GiventhatpointsPandQareinvertedtoP′andQ′inacircleofradiusrand

centerC,provethatP′Q′=(r2·PQ)/(CP·CQ).18.a)Provethatwhencirclesinvertsintocircles′,thenthecenterofs′isnot

theinverseofthecenterofs.b)Showthatwhencirclesinvertsintocircles′,thentheinverseofthecenterofsistheinverseofthecenterofinversionwithrespecttoinversionins′.

SECTION42 PROGRESSIONS,RATIOS,ANDPEAUCELLIER’SCELL

The theory of inversions is developed further in this section by presentingrelations between circles and inversion, progressions and inversion, and crossratiosandinversion.AlsodefinedhereisthePeaucelliercell,amechanicaltoolor linkage of the same sort as the straightedge and the compass. Where thestraightedgeallowsus todraw lines, and thecompasspermitsdrawingcircles,thePeaucelliercell isdesigned todraw the inverseofagivencurve.Now,wedrawcircleswith thecompass,andnotby tracingdisks.Furthermore, it isnotnecessary to have a circle in order to build a compass. But we trace a givenstraightlinewhenweuseastraightedge.Thusthetheoreticalquestionarisesastowherethe“first”straightlineevercamefrom.OneanswertothatquestionisgiveninExercise42.7,foraPeaucelliercellcanbeusedtoobtainalinewithoutrequiringanypreviouslydrawnlinesforitsconstruction.

42.1 Definition A transformation of the plane is called circular iff it mapscirclesandlinesintocirclesandlines.

42.2 Theorem Isometries, similarities, and inversions are circulartransformations.

42.3TheoremIfPandP′aredistinctpointsinverseinacirclec,thenanycirclesthroughPandP′isorthogonal(perpendicular)tocirclec.

LetCbethecenterofcirclec,anddrawthetangentCTtocircles(seeFig.

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42.3).

Figure42.3

ThenCP·CP′= (CT)2,byExercise6.18.ButalsoCP·CP′=r2,wherer is theradiusofcirclec,sincePandP′areinversepointsincirclec.ItfollowsthatCT=r,soTliesoncirclec,andthetwocirclesareorthogonal.

42.4Theorem If two circles are orthogonal, let a radius of one circle cut theothercircleintwodistinctpoints.Thenthesetwopointsareinversewithrespecttothefirstcircle(seeFig.42.3).

42.5TheoremTwopointsPandP′areinverseinacirclesiffthediameterABof s that passes through P and P′ is divided harmonically by P and P′ (seeDefinition3.11).

LetCbethecenterofcircles,asinFig.42.5.Thenthefollowingstatementsareequivalentonetotheother:

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andfinally

Thetheoremfollows.

Figure42.5

42.6 Definition Three numbers a, b, c are said to be in : 1. Arithmeticprogressioniffc–b=b–a.Wealsosaythatbisthearithmeticmeanofaandc.2.Geometricprogressioniffc/b=b/aanda>0,b>0,c>0.Wealsosaythatbisthegeometricmeanofaandc.

3.Harmonieprogressioniffl/a,l/b,l/careinarithmeticprogression,anda≠0,b≠0,c≠0.Wealsosaythatbistheharmonicmeanofaandc.

42.7CorollaryIfthetwopointsPandP′ofTheorem42.5areinversetoeachotherincircles,thenthesegmentsAP,AB,andAP′(withBlyingbetweenPand

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P′) are in harmonic progression, and the segments CP, CB, and CP′ are ingeometricprogression.

Wemustshowthat,inFig.42.5,

to prove thatAP,AB,AP′ are in harmonic progression; that is, thatAB is theharmonic mean of AP and AP′. It is easy to show that the two displayedequationsareequivalent.

Alsowemustshow,toverifythegeometricmeanproperty,that

whichistruebythedefinitionofinversepoints.Toprovethefirstrelation,startwith

and

Thetheoremfollows.

42.8Theorem IfPandP′aredistinct inversepointswithrespect toacircles,thenthechordjoiningthepointsofcontactofthetwotangentsdrawnfromthatpoint,sayP,externaltocirclespassesthroughtheotherpointP′.

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42.9 Definition The Peaucellier cell is a mechanical linkage defined asfollows.PintogetherarhombusAPBP′madefromfourrods(orpiecesofheavycardboard)ofequallengththatarehingedateachvertexinsuchawaythattherhombus is free to change its shape. In the same manner pin two congruentsegments,longerthanthefirstfour,oneeachtotheoppositeverticesAandBoftherhombus.Pintheother twoendsof theselongersegments togetherandpinthemtoa fixedpointC in theplane (seeFig.42.9).NowwhenP tracesoutacurvec,pointP′issaidtotracethePeaucellierimageofc.

Figure42.9

42.10TheoremThePeaucellierimageofacurveistheinverseofthecurveinacircleofcenterC,andwithpower(CA)2–(AP)2.

42.11DefinitionLetA,B,C,Dbefourpointslyingonacircle.Definethecrossratio(AB,CD)ofthesefourpoints,takeninthatorder,by

whereAC,CB,AD,DBarechordlengths.TheminussignistakenwhenpointsCandDseparatepointsAandB,andtheplussignistakenotherwise.(SeealsoDefinition 3.6.) 42.12Theorem Inversion preserves the cross ratio of fourpointsonalineoronacircle.

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LetObethecenterofaninversionthatmapsthefourpointsA,B,C,D,notwoofwhicharecollinearwithOandallfourofwhichlieonacircleoraline,toA′,B′,C′,D′,whichalsolieonalineorcirclebyTheorem42.2.ByTheorem41.9wehave(Fig.42.12)

Nowdirectsubstitution into theequationofDefinition42.11shows that | (AB,CD)|=|(A′B′,C′D′)|.SinceAandBseparateCandDiffA′andB′separateC′andD′,then(AB,CD)=(A′B′,C′D′).

Figure42.12

ExerciseSet42

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1.Provethattwocirclesareorthogonaliffaradiusofonedrawntoapointofintersectionofthetwocirclesistangenttotheother.

2.ProveTheorem42.2.3.ProvethattwocircleswithcentersCandDandradiirandsareorthogonaliffr2+s2=(CD)2.

4.ProveTheorem42.4.5.ShowthatthetwoequationsareequivalentineachdisplayedpairofequationsinthefirsttwoparagraphsoftheproofofCorollary42.7.

6.Showthatapairoforthogonalcircleswillinvertintoapairoforthogonalcircles,oracircleandoneofitsdiameters,ortwoperpendicularlines,andlocatethecenterofinversionforeachofthesecases.

7.Todrawastraightlinewetrace,usingastraightedge,astraightlinepreviouslyconstructedbysomeone.Thecompassisamechanicaldeviceforconstructingcircles;wedonotdrawacirclebytracingsomeone’spreviouslyconstructeddisc.a)Showhowtodrawthe“first”straightline.Thatis,showhowtoconstructastraightlineusingaPeaucelliercell,andassumingnoprevioususeofastraightline,noteventoconstructthePeaucelliercell.

b)MakeaPeaucelliercelloutofcardboardwithoutusingastraightedge,andusethecelltoconstructastraightline.

8.ProveTheorem42.8.9.ProvethatwhenA,B,C,Darefourdistinctcollinearpoints,then(AB,CD)<0iffAandBseparateCandD.

10.ProveTheorem42.10.11.GivenacirclesandapointP,constructtheinverseofPwithrespectto

circles.DosoforPinside,on,andoutsidethecircles.12.ProveTheorem42.12forthecaseinwhichtwoormoreofthepointsA,B,

C,DarecollinearwiththecenterO.13.GiventwopointsAandBlyinginsideacircles,provethatthereisexactly

onelineorcircle(accordingasAandBareorarenotcollinearwiththecenterofs)throughAandBandorthogonaltos.Showhowtoconstructthiscircleorline.

14.ProvethatifaradiusORofonecircleintersectsanothercircleattwopointsPandQsothatOP·OQ=(OR)2,theneachradiusORthatcutstheothercircleattwopointsPandQsatisfiesthatsameequation.

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15.AssumingthattwocirclesintersectattwopointsPandQ,andthateachcircleisorthogonaltoathirdcircles,provethatPandQareinversewithrespecttocircles.

16.Provethatiftwocurvesaretangentatapoint,thentheirinverseswithrespecttoagivencirclearetangentattheinverseofthepoint.Arethereanyexceptions?

SECTION43 INVERSIONANDCOMPLEXGEOMETRY

43.1 The representation of inversions in the Gauss plane is quite convenient.SinceCZ·CZ′=r2foraninversionwithcenterCandpowerr2,wehave

is real andpositive sinceZ,Z′, andC are collinear.This last conditioncanbewrittenintheform

Nowwehave

so

Conversely, assuming that this last equation is satisfied, the preceding onefollows.Sincer2isrealand|z–c|2isreal,then(z′–c)/(z–c)mustalsobereal.Nowtheargumentreversescompletely,toshowthatZ′istheinverseofZinthecirclewithcenterCandradiusr.Wehaveprovedthetheorem:43.2TheoremInversioninacirclewithcenterCandpowerr2hasthecomplexrepresentation

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43.3CorollaryInversioninthecircleofradiusrcenteredattheoriginhasthecomplexrepresentation

43.4TheoremInversionisinvolutoric.Let the inversionofTheorem43.2mapZ toZ′ and thenZ′ toZ″.Thenwe

have

ThusZ′=Z,soinversionisinvolutoric.

43.5TheoremInversionisacirculartransformation.UsingtheinversionofTheorem43.2,weshowthattheimageofthecircleor

linegivenbytheparametricformofTheorem38.3isanothersuchcircleorline.Thusletthelineorcirclehavetheparametricequation,withparametert,

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NowsubstitutethisvalueforzintotheinversionofTheorem43.2toobtain

anequationofthesameformasgiven,henceacircleorline.TheparametricformofTheorem38.3foracircleoraline,

mapsthereallineontoacircleoralineintheGaussplane.Thatis,thisequationrepresentsamapping,soifwelettbeanycomplexnumber,thenitisamappingoftheGaussplanetoitself.ThatitisatransformationoftheGaussplane,andaninterestingtransformation,isthesubjectoftherestofthissection.

43.6DefinitionAbilineartransformationorhomographyisthattransformationofthecomplexplanegivenby

43.7TheoremThe inverseof thebilinear transformationofDefinition43.6 isthebilineartransformation

43.8CorollaryThebilineartransformationofDefinition43.6isinvolutoriciffa=–d.

43.9TheoremThebilineartransformationisacirculartransformation.

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ThebilineartransformationofDefinition43.6maybewrittenas

Thus this transformation is the product of: a) a translation z1 = z + d/c, b) asimilarity(rotation-homothety)centeredat theoriginz2=(c2/(bc–ad))z1,c)areflectioninthex-axisz3= 2,d)aninversionintheunitcirclecenteredattheoriginz4=l/ 3,ande)anothertranslationz′=z4+a/c.Since each of these transformations is circular, then so also is the product acirculartransformation.

43.10 Theorem 43.9 can also be proved by substituting into the form for abilineartransformationtheequationforalineorcircle,aswasdoneintheproofof Theorem 43.5. Although such a proof is slightly shorter, it is far lessinstructiveas to justwhatabilinear transformationconsistsof.Thusabilineartransformation is direct and is a product of similarities and inversions. Inpassing,wenotewithoutproof that thebilinear transformationandthebilineartransformation preceded by a reflection in the x-axis are the most generalcirculartransformations.Theyhavetheequations

withad – bc ≠ 0.Clearly both transformations are circular.A proof that theyrepresent all circular transformations is given inHowardEves’Functionsof aComplexVariable,Vol.1,page39.

43.11Theorem The set of all bilinear transformations forms a transformationgroup.

43.12TheoremAbilineartransformationisdeterminedbythreedistinctpointsandtheirdistinctimages.Iftheimagesofa,b,carea′,b′,c′, thenthebilineartransformationhastheform

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Clearly ifz=a andz′=a′, then thegivendeterminant iszerobyTheorem36.17.Similarlyforz=borforz=c.ByDefinition36.12thedeterminanthastheexpansion

wherep,q,r,andsarethird-orderdeterminantsnotinvolvingzorz′Solvingforz′,weobtain

abilineartransformationwhenever–rq+ps≠0.If–rq+ps=0andp≠0,thenrq/p=s,andwehave

a constant. This is impossible, since the determinant equation is satisfied bythreedifferentvaluesforz′.

Finallyassumethatp=0andps–rq=0.Thenwemustalsohaveeitherr=0orq=0.Ifr=0,thenz′=–s/q,aconstant,againimpossible.Ifq=0,thenwehave

Thislaststatementimpliesthata=bora=corb′=c′contrarytohypothesis.It follows, then, that ps – rq ≠ 0, so the given determinant does indeed

representabilineartransformation.

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ExerciseSet43

1.Provethattheproductoftwoinversionswiththesamecenterisahomothetyinthatcenter.Finditsratio.

2.ProveCorollary43.3.3.Findwhenaproductoftwoinversionswithdistinctcentersiscommutative.4.ShowthattheinversionofTheorem43.2canbefactoredintotheproductof:(a)thetranslationz1=z–c,(b)theinversionattheoriginz2=r2/ 1,and(c)anothertranslationz′=z2+c.

5.Findanequationforthebilineartransformationthatmaps0,1,and∞toa)0,1,∞

b)0,1,ic)0,1+i,3–id)2,1+i,3–ie)1+i,1+3i,2–if)a,b,c6.ProveTheorem43.7.7.ProveCorollary43.8.8.Provethatahomographyisindeedatransformation.9.ProveTheorem43.11.10.Findthebilineartransformationthatmaps0,1,anditoeachofthesetsof

pointslistedinExercise43.5.11.Findtheimagesofeachofthefourpoints0,1,i,and∞underthebilinear

transformationa)

b)

c)

d)

12.Showthatanybilineartransformationthatleavestherealaxisinvariantcanbewrittenwithrealcoefficients.

13.a)Findabilineartransformationthatcarriesthecircle|z|=1intothecircle

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|z|=2.Isitunique?b)Findabilineartransformationthatsatisfiespart(a)andalsomapsito–2.Isitunique?

SECTION44 APPLICATIONSOFINVERSION

The usefulness of inversion was alluded to in the introductory paragraph ofSection41.Herewe shall demonstrate thatusefulness.Observe the transform-solve-transform method (see 18.1) in action: A problem is stated, thentransformedintoanewproblembyaninversion;thenewproblemissolvedbyonemeansoranother;andthesolutionistransformedbacktosolvetheoriginalproblem. Theorem 44.2 is one of several theorems whose proof is a clearexampleofthismethod:Aprobleminvolvingcirclesandlinesisinvertedintoaproblem involving a triangle and its altitudes. This latter problem has animmediatesolutionfromourearlierwork,andthissolvesthegivenproblem.

44.1TheoremAnythreecirclescanbeinvertedintothreecircleswhosecentersarecollinear.

SupposethatthecentersA,B,Cofthecircless,t,uarenotcollinear.Then(see Exercise 44.2) there is a circle v orthogonal to all three circles (see Fig.44.1).ChooseascenterofinversionanypointPoncirclevbutnotonanyoftheothercircles.Thenvinvertsintoalinev′andthecircless,t,uinvertintocircless′,t′u′.Sincevisorthogonaltoeachofs,t,andu,thenv′isadiameterofeachofthecircless′,t′,andu′.Nowthecentersofthesecirclesarecollinearonlinev′.

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Figure44.1

44.2TheoremLettwocirclessandtmeetatOandP,andleteachdiameterOSandOTofthetwocirclescuttheothercircleatAandB.ThenchordOPpassesthroughthecenterofcircleOAB(seeFig.44.2a).

Figure44.2a

%

Figure44.2b

InvertthefigureinpointO.LinesOP,OBT,andOAS invertintolinesOP′,OB′T′, andOA′S′, as inFig.44.2b.CirclesOAPT,OBPS, andOAB invert intolinesA′P′T′,B′P′S′,andA′B′.LetOP′andA′B′meetatX.SinceOSandOTarediameters, then linesOS′ andOT′ are perpendicular to linesB′S′P′ andA′T′P′.HenceOistheorthocenteroftriangleA′B′P′.NowlineP′OXisperpendiculartolineA′B′.Hence, in theoriginalfigure, lineOP isorthogonal tocircleOABby

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Theorem41.10.Thatis,lineOPpassesthroughthecenterofcircleOAB.

44.3TheoremPtolemy’sTheorem.Inaconvexcyclicquadrilateraltheproductofthediagonalsisequaltothesumofthetwoproductsofoppositesides.

LetABCD be the cyclic quadrilateral and invert in pointA (seeFig. 44.3).TheinversesB′,C′,D′ofB,C,Dlieonaline,byCorollary41.13,so

Lettingthepowerofinversionber2,byExercise41.17wehave

fromwhichitfollowsthat

Figure44.3

44.4LemmaIfapenciloffourlinesiscutbytwotransversalsatpointsA,B,C,DandA′,B′C′,D′respectively,andneithertransversalpassesthroughthevertexofthepencil,then

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SeeTheorem3.8.

44.5LemmaThetangentatA′totheninepointcircleoftriangleABCisparalleltothatcommoninternaltangenttotheequicirclescenteredatIandIathatdoesnotpassthroughA′.

Ofcourse,sideBCisonecommoninternaltangenttotheequicirclescenteredatIandIa.LettheothercommoninternaltangentbePQ,asshowninFig.44.5.ThenBC andPQ meet atU (which is collinear withA, I, and Ia). Since theninepoint circle is the circumcircle of triangleA′B′C′, which is homothetic totriangleABC, then the tangent tatA′ isparallel to theoppositesideDEof theorthic triangleof triangleABC,byTheorem7.9.ByCorollary7.10,DEmakesthe same angle with AB that AC does with BC, namely C. But by thesymmetry of Fig.AQCPB inmirrorAU, PQmakes this same anglewithAB.HenceDEisparalleltoPQ,sothetangenttisparalleltoPQalso.

Figure44.5

44.6TheoremFeuerbach’stheorem.Theninepointcircleofatriangleistangenttoeachoftheequicirclesofthetriangle.

Weshowthattheninepointcircleistangenttotheincircle,andtotheexcirclewhose center is Ia. We use Fig. 44.5, and the terminology of Lemma 44.5,addingtothatfigurethelinesshownbroken.ThusletXandXabethepointsof

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contactoftheequicircleswithsideBC,anddrawthealtitudeAD.Now(IIa,AU)=–1byExercise24.8,so(XXa,DU)=–lbyLemma44.4.TakeA′ascenterofinversion,andA′X( A′XabyTheorem6.22)asradiusofinversion.ByExercise41.15circlesIandIa,theequicircles,invertintothemselves.ByTheorem42.5,the inverseofD isU.LineBC is,ofcourse, self-inverse.Theninepointcircleinverts into a line throughU by Corollary 41.13. Since angles are preserved(Theorem41.10),theangleθmadewithlineBCbythetangenttotheninepointcircle at D is equal and opposite to the angle made by the inverse of theninepoint circlewith lineBC.By symmetry, this is the sameanglemadewithlineBCby the tangent to theninepointcircleatA′.ByLemma44.5, then, theinverseoftheninepointcircleislinePQ.SincePQistangentto(theinversesof)both equicircles, then (Theorem 41.10) the ninepoint circle is tangent to bothequicircles.Thetheoremfollows.

44.7 The reader is here reminded of the statements following Theorem 7.20,pointingouta totalof32equicirclesassociatedwithagiven triangleall32ofwhicharetangenttotheninepointcircle.

44.8 Theorem A homothety with positive ratio can be factored into twoinversions.

Weprovethetheorem,withoutlossofgenerality,forthehomothetyH(O,k)centered at the origin andwith ratio k > 0. Letα andβ denote the inversionsgivenby

Thenβαisgivenby

anequationforthedesiredhomothety.

44.9 Theorem Two inversionswith positive ratios and in the same centercommuteifftheirradiiareequal.

Ifthecirclesofinversionhaveradiirands,wherewearetakingtheoriginastheir commoncenter, then, lettingα andβ represent these transformations,wehave

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Nowtheproductsβαandαβhavetheequations

Thusthesetwomapsarehomothetieswithratioss2/r2andr2/s2.Theyareequaliffr=s.

44.10 Theorem Two inversions with positive ratios and distinct centerscommuteifftheircirclesareorthogonal.

LetthecirclesfortheinversionsαandβhavecentersOandAandradiirands.Theyareorthogonal,then,iffr2+s2=|a|2=a .Equationsforαandβare

sothatαβandβαaregivenby

and

Thesemapswillbeequalifftheirdifferenceiszero.Takingtheirdifferenceandsettingz=0therein,weobtain

Ifthisdifferenceiszero,thena –s2=r2.Conversely,ifa =r2+s2,thenwefindthat

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Itnowfollowsthatβα=αβiffthecirclesofinversionareorthogonal.

ExerciseSet44

1.Explainwhy,inTheorem44.1,thecirclevcouldnotbetakenasthecirclethroughthecentersofthethreegivencircles.

2.a)Provethatiftwocirclesintersect,thenthecenterofanycircleorthogonaltobothofthemmustlieonthelineoftheircommonchord.b)Assumingthattwocirclesintersect,provethatanycircleorthogonaltooneofthemandwhosecenterliesontheircommonchordisorthogonaltotheotheralso.

c)Showthatbothparts(a)and(b)remaintruewhen“intersecting”and“commonchord”arereplacedby“tangent”and“commontangentattheirpointoftangency.”

d)Provethatthetangentsdrawntotwointersecting(ortangent)circlesfromanypointonthecommonchord(orcommontangent)arecongruent.Thelocusofallthepointsfromwhichcongruenttangentscanbedrawntotwononconcentriccirclesiscalledtheirradicalaxis.Hencetheradicalaxisoftwointersectingcirclesistheircommonchord,andfortwotangentcirclesistheircommontangentdrawnattheirpointoftangency.

e)Giventhattwocirclessandtdonotintersectandarenotconcentric,thenshowthatitisalwayspossibletodrawacircleuintersectingbothcirclessandtinsuchawaythatthechordcommontouandintersectsthechordcommontouandtinanordinarypointP.

f)ShowthatitispossibletoconstructinfinitelymanydistinctpointsPinpart(e)bychoosingdifferentcirclesu.

g)ProvethatthelocusofallpointsPofpart(e)istheradicalaxisofcirclessandt.

h)Provethattheradicalaxisofpart(g)isastraightline.

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i)Provethattheradicalaxisisthelocusofthecentersofallcirclesorthogonaltotwocircles.

j)Provethatifthreecircleshavenoncollinearcenters,thenthethreeradicalaxesofpairsofthesecirclesconcur.Thispointiscalledtheirradicalcenter.

k)Showthattheradicalcenterofthreecircleswithnoncollinearcentersisthecenteroftheuniquecircleorthogonaltoeachofthethreegivencircles.

3.Showthatanythreecircleswithnoncollinearcenterscanbeinvertedintothemselves.

4.ExtendPtolemy’stheorem(Theorem44.3)tothecaseinwhichthequadrilateralisnotcyclicbyprovingthegeneralformula

5.LetAandBbeinversepointsincircles.InvertinacirclewhosecenterdoesnotlieonsoratAorBtoobtainA′,B′s′.ThenprovethatA′andB′areinversewithrespecttocircles′.

6.InExercise44.5,whathappenswhenthecenterofinversionliesoncircles?WhatistherelationofA′andB′tos′then?Thisexplainswhyreflectioninalineissometimescalledinversioninaline,andinversionissometimescalledreflectioninacircle.

7.Findtheradiusofthecircleinversetoagivencirclewithrespecttoagiveninversion.

8.Showthatanytwononconcentricnoncongruentcirclescanbeinvertedonetoanother.

9.ShowthattheinversionwithcenterOandradiusrhastheCartesianequations

10.UsingExercise44.9,writeanequationfortheinverseinthecircleofradius1centeredattheoriginofeachlistedequation:a)Thelinex=lb)Thelineax+by+c=0c)Thecirclex2+2x+y2=0d)Thecirclex2+y2+ax+by+c=0

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e)Theellipseb2x2+a2y2=a2b2istheinversecurvealsoanellipse?Sketchitsgraph.

f)Theparabolay=x2;sketchthegraph.g)Thepairofhyperbolasx2y2=1;sketchthegraph.

11.Circlestanduareinversewithrespecttocircles.Provethattanduaremappedintocongruentcirclesbyinversionwithrespecttoanypointoncircles.

12.LetA,B,C,Dbefourconcyclicpoints.Showhowtoinvertthemintotheverticesofarectangle.

13.Invertthetheorem“anangleinscribedinasemicircleisarightangle”:a)inthecenterofthesemicircle,b)inoneendofthediameterofthesemicircle,c)inthevertexoftherightangle.

14.Whatistheimageofthelineofcentersoftwogivencirclesunderaninversion?

15.ProvePappus’ancienttheorem:Inthefigureforthisexercise,OA,OB,andABarediametersofcirclest,u,ands0.ThencircleS2isplacedtangenttocirclest,u,ands0,circles2istangenttocirclest,u,ands1…,circlesnistangenttocirclest,u,andsn–1.ProvethattheheightofthecenterofcirclesnabovelineOBisntimesthediameterofsn.

Exercise44.15

16.Giventhatthreecirclesconcuratapoint,provethattherearefourdistinctcircleseachofwhichistangenttoallthreecircles.

17.WhenpointsAandBareinverseincircleshowthattheinverseofBisthecenteroftheinverseofswhenAisthecenterofinversion.

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6 ISOMETRIESINTHEPLANE

SECTION45 WHATNEXT?

45.1Thetitleofthissectionrefersnottopossiblefuturetrendsinthetheoreticaldevelopmentofgeometry,butrathertothedriftofgeometryinthehighschools.Evenmoreimportant,perhaps,thequestionWhatnext?maybetakenpersonally,especiallybythoseintendingtoteachgeometry.45.2Overthelastquartercentury,analyticgeometryhasgainedever-increasingprominence in high school geometry textbooks, while solid geometry hasdecreased in popularity. These trends are partially due to a lack of emphasisplaced on geometry in colleges and normal schools until quite recently. It iseasier and requires less skill to prove a theoremby analytic geometry thanbysynthetic methods. And very few teachers feel comfortable beyond the mostelementaryworkinspace.45.3Thesetrendswillcontinue,foranalyticgeometryprovidesapowerfultoolin solid geometry, one that should not be overlooked. Its inclusion in thetraditionalsophomoregeometrycourseistobeencouraged.Studentsshouldbeallowed, even encouraged, to use whatever tools are available to solve theirproblems.Thisplacesaheavierburdenontheteacher:Hemustbequalifiedtoevaluateastudent’sworknomatterwhichtoolsheuses.45.4Itiswithasighofregretthatthepassingofasemesterofsolidgeometryinthesenioryearisobserved,notasthecoursegenerallyhasbeentaught,butasitshouldhavebeentaught.Theapparentreasonfordiscontinuingthiscoursewasthat spatial perception would be taught in the sophomore geometry coursesimultaneously with plane geometry. It rarely is. But the real reason was thegreat pressure from teachers who did not have the training to teach solidgeometry.Very fewhighschoolgeometry teachershavesignificant training ingeometry beyond their own sophomore plane geometry courses. This meagertraining is not enough. At least one significant college-level geometry courseshould be the minimum requirement for teaching geometry. More than thisminimumisstronglyrecommended.

The effects of inadequate preparation of geometry teachers are felt at thecollegelevel.Youmayrecall thatstudentsof thecalculushavegreatdifficultyworkinginspace.Manycannotvisualizesurfacesandsolidsatall,andsketchingsolidsandtheirplanesectionsisadifficulttaskforallbutaveryfewstudents.Nonetheless, it isdoubtful that a course in solidgeometrywill regain favor inhighschoolsoreven incolleges. It is recommended thataunit (ofat least sixweeks’ duration) on spatial perception and sketching graphs of solids be

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includedinthehighschoolprogramafter(orduring)sophomoregeometry.Andit is here that analytic geometry proves to be especially useful—in solidgeometry.45.5 Isometries and similarities are mentioned briefly in current high schoolgeometry texts.And texts on this subject at the college level are appearing ingreat profusion, this book being one such.As observed throughout thiswork,these mappings provide another tool of the order of analytic geometry withwhich to attack problems in geometry. Hence isometries and similarities willgain increasing favor in the high school, and will begin to appear early intextbooks in order that these transformations may be used meaningfullythroughoutthestudent’sgeometrycourse.45.6Theunderstandingofisometriesclarifiesandrigorizesthehazymethodofsuperposition thatwas formerly used as a congruence axiom (seeAxiom 4 inAppendixA). Itwasneverclear justhowonewas topickupone triangleandplaceitonanother.Shouldyouthinkofthetriangleasapieceofcellophanetapetobepeeledofftheplane?Ifso,thenitcouldbeplacedonthelateralsurfaceofacylinder.Ontheotherhand,itcannotjustbeslid(translatedandrotated)ifthecongruence is opposite. The problem is, “How do you accurately define justwhatarigidmotionis?”Oneansweris,“Byisometries.”CurrentgeometrytextspostulatetheSAScongruenceconditioninordertoavoidthenecessityofstatingconsiderable background material in isometries before being able to discusscongruence.Itseemsbetterpedagogicallytobeabletoprovebasiccongruencetheoremsearly,andtheSASpostulateaccomplishesthisresult.Thisisnottosaythatisometriesshouldbedismissed;theyshouldcomeearlyinthecourse.But,as in this text, perhaps isometries should be treated after a preliminaryfamiliaritywithbasicgeometricconceptshasbeendeveloped.45.7NowletusturntothepersonalaspectofthequestionWhatnext?Whatareyouplanningtodotomakegeometrymeaningfultoyourstudents?

Insomewaysthosewhoteachgeometryaremorefortunatethanteachersofalgebra. The student who never really understood arithmetic tends to haveinsurmountableproblemswithalgebrauntilhisdeficienciesareuncoveredandcorrected.But ifwecangivehima rightattitude towardgeometry,his earlierdifficulties will have a much less detrimental effect upon his progress ingeometry.

Theexperienced teacher iswellawareof theaverageorbrightstudentwhoscores high on mathematical ability tests, but is low on mathematicalachievementtests.Asteachersofgeometry,wehavearatheruniqueopportunitytobreak through thepsychologicalbarrierof sucha student,help todispelhisdreadofmathematics,andshowhimthathecan findgeometryinterestingand

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withinhisgrasp.Theobservant teacherwill use thisopportunity to locate andhelpcorrecthisdeficienciesinarithmetic.

Butnotbysimplylistingtheorem,proof,theorem,proof,theorem,proof,etc.,allcopiedneatlyrightoutofthebook.45.8A student ismost ready for an answer to a questionwhen he feels therereally is aquestion and hewants to know its answer. If a student looks at anisoscelestriangleandimmediatelyknowsthatitsbaseanglesarecongruent,thenaproofofthatfactmaybesimplyredundantandmeaninglesstohim.45.9Inbeginningadiscussionofisoscelestriangles,forexample,afterdefiningtheterm,onemightaskwhatelseistrueaboutisoscelestriangles,otherthanthattwo sides are congruent. When a student answers that the base angles arecongruent,oneshouldask,“Always?”and“Howdoyouknowthis?”Also,“Isitnecessarytoproveit?Ifso,why?”

Suchquestionsand the resultingdiscussionscause the students to think forthemselves, and to develop the curiosity necessary to appreciate deductivemathematics.Of course, indiscussing isosceles triangles, propertiesother thanthecongruenceofthebaseanglesshouldenternaturallyintothediscussion,andshouldbetreatedastheyoccur.45.10 In studying right triangles, tell the students to draw right triangles (forhomework,perhaps)havinglegsoflengths3and4,8and15,10and24,12and35,and16and63,forexample.Thestudentsshouldformatableindicatingtheleg lengths and the corresponding hypotenuse lengths which they carefullymeasure.Theyshouldlookforgeneralizationsfromthedatainthetable.Havethemformanothercolumnforthesumofthesquaresoftheleglengthsforeachtriangle,andanothercolumnforthesquareofthehypotenuselength.Askthemtolookforapparenttruths.Ontheirdrawingstheyshouldlocatethemidpointofeach hypotenuse andmeasure and record its distances from the three vertices.Ask what appears to be true. Ask if they can prove it. This is the discoverymethod.45.11Thediscoverymethodrequiresmoretimethanlecturing,andinthatsenseitmightbecalledinefficient.Butcertainlytheinitialpartsofageometrycourseshould be especially well motivated. The student should learn to hunger andthirst after geometric knowledge.Later, hewill bemore receptive to a formalchainoftheoremswhentimerequiresthatmaterialbecoveredmorerapidly.Butclasstimespentworkingoutoriginalproblemstogetheristimewiselyspent.Itiseasytosay,“Hereisthesolution.”Itisfarmoreinstructivetosay,“Howcanweobtainasolution?Letusworkitouttogether.”

Inshort,motivate.45.12 Should this course end your geometric studies ? No. Your study of

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geometry has just begun. Now you are starting to do geometry, and yourgeometriceducationshouldbecontinued,bothformallyandinformally.Anditshould be reasonably continuous. An occasional course in some aspect ofgeometry will keep you active and help prevent staleness, which can easilydevelop, especially in a small school using the same textbook year after year.Specifically, youmaywish to study such topics as projective geometry, non-Euclidean geometries, analytic geometry, foundations of geometry, or others.Thetopicsingeometrythatareopenedtoexplorationbystudentscompletingabasictextsuchasthisonearelegion.Workingproblemsregularlyfromabooksuch as this one or Horblit and Nielsen’s Plane Geometry Problems withSolutions(CollegeOutlineSeries),orRich’sPrinciplesandProblemsofPlaneGeometrywithCoordinateGeometry (Schaum’sOutlineSeries) also improvesyour ability to work original problems. More important, such practice willimproveyour ability to judgea student’swork. Ishisoriginalproofvalid ? Ifnot,wheredidhegoastray?Justbecausehisisnottheproof“inthetext”doesnotmeanthatitisincorrect.

Subscribe to and read journals in your field, such as The MathematicsTeacher, The Arithmetic Teacher, Mathematics Magazine, The Two-YearCollegeMathematics Journal,etc. From time to timeyouwill find interestingitems in the literature to bring up in class, or to give to bright students asenrichment material. You will have a better knowledge of what is new inmathematicalthinkingandwillbebetterabletoevaluatenewideas.45.13Byyourloveformathematicsandyourinterestinyourstudents,letyourclassbeoneinwhichthissign(Fig.45.13)isproudlydisplayed.ExerciseSet451.Drawtherighttrianglessuggestedin45.10,andnotethateachhypotenuseis2unitsgreaterinlengththanoneofthelegs.IsthistrueforallPythagoreantriangles?

2.Drawrighttriangleswithleglengths5and12,7and24,20and21,and7and10.Measurethelengthofeachhypotenuse.Ifthelegsareofintegrallength,willthehypotenusealsobeofintegrallengthalways?

3.Showthatwhenpandqareintegerswithp<q,thena=2pq,b=p2–q2,andc=p2+q2formaPythagoreantriple;thatis,a2+b2=c2anda,b,andcarepositiveintegers.

4.DecideiftheformulasofExercise45.3giveallPythagoreantriples.SeeDodge’sNumbersandMathematics,pages243–245.

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Figure45.13SECTION46 INTRODUCTIONTOTHREEDIMENSIONS

46.1InChapter2wesawthatthereflectioninalineisthebasicisometryoftheplane;allotherisometriescanbewrittenasproductsofthisbasicisometry.Thesituationinspaceisquitesimilar.Herethebasicisometryisthereflectioninaplane(seeFig.46.1a)whichisdefinedinjustthemannerthatonewouldexpect(seeDefinition47.4).Thenatranslation(Definition48.1)istheproductoftworeflections in parallel planes, and a rotation through angle θ about a line m(Definition48.8)istheproductofreflectionsintwomirrorsthroughlinem,andintersectingatangleθ/2.Figure46.1bshowsthetranslation,andP'ig.46.1cthe

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rotation, asproductsof reflections inplanesΠ and∆. In eachcase,pointP ismappedtopointP′.Comparethesefigureswiththeirplanecounterparts,showninFigs.10.6and10.8.Arotationaboutlinemthrough180°iscalledahalfturnaboutlinem,andistheproductofreflectionsinanytwoperpendicularmirrorsthroughm,asshowninFig.46.1d,whereagainpointPismappedtopointP′byreflectionsinplanesΠand∆.

Figure46.1a

%

Figure46.1b

Figure46.1c

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Figure46.1d46.2Onemorebasicisometryoccursinspace.AcentralinversioninapointOisthatinvolutoricisometrythatmapseachpointPinspacetopointP′sothatOis the midpoint of segment PP′ It can be factored into the product of threereflections in any threemutually perpendicular planes throughO. Figure 46.2shows a central inversion in pointO,mapping pointP toP′ by reflections inplanes Π, ∆, and Γ. Reflections and central inversions are opposite spaceisometries, reversing the sense of a tetrahedron (see Definition 47.10);translationsandrotationsaredirect.

Figure46.246.3Any isometry in space iscompletelydeterminedby the fourverticesofatetrahedron and their images, so each such isometry is the product of atmostfour reflections in planes, three when there is at least one fixed point (seeTheorem47.9andcompareitwithTheorem13.13fortheplanecase).Variouscombinationsofproductsofthreeorfourreflectionsyieldthreemorenewspaceisometrieswhichwemight liken to the glide-reflection in the plane.A screw

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displacement(Definition49.1)istheproductofarotationaboutalinemandatranslationalonglinem(seeFig.49.1).Aglide-reflection(Definition49.6)istheproductofareflectioninamirrorΠandatranslationalonganyvectorlyinginplaneΠ (see Fig. 49.6). It factors into a product of three reflections, two inparallelplanesandthethirdinaplaneperpendiculartothefirsttwo.Finally,arotatory reflection (Definition49.7) is theproductofa reflection inaplaneΠandarotationaboutalinemperpendiculartoΠ(seeFig.49.7).Itisalsocalledarotatoryinversion(seeTheorem49.9),sinceitcanbefactoredintoaproductofacentralinversioninapointOandarotationaboutalinemthroughO.Thescrewdisplacement(Whyisitcalledthis?)isdirect;theotherisometriesdefinedinthisparagraphareopposite.46.4Weshallshowinthischapterthateachoppositeisometrywithaninvariantpointisarotatoryinversion,ofwhichareflectionisaspecialcase.Anoppositeisometrywithnofixedpointsisaglide-reflection.Eachoppositeisometryistheproductofareflectionandahalfturn.46.5Eachdirectisometryistheproductoftwohalfturns(seeTheorem49.2).Ifithasafixedpoint,thedirectisometryisarotation.Ifithasnofixedpoints,thenitisatranslationorascrewdisplacement,theformerbeingaspecialcaseofthelatter.ExerciseSet461.Showthataproductoftworeflectionsinparallelplanesisatranslationthroughtwicethenormalvectorfromthefirstplanetothesecond.

2.Showthataproductoftworeflectionsinintersectingplanesisarotationthroughtwicetheanglefromthefirstplanetothesecond.

3.Showthatreflectionsintwoperpendicularplanescommute.4.Therearesixdifferentordersinwhichtowriteaproductofreflectionsinthreeplanesr,∆,andΠ,andtwowaystoassociateeachsuchorder.Showthatwhenthethreeplanesaremutuallyperpendicular,thenalltwelvesuchproductsareequaltothesamecentralinversion.

5.Findaspecificisometryinspacethatcanbewrittenas(aproductof)nolessthantheindicatednumberofreflections,andwriteitassuchaproduct:a)1b)2c)3d)4

6.Statethesmallestnumberofreflectionsinplanesnecessarytowriteeachindicatedspaceisometryasa(productof)suchreflection(s).a)reflectionb)theidentitymapιc)translationd)centralinversione)rotationf)screwdisplacementg)

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halfturnh)rotatoryreflectioni)glide-reflectionj)rotatoryinversion7.StatewhichisometriesinExercise46.6aredirectandwhichareopposite.

8.Fashionasuitabledefinitionfordirectandoppositespacetransformations.9.Statewhatmustbetrueoftheaxesoftwohalfturnsinorderthattheirproductrepresentsthefollowing:a)arotationb)atranslationc)ascrewdisplacementd)theidentitymap10.Statewhichpoints,lines,andplanesarefixedundereachoftheisometrieslistedinExercise46.6.

11.Showthataspaceisometrypreserves:a)linesegmentsa)linesc)planesd)anglesbetweenlinese)anglesbetweenplanes12.Showthateach.isometrycanbewrittenasaproductofatmostfourreflectionsinplanes.

13.Showthataproductofthreehalfturnsaboutparallellinesisahalfturnaboutalineparalleltoeachofthethreelines.Furthermore,showthatwhenanythreeofthesefourlinesarecoplanar,thenallfourarecoplanar.

14.Showthatarotatoryreflectionandarotatoryinversionareindeedequivalentspaceisometries.

SECTION47 ReflectioninaPlane

47.1 The entire theory of transformations given in Section 12 holds with theword"plane"replacedby“space.”Suchreplacementisassumedforthischapter.The student shouldnow rereadSection12 toverify that all its definitions andtheorems do indeed hold in space. Similarly, the definition of isometry isunaltered.Infact,theentirespacedevelopmentparallelsquitecloselythatgivenfortheplaneinChapter2.47.2 Definition An isometry in space is a map of the points in space tothemselvesthatpreservesdistance.47.3Theorem IfPQRS is a tetrahedron, andα andβ are isometries such thatα(P)=β(P),α(Q)=β(Q),α(R)=β(R)andα(S)=β(S),thenα=β.

Weneed showonly that any fifth point in space is uniquely located by itsdistancesfromP,Q,R,andS.Tothatend,letp,q,r,sdenotethedistancesofapointXfrompointsP,Q,R,S.ThenthereisasphereofpointsYatdistancepunitsfromP.SimilarlythereisasphereofpointsYatdistanceqfromQ.Thesetwospheresintersectin(atmost)acirclecofpoints.SinceRdoesnotlieonlinePQ, thesphereofpointsrunitsfromRcutscirclec in(atmost)twopointsY1andY2,whichareequidistantfromplanePQR(seeFig.47.3).SinceSdoesnotlieinplanePQR, thenonlyoneofitsdistancestoY1andY2canbeequaltos.ThuspointXisuniquelydeterminedbyitsdistancesfromthefourverticesofthetetrahedron.Thetheoremnowfollows.

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Figure47.347.4DefinitionAspacetransformationisareflectioninaplaneΓ,denotedbyσΓ, iff whenever B = σΓ(A) for points A and B, then Γ is the perpendicularbiscctorofsegmentAB,orA=BandA∈Γ.47.5Rememberthatwehaveagreedtouseupper-caseGreeklettersΓ,∆,Π,…todenoteplanes.47.6TheoremAreflectioninaplaneisanisometry.

LetσΓmapAtoA′andBtoB′.Thenthereisaplane∆,uniqueifA,A′,B,B′andnotcollinear,perpendiculartoplaneΓandcontainingA,A′,B,B′(seeFig.47.6).Inplane∆,A′andB′are(byDefinition13.9)thereflectionsofAandBinthe(plane)reflectioninlinem,thelineofintersectionofthetwoplanesΓand∆.ThetheoremthenfollowsfromTheorem13.10,theplaneanalogofthistheorem.

Figure47.6

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47.7CorollaryEachproductofreflectionsisanisometry.47.8TheoremForeachreflectionσΓ,σΓ–1=σΓ.47.9TheoremEachisometryinspaceistheproductofatmostfourreflectionsinplanes.

LettheisometryαmaptetrahedronABCDintothecongruenttetrahedronA′B′C′D′.Weconsider5cases.Case1.IfA=A′,B=B′,C=C′andD=D′,thenαistheidentitymap,whichcanbewrittenasthesquareofareflectioninanygivenplane.Case2.IfA=A′,B=B′,C=C′,butD≠D′,thenplaneABCistheperpendicularbisectorofsegmentDD′ (seeExercise47.11),soa isa reflection in thatplane(seeFig.47.9a)

Figure47.9a

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Figure47.9b

Case3.IfA=A′andB=B′,butC≠C′thenAandBbothlieontheplaneΠ,whichistheperpendicularbisectorofsegmentCC′.AreflectioninplaneΠthenreducesthiscasetoeitherCase1orCase2(seeFig.47.9b).Case4.IfA=A′butB≠B′,thenliesontheperpendicularbisectorplaneΠofsegmentBB′ so a reflection in that plane reduces this case to oneof the threepreviouscases(seeFig.47.9c).

Figure47.9c

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Figure47.9d

Case5.GiventhatA≠A′reflecttetrahedronABCDintheperpendicularbisectorplaneofsegmentAA′,reducingCase5tooneofthefourearliercases(seeFig.47.9d).

Ofcourse,ifanyoneofCases2to5reducestoCase1,thenafactoroftheidentitymap(aproductoftworeflectionsinthesameplane)issimplyomitted.Thusweseethateachisometrycanbewrittenasareflectioninaplaneorasaproductofatmostfoursuchreflections.

47.10DefinitionThesenseofatetrahedronABCDcanbedefinedasrightorleftaccording as a right-hand screwmoves toward or away from vertexD whenpiercingtheplaneABCalongthealtitudetoD,andwhenrotatedinthedirectionA–B–C–A (see Fig. 47.10). An isometry that maps tetrahedron ABCD totetrahedron A′ B′ C′ D′ is called direct or opposite according as the twocongruenttetrahedronshavethesameordifferentsenses.

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Figure47.10

47.11We are not as concerned herewithmemorizing specific details of rightand left senses for a tetrahedron as we are with understanding direct andopposite isometries. Of course, it is quite useful to be able to apply thisdefinition of sense to a specific situation in order to ascertain whether theisometryisdirectoropposite.

47.12TheoremAreflectioninaplane,oraproductofanoddnumberofsuchreflections,isanoppositeisometry;aproductofanevennumberofreflectionsisadirectisometry.

47.13TheoremThereareexactly twoisometries,onedirectandoneopposite,thatcarryagiventriangleABCintoacongruenttriangleA′B′C′.

ThemethodofTheorem47.9showsthatatmostthreereflectionsareneededtomaptriangleABCtotriangleA′B′C′.Letanysuchisometrybedenotedbyα.IfΠdenotesplaneA′B′C′,thenσΠalsoisanisometrycarryingtriangleABCtoA′B′C′.AndofαandσΠ,oneisdirectandtheotherisopposite.

Given any fourth pointD in space but not in planeΠ, there are only twotetrahedrons A′ B′ C′ D′ and A′ B′ C′ D″ on triangle A′ B′ C′ congruent totetrahedron ABCD (see Fig. 47.13). Hence there can be no more than twoisometriestakingtriangleABCtotriangleA′B′C.

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Figure47.13

ExerciseSet47

1.RereadthetransformationtheoryinSection12intermsofspacetransformations.

2.ProveCorollary47.7.

3.ProveTheorem47.8.

4.Makeacardboardmodelofaright-sensedtetrahedron.Cantheverticesberelabeledtochangeitssense?Explain.

5.ProveTheorem47.12.

6.FindthetwoisometriesthatcarrytriangleABCtoA′B′C′whenthecoordinatesofthepointsareA(0,0,0),B(1,0,0),C(0,2,0),A′(0,0,3),B′(1,0,3),C′(0,2,3).

7.Provethatareflectioninaplaneisinvolutoric.

8.Provethateachisometryiseitherdirectoropposite,butnotboth.

9.HowmanyisometriescanbefoundthatmapagivensegmentABtoacongruentsegmentA′B′

10.WritethecoordinatesfortheverticesofatetrahedronA′B′C′D′,intowhichtetrahedronABCD,whereA(0,0,0),B(1,0,0),C(0,1,0),andD(0,0,1),mapsunderanisometrythatismadeupofexactly:a)onereflectionb)two

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reflectionsc)threereflectionsd)fourreflections11.GiventhatABCisatriangleandAD≅AD′BD≅BD′andCD≅CD′fortwodistinctpointsDandD′provethatplaneABCistheperpendicularbisectorofsegmentDD′

12.WhenPABisatrianglewithPA≅PB,provethatPliesontheplanethatistheperpendicularbisectorofsegmentAB.

SECTION48 BASICSPACEISOMETRIES

48.1DefinitionAspacemapαiscalledatranslationthroughvector ifffor

eachpointA,α(A)=Bwhere = .

48.2TheoremEachproductof reflections in twoparallelplanesΠ and≠ is atranslationthroughtwicethenormalvectorfromplaneΠtoplane≠.Conversely,each translation can be factored into such a product of reflections in parallelplanes,bothofwhichareperpendicular to thevectorof translationandhalf itslengthapart.Henceeachtranslationisanisometryandisdirect(seeFig.46.1b).

48.3TheoremTheinverseofatranslationthroughvector isthetranslation

throughvector .Ifthetranslationαisgivenbyα=σΠσ∆,thenα–1=σ∆σΠ.

48.4TheoremTranslationscommute.

48.5TheoremTheproductoftwotranslationsisatranslation.

48.6Thustranslationsbehaveexactlythesameinspaceastheydointheplane.The proofs of these first theorems in this section are quite analogous to thosegiven for the corresponding plane theorems (see Section 14), so they are notrepeatedhere.

48.7A rotationabouta line inspace isalsoquite similar toa rotationaboutapoint in the plane, as attested by items 48.8 through 48.13. There are somedifferences,however,inthatproductsofrotationsaboutskewlinesdonotyieldeitherrotationsortranslations,asnotedinTheorem49.3.

48.8DefinitionLetAbeanypointandmanylineinspace,letΠbetheplanecontainingA and perpendicular tom, and letm andΠ intersect in pointO.Aspacemapαiscalledarotationaboutlinemthroughangleθiffforeachsuch

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pointA,αisthepalnerotationofplaneΠthroughangleθaboutpointO.Linemiscalledtheaxisoftherotation(seeFig.48.8).

Figure48.8

48.9TheoremEachproductofreflectionsintwoplanesΠand≠intersectingina linematanangleθ isa rotationofangle2θabout linem.Conversely,eachrotationcanbefactoredintosuchaproductofreflectionsinplanesintersectingon itsaxisathalf theangleof the rotation.Henceeachrotation isan isometryandisdirect(seeFig.46.1c).

48.10Theorem The inverse of a rotation about linem through angleθ is therotationaboutlinem throughangle–θ.If therotationα isgivenbyα=σΠσ∆,thenα–1=σ∆σΠ.

48.11TheoremTworotationsaboutthesameaxiscommute.

48.12TheoremAnisometrywithaninvariantpointisareflectioninaplaneoraproductofatmostthreereflections.

IntheproofofTheorem47.9,wemayassumethatpointAisthefixedpoint,soCase 5 is eliminated.Hence the isometry is a reflection or a product of at

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mostthreereflectionsinlines.

48.13TheoremAdirectisometrywithaninvariantpointisarotation.Sincetheisometryisdirect,itisaproductofanevennumberofreflections,

henceof just tworeflectionsbyTheorem48.12.Buta translationhasnofixedpoints, so the isometry must be a product of reflections in two intersectingplanes,arotation.

48.14DefinitionArotationof180°aboutalinemiscalledahalfturnaboutlinemdenotedbyσm(seeFig.46.1d).

48.15TheoremAhalfturnabout linem is aproductof two reflections inanytwoperpendicularplanesthroughm.

48.16CorollaryAproductofreflectionsintwoperpendicularplanescommutes.

48.17CorollaryAhalfturnaboutalineisinvolutoric.

48.18DefinitionAspacemapaiscalledacentralinversioninapointPcalleditscenteriffα(P)=P,and,foreachpointA≠P,α(A)=B,whereBistakensothatP is themidpoint of segmentAB.This central inversion is denotedbyσP(seeFig.46.2).

48.19 Theorem The product of three reflections in mutually perpendicularplanesintersectingatapointPisacentralinversioninpointP.Conversely,eachcentralinversioninapointPcanbewrittenasaproductofthreereflectionsinany three mutually perpendicular planes passing through P. Hence a centralinversionisanisometryandisopposite.

48.20TheoremAcentralinversionisinvolutoric.Theterm“central inversion”issomewhatunfortunate,sincethismapisnot

tobeconfusedwithaninversioninacirclestudiedinChapter5.Itisareflectioninapoint,andsomeauthorsusethatterm,butthentheterm“reflection”tendstobeoverworked.Weshallholdtotheterm“centralinversion”here.

ExerciseSet48

1.Indicatethetwoplanesofreflectionintowhichαtranslationacanbefactoredifαcarriesthepoint(0,0,0)into(1,1,1).

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2.ProveTheorem48.2.

3.ProveTheorem48.3.

4.ProveTheorem48.4.

5.ProveTheorem48.5.

6.Defineaspacetranslationintermsofaplanetranslation,justasDefinition48.8doesforaspacerotation.

7.Showthataproductoftworotationsaboutparallelaxesisarotationoratranslation.

8.Statewhenaproductoftwohalfturnsisatranslation.

9.ProveTheorem48.9.10.ProveTheorem48.10.11.ProveTheorem48.11.12.Showthataproductoftwocentralinversionsisatranslation.13.ProveTheorem48.15.14.ProveCorollary48.16.15.ProveCorollary48.17.16.Whatisometryistheproductofthreecentralinversions?17.ProveTheorem48.19.18.ProveTheorem48.20.

SECTION49 MORESPACEISOMETRIES

49.1DefinitionAscrewdisplacement is theproductofa rotationabouta lineand a translation through a vector parallel to the axis of the rotation (see Fig.49.1).

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Figure49.1

49.2TheoremEachdirectisometrycanbewrittenasaproductoftwohalfturns.A translation or rotation α can be factored into a product σ∆ σΠ of two

reflectionsinplanesΠand∆.TakeaplaneΓperpendiculartobothgivenplanes.Thenα=(σ∆σΓ)(σΓσ∆)aproductoftwohalfturns.

We need consider only a productα of four reflections. If any three of themirrorsformapencil,thentheisometryreducestoaproductoftworeflections(Exercise49.1).Similarly,aproductoftwotranslationsreducestoatranslation(Theorem48.5).Ifonlythefirstandsecondplanesoronlythethirdandfourthplanes are parallel, then the second and third planes intersect. In any casewemayassumethatthesecondandthirdplanesdointersect,sowemayrotatethemabout their lineof intersectionsothatαbecomesaproductof tworotations.Ifthe rotations are about the same or parallel axes, they may be replaced by asingle rotation or translation (Exercise 48.7). Thus we need consider only aproductoftworotationsinskeworintersectinglines.

Letα=σ4σ3σ2σ1whereplanes1and2meetinlinem,andplanes3and4meetinlinen,andlinesmandnareeitherintersectingorskew.Wemayassumewithout lossofgenerality thatplane3cuts linem,and thatplanes2and3areperpendicularalonga linep (seeFig.49.2a).Nowrotateplanes2and3aboutlinepinto2′and3′sothatplanes1and2′areperpendicular.Ofcourse,planes2′and3′remainperpendicular(seeFig.49.2b).Thenwehave

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Since both planes 1 and 3′ are perpendicular to plane 2, then their line q ofintersectionisperpendiculartoplane2.Thenrotateplanes1and3′aboutlineqintoplanes1′and3″sothat3″isperpendiculartoplane4.Alsoplanes1′and3″areperpendiculartoplane2′(sincelineqisperpendiculartoplane2′).Nowwehave

aproductoftwohalfturns.

Figure49.2a

Figure49.2b

49.3TheoremAproduct of two rotations a) about the same line is a rotationaboutthat line;b)aboutparallel linesisarotationaboutanaxisparallel tothe

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two lines,ora translationwhosevector isperpendicular to thedirectionof thetwo lines;c)about two lines intersectingatpointO isa rotationaboutanaxisthroughO;d)abouttwoskewlinesisascrewdisplacement.

Parts(a)and(b)areanalogoustothecorrespondingplanetheorem(Theorem14.15).Part(c)isestablishedbyTheorem48.13,sincepointO isafixedpointbecauseitisinvariantunderbothrotations.

Forpart(d),inviewofTheorem49.2,weneedshowonlythattheproductoftwohalfturnsaboutskewlinesisascrewdisplacement.Letα=σnαm=σ4σ3σ2σ1besuchanisometry,wheremandnareskewlineswithplanes1and2andplanes 3 and 4 perpendicular.Wemay also assumewithout loss of generalitythatplane1isparalleltolinenandthat4isparalleltom.Thenplanes2and3areeachperpendiculartobothofplanes1and4.Thus

Figure49.3

a screw displacement, since planes 1 and 4 are parallel and the line p ofintersectionofplanes2and3isperpendiculartoplanes1and4(seeFig.49.3).

49.4 Corollary The axes of the two halfturns of a direct isometry with noinvariantpointareeitherparallel(atranslation)orskew(ascrewdisplacement).

49.5 Corollary Each direct isometry is a rotation, a translation, or a screwdisplacement.

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49.6DefinitionAglide-reflectionistheproductofareflectioninaplaneandatranslationthroughavectorparalleltothemirrorofthereflection(Fig.49.6).

Figure49.6

49.7DefinitionArotatoryreflectionistheproductofareflectioninaplaneanda rotation about an axism perpendicular to the mirror of the reflection (Fig.49.7).

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%

Figure49.7

49.8 Theorem Each opposite isometry is a reflection, a glide-reflection, or arotatoryreflection.

An opposite isometry a is a reflection or a product of three reflections. Soconsidertheproduct

This product reduces to a single reflection if themirrors forma pencil, soweassume that they do not.Wemay also assumewithout loss of generality thatplanesΓand∆meetinalinem.Case1.IflinemisparalleltoplaneΠ(seeFig.49.8a),thenwemayassumethatΓisparalleltoΠ,anditfollowsthattheisometryisaglide-reflection.TheproofisanalagoustothatforTheorem16.4.

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Figure49.8a

Case2.LetlinempierceplaneΠatapointP(seeFig.49.8b).FirstrotateΓand∆aboutlinem intoΓ′and∆′,sothat∆′ isperpendiculartoΠ(seeFig.49.8c).Nextrotate∆′andΠabouttheirlinenofintersectioninto∆″andΠ″sothat∆″isperpendiculartoΓ′(seeFig.49.8d).Ofcourse,Δ″isperpendiculartoπ′.Then

arotatoryreflection,sinceΔ″isperpendiculartobothπ′andΓhencetotheirlineofintersection.

Figure49.8b

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Figure49.8c

%

Figure49.8d

49.9TheoremArotatory reflection is theproductofacentral inversionandarotationwithaxisthroughthecenterOofthecentralinversion,henceitiscalledalsoarotatoryinversion.

Lettherotatoryreflectionbe

whereIIisperpendiculartothelinemofintersectionofplanesΓandΔ.LetΩbethe plane through m perpendicular to plane Γ (see Fig. 49.9). Since II isperpendiculartobothplanesΔandΩ,thenρΔcommuteswithbothandρΩ,and

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wehave

arotatoryinversion.

Figure49.9

49.10TheoremAn isometrywith three noncollinear fixed points is either theidentityorareflectioninaplane.

49.11 Theorem An opposite isometry with no invariant point is a glide-reflection.

SuchanoppositeisometryαistheproductofthreereflectionsinplanesΓ,Δ,and II that donot formapencil anddonot concur at a point (Exercise 49.9).Thusweassume,withoutlossofgenerality,thatthethreelinesofintersectionofthepairsoftheseplanesareparallel.Byrotatingpairsofplanesappropriately,aswas done for lines in Theorem 16.4, we obtain α = σII′ σΔ′ σΓ′ where Δ′ isperpendicular tobothΓ′andII′,andtheselast twoplanesareparallel(seeFig.49.8a).Thusα=(σII,σΓ′)σΔ′glide-reflection.

49.12 Theorem An opposite isometry with an invariant point is a rotatoryreflection(ofwhichacentralinversionandareflectionarespecialcases).

The threemirrorswhichmakeup suchan isometry areparallel, or all passthroughitsfixedpointP(Exercise49.12a).NowpairsofplanesmayberotatedabouttheirlinesofintersectiontoobtainthreeplanesII,Γ,andΔ,withthefirst

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twoplanesbothperpendiculartoplaneΔ(asinFig.49.8d).ThenσΔσΓσII isarotatoryreflection,

49.13 Theorem A direct isometry with no invariant point is a screwdisplacement(ofwhichatranslationisaspecialcase).

49.14Weterminateourdevelopmentofatheoryofspaceisometryhere,havinggiven a sufficient background for our purposes. The last two sections of thischapter contain applications of space isometries to elementary solid geometryand analytic equations for these transformations. Certainly the algebra ofisometries could be pursued much further, as it was for plane isometries inSections 16 and 17.You should now have sufficient preparation to undertakesuchstudieswhenyoufeeltheneedtodoso.Ifso,thenourworkiscomplete.

ExerciseSet49

1.Provethataproductofthreereflectionsinplanesthatformapencil(theyareallparallelorallpassthroughaline)isareflectioninanotherplaneofthesamepencil.

2.MakeamodelofeachofFigs.49.2a,49.2b,and49.3bygluingortapingtogether3-by-5cards.Labelimportantpoints,lines,andplanes.

3.a)ProveTheorem49.3,part(a).b)ProveTheorem49.3,part(b).c)ProveTheorem49.3,part(c),byfactoringtherotationsintoappropriatelychosenreflectionsinplanes.

4.ProveCorollary49.4.5.ProveCorollary49.5.6.MakeamodelofeachofFigs.49.8b,49.8c,and49.8dbygluingortapingtogether3-by-5cards.Labelimportantpoints,lines,andplanes.

7.Whatisometryisthesquareofarotatoryreflection?8.ProveTheorem49.8,Case1.9.Showthatanoppositeisometrywithnofixedpointisaproductofthreereflectionsinplanesthatdonotconcuratapoint,anddonotformapencil.

10.ProveTheorem49.10.11.CompletetheproofofTheorem49.11.12.a)Provethatthethreemirrorsofthereflectionsthatmakeupanopposite

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isometrywithafixedpointareparallelorallpassthroughthefixedpoint,b)SupplythedetailsintheproofofTheorem49.12.

13.ProveTheorem49.13.14.Showthatthetranslationandreflectionofaglide-reflectioncommute.15.Showthattherotationandreflectionofarotatoryreflectioncommute.16.Showthattherotationandcentralinversionofarotatoryinversion

commute.17.Showthatareflectionandacentralinversionarebothspecialcasesofa

rotatoryreflection.18.Showthatatranslationisaspecialcaseofascrewdisplacement.

SECTION50 SOMEAPPLICATIONS

50.1TheoremVerticaldihedralangles(anglesbetweenplanes)arecongruent.

50.2 Theorem Parallel planes cut by a transversal plane form congruentcorrespondingandalternatedihedralangles.

The lines of intersection of the two planes with the transversal plane areparallel, since they lie in the same plane (the transversal) and do not meet.Translate one of the lines of intersection along the transversal plane at rightanglestoitselfintotheotherlineofintersection.Thetransversalplanemapsintoitselfandoneparallelplanemapsintotheother.Thusthisisometrymapsonesetofverticaldihedral angles into theother.Hence theyarecongruent. (ComparethisproofwiththatgivenforTheorem18.7.)

50.3 Theorem The opposite faces of a parallelepiped are congruentparallelograms.

50.4Theorem The diagonals of a parallelepiped concur at a point (called thecenteroftheparallelepiped)thatbisectseachdiagonal.

LetADandBCbeapairofoppositeedges.Theyareparallelandcongruent.Thus ABCD is a parallelogram whose diagonals meet at their midpoints byTheorem18.9.Thetheoremfollows(seeFig.50.4).

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Figure50.4

50.5 Theorem The diagonals of a rectangular parallelepiped (a box) arecongruent.

50.6TheoremAparallelepipedissymmetricwithrespecttoitscenter;thatis,aparallelepipedisinvariantunderacentralinversioninitscenter.

50.7 Corollary Any segment through the center of a parallelepiped joiningpointsonoppositefacesisbisectedbythecenter.

50.8CorollaryAplanethroughtwooppositeedgesofaparallelepipeddividestheparallelepipedintotwocongruenttriangularprisms.

50.9 Theorem The sides of an isosceles trianglemake congruent angleswithanyplaneinwhichthebaselies(seeFig.50.9).

50.10 Theorem The following statements are equivalent for any three givenplanesΓ,Δ,andII:a)Δ=σΓ(II),b)ΓliesmidwaybetweenΔandII(orbisectstheirangle),c)σΔ=σΓσIIσΓ.

CompareTheorem50.10withitsplanecounterpart,Theorem17.5.

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%

Figure50.9

50.11Theorem The three planes that bisect the dihedral angles of a trihedralangle(formedbythreeplanesthatconcuratapoint)concuralongaline.

50.12DefinitionThevolumeofarectangularparallelepipedistheproductofthelengthsofitsthreeedges.

50.13TheoremThevolumeofaparallelepipedistheproductoftheareaofanyfaceasbaseandthelengthofthealtitudetothatbase.

50.14CorollaryThevolumeofatriangularprismistheproductoftheareaofitstriangularfaceasbaseandthelengthofthealtitudetothatbase.

50.15 Corollary The volume of any prism is the product of the area of its(polygonal)baseandthelengthofthealtitudetothatbase(seeFig.50.15).

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Figure50.15

50.16TheoremAnobliqueprismhasthesamevolumeasarightprismwhosebaseisarightsectionoftheobliqueprism,andwhosealtitudeisalateraledgeoftheobliqueprism(seeFig.50.16).

Figure50.16

50.17TheoremAsphereissymmetricwithrespecttoitscenter,withrespecttoanylinethroughitscenter,andwithrespecttoanyplanethroughitscenter.

BecauseanyisometrythatleavesthecenterOofthespherefixedmapseachpointPon the sphere intoapointP′ the samedistance (the radius) fromO, itfollowsthatP′liesonthesphere,

50.18 Theorem A right circular cylinder is symmetric with respect to a) themidpointofitsaxis,b)itsaxis,c)thenormalplanetoitsaxisatitsmidpoint,d)anyplanethroughitsaxis.

50.19Corollary A right circular cylinder is generated by rotating a rectangleaboutonesideasaxis.

50.20DefinitionLetPbethecentroidoffaceABCoftetrahedronABCD.ThensegmentDPiscalledamedianofthetetrahedron(seeFig.50.20).

50.21TheoremThefourmediansofatetrahedronconcuratapointGcalledthecentroidofthetetrahedron.

First,centralinversionsinspacebehavequitelikehalfturnsintheplane.Thatis,1)acentralinversionσxisinvolutoric(Theorem48.20),2)aproductoftwo

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central inversions is a translation (Exercise 48.12) and translations commute(Theorem48.4),and3)aproductσCσBσAofthreecentralinversionsisacentralinversion(seeExercise48.16),soσCσBσA=σAσBσC

LetABCDbethetetrahedronhavingmediansAQandDPandwithA′,B′,C′themidpointsofthesidesoftriangleABC,asshowninFig.50.20.Bya“shrewdguess,”letusplacepointGthree-fourthsofthewayfromDtoPalongmedianDP (assume that the theorem is true and applyMenelaus’ theorem to triangleDA′Ptoobtainthe“shrewdguess”).Thenwehave

SincePandQarethecentroidsoftrianglesABCandDBC> thenPandQaretwo-thirdsofthewayalongthetrianglemediansAA′andDA′.Hence

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Figure50.20

ToprovethetheoremitwillsufficetoshowthatGnowliesthree-fourthsofthewayfromAtoQ;thatis,wemustshowthat(σAσG)(σQσG)3= .Tothatend,wetake

Similarly,Gliesthree-fourthsofthewayalongtheothertwomedians.

ExerciseSet50

1.ProveTheorem50.1.2.Provethateachproductoftwohalfturnsaboutaxesintersectingatanangleθisarotationthroughangle2θ.

3.ProveTheorem50.3.4.Provethateachtranslationcanbewrittenasaproductoftwocentralinversions,ortwohalfturnsinparallelaxes,ortworeflectionsinparallelmirrors.

5through11.ProveTheorems(andCorollaries)50.5through50.11.12.Provethateachoppositeisometryistheproductofareflectioninaplane

andahalfturnaboutaline.13through16.ProveTheorems(andCorollaries)50.13through50.16.17.Provethateachrotationaboutalinecanbefactoredintoaproductoftwo

halfturnsaboutintersectinglines.18.ProveTheorem50.18.19.ProveCorollary50.19.

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20.Provethataproductoftwohalfturnsabouttwoskewlinesatrightanglesisascrewdisplacement,namelytheproductofahalfturnaboutthelineofshortestdistancebetweentheaxesofthehalfturnsandatranslationthroughtwicethisshortestdistance.

21.Nametheisometrythatmapseachpoint(x,y,z)inspaceintoa)(x,y,–z)b)(‒y,x,z)c)(x,y,z+l)d)(‒y,x,z+1)e)(–x,y,z+1)f)(–y,x,–z)

Section51 ANALYTICREPRESENTATIONS

51.1LetusdenotethevectorvfromtheoriginO(0,0,0)toapointA(h,k,l)byv=(h,k,l).

51.2TheoremThetranslationthroughvectorv=(h,k,l)thatcarriespointP(x,

y,z)topointP′(x′,y′,z)isdeterminedby

51.3TheoremThe rotationabout thez-axis throughangleθ (seeFig.51.3) isgivenby

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Figure51.3

51.4 Although we could develop equations for other rotations, we shall becontentwithequationsof rotationsabout just thecoordinateaxes. Inanycase,anyrotationcanbewrittenastheproductofatranslation,onetothreerotationsabout thecoordinateaxes,and then the inverseof the translation.Toillustrate,letusperformthetranslationfromA(h,k,l)toO(0,0,0),thenrotationsthroughanglesθ,ϕ,andλaboutthez-,y-,andx-axes,thenthetranslationfromObacktoA,toarriveattheequations

Clearlytheseequationsarefartoocomplicatedtobeofpracticalusetous.

51.5TheoremThereflectioninthexy-plane(seeFig.51.5)isgivenby

Figure51.5

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51.6Similarequationsdeterminereflectionsintheothercoordinateplanes.ThusweobtainthenexttwoimmediatecorollariesfromTheorems48.15,48.19,and51.5.

51.7CorollaryAhalfturnaboutthez-axis(seeFig.51.7)isgivenby

51.8CorollaryAcentralinversionintheorigin(seeFig.51.8)isgivenby

51.9TodeveloptheequationsforareflectioninthegeneralplaneII,letuswritetheequationofIIinthestandardform

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Figure51.7

Figure51.8

LettingσIImapP(x,y,z) toP′(x′,y′,z′),wehavethelinePP′noemal toII,sothatthecoordinatesofPandP′satisfytheequation

or

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wheretisarealconstant.NowthedistancefromPtoIIisequaltothatfromP′toII.Recallingthatthe

distancefromPtoIIisgivenby

TheplusorminussignbeingtakenaccordingtowhichsideoftheplaneP lieson,wehave,sincePandP′lieonoppistesidesoftheplaneII,

whichwerewriteas

Substituting the parametric equations of the preceding paragraph into theequationabove,weobtain

fromwhich

Thisvaluefortmaynowbesubstitutedintotheparametricequationstoobtainequations fora reflection in theplane II.Hencewehaveproved the followingtheorem.

51.10TheoremAreflection in theplaneAx+By+Cz+D=0, inwhichwehaveA2+B2+C2=1,hastheequations

51.11Sinceeachisometryinspaceisaproductofreflections,itisnowpossibleto write a set of equations for any isometry. The calculations become quite

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complicatedveryquickly,soweshallnotpursuethematterfurther.

ExerciseSet51

1.ProveTheorem51.2.2.ProveTheorem51.3.3.ProveTheorem51.5.4.ProveCorollary51.7.5.ProveCorollary51.8.6.ShowthatthetranslationofTheorem51.2canbeaccomplishedbyaproductoftworeflectionsinplanes,asgiveninTheorem51.10.

7.Showthataproductoftworeflectionsinintersectingplanes(asgiveninTheorem51.10)isarotation.

8.Findequationsforacentralinversionaboutanygivenpointascenter.9.ShowthatTheorem51.5isaspecialcaseofTheorem51.10.10.Findequationsforahalfturnaboutagivenlinebytakingtheproductofa

reflectioninaplaneperpendiculartothelineandacentralinversioninthepointofintersectionofthelineandtheplane.

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APPENDIXESAPPENDIXA ASUMMARYOFBOOKIOFEUCLID’SELEMENTS

TheAxioms(CommonNotions)1.Thingsequaltothesamethingarethemselvesequal.2.Whenequalsareaddedtoequals,thenthesumsareequal.3.Whenequalsaresubtractedfromequals,thenthedifferencesareequal.4.Thingswhichcoincidewithoneanotherareequal.5.Thewholeisgreaterthananypartofthewhole.ThePostulates1.Alinesegmentcanbedrawnbetweenanytwopoints.2.Alinesegmentcanbeextendedindefinitely.3.Acirclecanbedrawnhavinganypointascenterandpassingthroughanyotherpoint.

4.Allrightanglesarecongruent.5.Ifatransversalcutstwolinessothatthesumofthetwointerioranglesononesideofthetransversalislessthantworightangles,thenthetwolineswillmeetonthatsameside.

Selectionsfromthe48PropositionsofBookI1.Toconstructanequilateraltriangleonagivensegment.2.Todrawasegmentcongruenttoagivensegmentfromagivenpointasendpoint.

3.Tocutofffromagivensegmentasegmentcongruenttoasmallergivensegment.

4.TwotrianglesthatsatisfytheSASconditionarecongruent.5.Thebaseanglesofanisoscelestrianglearecongruent.6.Iftwoanglesofatrianglearecongruent,thenthesidesoppositetheseanglesarecongruent.

7.Onlyonetriangledirectlycongruenttoagiventrianglecanbeconstructedonagivensideofagivensegmentcongruenttothebaseofthegiventriangle.?

8.TwotrianglesthatsatisfytheSSSconditionarecongruent.9.Tobisectagivenangle.10.Tobisectagivensegment.11.Toerectaperpendicularatagivenpointonaline.12.Todropaperpendicularfromapointtoaline.13.Ifarayemanatesfromapointonaline,thenthetwoanglesformedareright

anglesorhaveasumequaltotworightangles.14.Iftworaysemanatingfromoppositesidesofapointonalinemaketheir

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adjacentangleshaveasumequaltotworightangles,thentheylieonaline.15.Verticalanglesformedbyintersectinglinesarecongruent.16.Anexteriorangleofatriangleisgreaterthaneitheroppositeinteriorangle.17.Thesumofanytwoanglesofatriangleislessthantworightangles.18.Inatrianglethegreatersideliesoppositethegreaterangle.19.Inatrianglethegreaterangleliesoppositethegreaterside.20.Inatrianglethesumofanytwosidesisgreaterthanthethirdside.22.Toconstructatrianglefromthreegivensegments.23.Toconstructatapointonalineananglecongruenttoagivenangle.24.Iftwotriangleshavetwosidescongruenttotwosidesrespectivelybutthe

includedangleofthefirstgreaterthanthatofthesecond,thentheoppositesideofthefirstisgreaterthanthatofthesecond.

25.Iftwotriangleshavetwosidescongruenttotwosidesrespectively,butthethirdsideofthefirstisgreaterthanthatofthesecond,thentheangleofthefirstincludedbetweenthetwosidesisgreaterthanthatofthesecond.

26.TwotrianglesthatsatisfyeithertheASAortheAASconditionarecongruent.

27.Ifatransversalcutstwolinessothatalternateinterioranglesarecongruent,thenthetwolinesareparallel.

29.Atransversalcuttingparallellinesmakescongruentalternateandcorrespondingangles.

30.Parallelismoflinesistransitive.31.Todrawalinethroughapointparalleltoagivenline.32.Anexteriorangleofatriangleisequaltothesumofthetwooppositeinterior

angles.Thesumoftheanglesofatriangleisequaltotworightangles.33.Thesegmentsjoiningcorrespondingendpointsoftwocongruentsimilarly

directedparallelsegmentsarethemselvescongruentandparallel.34.Theoppositesidesandoppositeanglesofaparallelogramarecongruent,and

eachdiagonalbisectsthearea.36.Parallelogramsoncongruentbasesandcontainedwithinthesameparallels

haveequalareas.38.Trianglesoncongruentbasesandwithcongruentaltitudeshaveequalareas.40.Equaltrianglesoncongruentbasesandonthesamesideofthebaselineare

alsowithinthesameparallels.41.Ifaparallelogramhasthesamebaseandliesbetweenthesameparallelsasa

triangle,thentheparallelogramhastwicetheareaofthetriangle.46.Todescribeasquareonagivensegment.47.Inarighttrianglethesquareonthehypotenuseisequaltothesumofthe

squaresonthelegs.

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48.Ifthesquareononesideofatriangleisequaltothesumofthesquaresontheothertwosides,thenthetriangleisarighttrianglewiththerightangleoppositethefirstside.

APPENDIXB BASICRULERANDCOMPASSCONSTRUCTIONS

B.lNotationWedenoteacirclewithcenterAandradiusr=m(CD)byA(r)orA(CD).ThecirclewithcenterAandpassingthroughpointBisdenotedbyA(B).Unlessstatedotherwise,throughoutthisappendixrandsshalldenotearbitraryconvenientlengths.B.2Theconstructionslistedherebynomeansformanexhaustivelist.Theyarebasic constructions that every serious student of geometry shouldhave readilyavailablewheneverneeded.Itissuggestedthatyoupracticetheseconstructions,keepinginmindthatprecisionisrequired.Thepencilsyouuseshouldbealwayskeptfanaticallysharp,andanerasershouldneverbeused.Besureyourlinesandcircles pass precisely through the intended points, not just close to them.Andnever put a large blob on a constructed point to indicate its location! Ifnecessary,drawasmallarrowpointingtowardthatpoint,butdonotobliterateorcoverit.Observehowthesetechniquesarefollowedintheconstructionsbelowanddolikewise.Tobesureyourpracticeconstructionsareaccurate,checkyourangleswithaprotractorandmeasureyourdistanceswitharuler.B.3ConstructionDrawatriangleABCgiventhelengthsofitsthreesidesa,b,c.FromapointionalinemdrawcircleA(c)tocutlinematpointB.ThendrawcirclesA(b)andB(a)tomeetatC(seeFig.B.3).

FigureB.3B.4ConstructionDraw an angleC′A′B′ congruent to a given angleCAB at apointA′ onagiven linem.DrawcircleA(r) to eutAB atD andAC atE, andcircleA′(r)toeutlinematB′(seeFig.B.4).DrawcircleB′(DE)toeutcircleA′(r)atC′.

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FigureB.4B.5ConstructionBisectagivenangleCAB.DrawcircleA(r)tocutABatDandACatE.DrawcirclesD(s)andE(s)tomeetatT.ThenATisthedesiredbisector(seeFig.B.5).

FigureB.5B.6ConstructionBisectagivensegmentAB,orerectitsperpendicularbisector.DrawcirclesA(r) andB(r) tomeet atC andD.ThenCD is theperpendicularbisectorofsegmentABmeetingABatitsmidpointM(seeFig.B.6).

FigureB.6B.7ConstructionDropaperpendicularfromagivenpointPtoagivenlinem.DrawcircleP(r)tocutlinemintwopointsAandB.DrawcirclesA(s)andB(s)tomeetatCasinFig.B.7.ThenlinePCisthedesiredperpendicular.B.8ConstructionErectaperpendicularatapointPonagivenlinem.Firstmethod.DrawcircleP(r) tocutmatAandB.SeeFig.B.8a.Withs>r,drawcirclesA(s)andB(s)tomeetatC.ThenCPisthedesiredperpendicular.Second method. Choose any point Q not on m and not on the desiredperpendicular, as in Fig.B.8b.Draw circleQ(P) to cut linem again atR anddrawdiameterRQT.ThenPTisthedesiredperpendicular.

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FigureB.7

FigureB.8a

FigureB.8bB.9ConstructionDrawalinenparalleltoagivenlinemandpassingthroughagivenpointP.Drawanyline throughP tocut linematA.UsingConstructionB.4,drawanangleatPcongruenttotheangleatAandontheoppositesideoflinePA.Thenlinenistheterminalsideofthisconstructedangle.(SeeFig.B.9.)

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FigureB.9B.10ConstructionDivideagiven segmentAB internally in agiven ratioa:b.OnaconvenientrayADnotlyingonlineAB,markAC′oflengthaandC′B′oflengthbasshowninFig.B.10.DrawBB′andalinethroughC′paralleltoBB′andcuttingABatC,thedesiredpoint.

FigureB.10

B.llConstructionB.10iseasilyalteredtoprovidethepointCwhichdividesABintheratioa:bexternally.Ifa>b,thenletpointB′liebetweenAandC′insteadofC′betweenAandB′.Therestoftheconstructionisunaltered.Ifa<b,theninterchange the roles played byA andB, and by a and b before starting theconstruction.

B.12ConstructionLocatethecenterofagivencircles.DrawanychordABandletitsperpendicularbisector(ConstructionB.6)meetthecircleatCandD.ThemidpointofCD(ConstructionB.6again)isthecenterOofcircles(Fig.B.12).

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FigureB.12

B.13ConstructionDrawatangenttoacirclesatagivenpointTonthecircle.Assuming the centerO of the circle is given, draw the radiusOT. The lineperpendiculartoOTatT(ConstructionB.8)isthedesiredtangent.

B.14ConstructionDrawa tangent toacircles fromapointPexternal to thecircle. Assuming the centerO of the circle is given, drawOP and locate itsmidpointM(ConstructionB.6).DrawcircleM(O)tocutcirclesatpointT.ThenPTisthedesiredtangentline(Fig.B.14).

FigureB.14

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BIBLIOGRAPHYAdler,C.F.,ModernGeometry.NewYork:McGraw-Hill,1958Ahlfors,L.V.,ComplexAnalysis.NewYork:McGraw-Hill,1953Anderson, R. D., J. W. Garon, and J. G. Gremillion, School MathematicsGeometry.Boston:HoughtonMifflin,1969Bachmann, F., Aufbau der Geometrie aus dem Spiegelungsbegriff. Berlin:Springer-Verlag,1959Ball, W. W. R., A Short Account of the History of Mathematics New York:DoverPublications,1960Barry, E. H., Introduction to Geometrical Transformations. Boston: Prindle,Weber&Schmidt,1966Benson,R.V.,EuclideanGeometry andConvexity.NewYork:McGraw-Hill,1966Bieberbach,L.,AnalytischeGeometrie.Leipzig:Teubner,1930Blumenthal,L.M.,AModernViewofGeometry.SanFrancisco:W.H.Freeman,1961Bohuslov,R.,AnalyticGeometry.NewYork:Macmillan,1970Bryant,S.J.,G.E.Graham,andK.G.Wiley,NonroutineProblems.NewYork:McGraw-Hill,1965Cajori,F.,AHistoryofMathematicalNotations,2vols.Chicago:OpenCourt,1928Cajori,F.,AHistoryofMathematics,2ndedition.NewYork:Macmillan,1919Choquet,G.,GeometryinaModernSetting.Boston:HoughtonMifflin,1969Chrestenson,H.E.,MappingsofthePlane.SanFrancisco:W.H.Freeman,1966Churchill,R.V.,ComplexVariablesandApplications,2ndedition.NewYork:McGraw-Hill,1960Court,N.A.,CollegeGeometry.Richmond:Johnson,1925Coxeter,H.S.M.,IntroductiontoGeometry.NewYork:JohnWiley,1961Coxeter,H.S.M.,ProjectiveGeometry.NewYork:Blaisdell,1964Davis, D. R., Modern College Geometry. Reading, Mass.: Addison-Wesley,1949Deaux,Roland, Introduction to theGeometry ofComplexNumbers, translatedbyH.Eves.

NewYork:FrederickUngar,1956Dodge,C.W.,TheCircularFunctions. EnglewoodCliffs,N.J.: Prentice-Hall,1966Dodge,C.W.,NumbersandMathematics.Boston:Prindle,Weber&Schmidt,

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1969Dodge, C.W., Sets,Logic and Numbers. Boston: Prindle,Weber& Schmidt,1969Eves,H.W.,AnIntroductiontotheFoundationsandFundamentalConceptsofMathematics,revisededition.NewYork:Holt,RinehartandWinston,1965Eves,H.W.,An Introduction to theHistoryofMathematics,3rdedition.NewYork:Holt,RinehartandWinston,1969Eves,H.W.,FunctionsofaComplexVariable,2vols.Boston:Prindle,Weber&Schmidt,1966Eves,H.W.,FundamentalsofGeometry.Boston:AllynandBacon,1969Eves, H. W., In Mathematical Circles, 2 vols. Boston: Prindle, Weber &Schmidt,1969Eves, H. W., Mathematical Circles Revisited. Boston: Prindle, Weber &Schmidt,1971Eves,H.W.,ASurveyofGeometry,2vols.Boston:AllynandBacon,1963Fink,K.,ABriefHistoryofMathematics,translatedbyW.W.BemanandD.E.Smith.Chicago:OpenCourt,1900Forder,H.G.,Geometry.NewYork:Harper&Brothers,1962Fujii,J.N.,GeometryandItsMethods.NewYork:JohnWiley,1969Gans, D., Transformations and Geometries. New York: Appleton-Century-Crofts,1969Guggenheimer,H.W.,PlaneGeometryandItsGroups.SanFrancisco:Holden-Day,1967Hall,D.W.,andS.Szabo,PlaneGeometry,anApproachThroughIsometries.EnglewoodCliffs,N.J.:Prentice-Hall,1971Hemmerling, E. M., Fundamentals of College Geometry. New York: JohnWiley,1964Horblit, M., and K. L. Nielsen, Plane Geometry Problems with Solutions.CollegeOutlineSeries.NewYork:Barnes&Noble,1947Hyatt,H.R., andC.C.Carico,ModernPlaneGeometry forCollegeStudents.NewYork:Macmillan,1967Jolly,R.F.,SyntheticGeometry.NewYork:Holt,RinehartandWinston,1969Jurgensen, R. C., A. J. Donnelly, and M. P. Dolciani, Modern SchoolMathematics:Geometry.Boston:HoughtonMifflin,1969Kay,D.C.,CollegeGeometry.NewYork:Holt,RinehartandWinston,1969Kenner, M. R., D. E. Small, and G. N. Williams, Concepts of ModernMathematics,Book2.NewYork:AmericanBook,1963Lathrop, T. G., and L. A. Stevens, Geometry, A Contemporary Approach.Belmont,Cal.:Wadsworth,1967

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Levi,H.,TopicsinGeometry.Boston:Prindle,Weber&Schmidt,1968Levy,L.S.,Geometry:ModernMathematicsvia theEuclideanPlane.Boston:Prindle,Weber&Schmidt,1970Meder,A.E., Jr.,Topics from InversiveGeometry.Boston:HoughtonMifflin,1967Meserve, B. E., and J. A. Izzo, Fundamentals of Geometry. Reading, Mass.:Addison-Wesley,1969Miller,L.H.,CollegeGeometry.NewYork:Appleton-Century-Crofts,1957Moise, E. E., Elementary Geometry from an Advanced Standpoint. Reading,Mass.:Addison-Wesley,1963Moser,J.M.,ModernElementaryGeometry,EnglewoodCliffs,N.J.:Prentice-Hall,1971Newman, J. R., The World of Mathematics, 4 vols. New York: Simon andSchuster,1956Osgood,W.F.,andW.C.Graustein,PlaneandSolidAnalyticGeometry.NewYork:Macmillan,1921Prenowitz, W., and M. Jordan, Basic Concepts of Geometry. New York:Blaisdell,1965Rees,P.K.,AnalyticGeometry,3rdedition.EnglewoodCliffs,N.J. :Prentice-Hall,1970Rich, B., Principles and Problems of Plane Geometry with CoordinateGeometry.Schaum’sOutlineSeries.NewYork:McGraw-Hill,1963Rosskopf,M.F.,J.L.Levine,andB.R.Yogeli,Geometry:APerspectiveView.NewYork:McGraw-Hill,1969Shively,L. S.,An Introduction toModernGeometry.NewYork: JohnWiley,1939Sigley,D.T., andW.T.Stratton,SolidGeometry, revisededition.NewYork:DrydenPress,1956Smith,D.E.,ASourceBookinMathematics.NewYork:McGraw-Hill,1929Spiegel,M.R.,TheoryandProblemsofVectorAnalysisandanIntroductiontoTensorAnalysis.Schaum’sOutlineSeries.NewYork:Schaum,1959Tryon,C.W.,ElementaryGeometryforCollege.NewYork:Harcourt,Brace&World,1969Tuller,A.,AModernIntroduction toGeometries.Princeton:D.VanNostrand,1967Yeblen,O.,andJ.W.Young,ProjectiveGeometry,2vols.NewYork:Blaisdell,1938Wells,W.,andW.W.Hart,ModernSolidGeometry.Boston:D.C.Heath,1927Wylie,C.R.,Jr.,FoundationsofGeometry.NewYork:McGraw-Hill,1964

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Wylie,C.R.,Jr.,IntroductiontoProjectiveGeometry.NewYork:McGraw-Hill,1970Yaglom,I.M.,ComplexNumbersinGeometry,translatedbyE.J.F.Primrose.NewYork:AcademicPress,1968

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HINTSFORSELECTEDEXERCISESInthedouble-numberingsystem,thefirstnumberreferstothesectioninwhichtheexerciseappears,andthesecondnumberreferstotheexercisenumber.Thus1.3(below)givesahintforExercise3ofSection1.1.1Dropanaltitudefromtheapexoftheisoscelestriangleandapplythe

Pythagoreantheorem.1.2Set andsolveforπ1.3DrawdiagonalsanduseTheorem2.17.1.4LettheoriginalpyramidhavealtitudeH.ShowthatH/(H–h)=B/band

solvethisequationforH.Thensubtractthevolumeofthesmallcut-offpyramidfromthatoftheoriginalpyramid.

1.5Considerthenumericalvaluesoftheareas.1.6Somepersonsfeelthe“moltensea”wasoval-shaped,accountingforits

circumferencebeingonlythreetimesits(maximum)diameter.1.7c)Lety/2=x.

d)Let3y=x.2.2Considerthat|d(AB)|=|d(BA)|butABandBAareoppositelydirected.2.4DrawthealtitudehctosideAB.2.5UseTheorem2.19.2.6ApplyTheorem2.11,usingd(AM)=d(MB).2.9ReplaceAB,BC,CAintermsofDA,DB,DC.2.10LetEbethefootoftheperpendicularfromDtolineABC.ByExercise2.9,

theformulaholdsforA,B,C,andE.SubtractthisformulafromthedesiredformulaandusethePythagoreantheoremtoshowthatthedifferenceiszero.

2.12InStewart’stheorem,replaceA,B,C,DbyB,L,C,A,respectively,intriangleABCformedianAL.RecallthatBL=LC.

2.13SeeHint2.12.InthiscaseBL/LC=c/bbyExercise2.5.2.16ShowthattheareasrepresentedbythetermsofEuler’stheoremadd

togetherastheyshould.3.1b)Considertheidealplane.3.2b)Considertheidealline.

d)LetmbetheideallineandPanordinarypoint.3.3d)Threenoncollinearidealpointsoccuronlyinextendedspaceintheideal

plane.Itcannotbeshowninafigure.3.6Onemustviolatethewordpreceding“numerators”intheproofofTheorem

3.5.3.8ApplyTheorem2.19twicetotriangleVAB.

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3.12SeeTheorem3.9.4.1Use2.18.4.2Wemustagreethat,ifCistheidealvertexoftriangleABC,then0·CX=0

foranypointX;ifLandMareordinarypoints,thenLC–CM;andifL(orM)isatC,thenLC(orCM)iszero.

4.4StartwithTheorem4.2anduseTheorem2.19.4.6UseExercise4.5sixtimes.4.7UseExercise4.2.4.8UsethetrigonometricformofMenelaus’theorem,showingthat,whenAL

istheexternalanglebisectorofangleBAC,thenanglesBALandLACaresupplementaryinmagnitudeandoppositeinsense.

4.10UseTheorem4.2.ShowthatBL=L’C,etc.4.11UseTheorem4.3andideasanalogoustothoseofHint4.10.4.12Youmustshowthat,ifthefourgivenpointsarecollinear,thenthegiven

equationholds.DrawdiagonalACtocutlineLMNOatpointX.ThenapplyMenelaus’theoremtothetwotrianglesABCandACD.

4.13Tangentsfromapointtoacirclearecongruent.4.14UsetriangleABA’cutbylineCPQ.4.16SeeAnswer4.15.4.17SeeAnswer4.15.5.1SeetheproofoftheconversetoTheorem4.2.5.2AssumeExercise4.1.5.3UseTheorem2.19.5.4Show,forexample,thatBL/OL=AS/AO,CM/MA=BC/AS,andtwosimilar

relations.5.6SeeTheorem5.9.Notethatthisconstructionrequiresonlythestraightedge.5.7UseTheorem5.9.5.10LetthecommonratioAO/OL=BO/OM=CO/ON=r.Thenusetriangle

BOCandceviansBN,CM,andOL.5.11UsethetrigonometricformofCeva’stheorem.5.12SeeHint5.11.5.13SeeExercise4.11.5.14UseTheorem5.2andrecallthattheproductofasecanttoacirclefroman

externalpointPanditsexternalsegmentisaconstant.Thus,forexampleAM·AM′=AN·AN′(seeExercise6.18).

6.4LetObethemidpointofdiagonalACinparallelogramABCD.ShowthattrianglesABOandCDOarecongruent.

6.6InFig.6.3,showthatAA′andB′C′bisectoneanother.6.8SeeTheorem6.10.

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6.12SeeTheorem6.15.6.14UseTheorem6.20.6.16b)LetPTbeatangentandPABasecanttoagivencircleanddrawlineAT.

Then TABisanexternalanglefortrianglePTA.Nowapplypart(a).6.18SeeHint6.16(b).6.20UseTheorems6.15and6.16.6.21UseExercise6.20,Theorem6.15,andCorollary6.19.6.22UseTheorem6.20.7.2UsethelastparagraphintheproofofTheorem7.2.7.3UseCorollary7.4.7.8SeeTheorem29.4.7.10UseTheorem7.17.7.11Drawanaccuratefigure.7.13Showtheyarediagonalsofaparallelogram.7.14UserighttriangleBA’OinFigure7.16.7.18UseExercise6.9.7.20Startwiththeright-handsideoftheequationandusetheindicated

theorems.8.1a)DrawAXandBX.

b)Usesimilartriangles.f)DrawAC,CQ,CP,PD,BD,andDQ.g)WhatkindoftriangleistrianglePQC?

8.2UseProblem8.12.8.4Formanisoscelestriangleonthegivenangleasabaseangleandthe

congruentsidesthesamelengthasthecompassopening.TransferbyProblem8.19thebasedistancetothegivenpointonthegivenline,etc.

8.5OnalinemarkAPandPBoflengthsaand1.OnaparallellinedrawasemicircleA′B′usingthegivencompassopeningasradius.Useproportion.

8.6Thelineardimensionswillbetwo-thirdsofthegivendimensions.8.10Placetheisoscelestrianglesothatoneofitscongruentsidesisthebase.8.11Firstdrawafigureshowingthecompletedtriangleonwhichthegivenparts

havebeenmarked.8.14Thefirststatementinthe“proof”isnottrue.8.16h)Withthesetoolsonecanconstructonlytheintersectionsofcirclesand

lines.9.2SeeEves,SurveyofGeometry,vol.1,pages44,50,233,and236,for

example.9.3b)Startwithahexagonandapplytheformulaofpart(a)fourtimes.9.4Theformulaofpart(a)isreadilyfound,butthatofpart(b)isfartoo

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complicatedtowriteexplicitly.Theperimeterofthecircumscribedhexagonbeing2 ,thatofthe96-sidedpolygonis3.1427.

10.4Recallvectoraddition(seeDefinition33.6).10.6Ifthecorrespondingsidesofthetrianglesarenotparallelandsimilarly

directed,thecenterofrotationistheintersectionoftheperpendicularbisectorsofthesegmentsjoiningcorrespondingverticesofthetriangles.

10.8UseExercise10.1.11.8See11.8.11.9See11.8.11.10LetthetranslationcarrypointAtopointB,andletanarbitraryrotation

throughangleθcarryAtoapointC.Findtherotationthroughangle–θthatcarriesCtoB.SeeExercise10.6.

11.11Arectangleisformed.Whatistrueofitsdiagonals?12.3Tryatranslationandareflectionorrotation.12.5Boththesemapsareinversetoα–1.12.6Boththesemapsareinversetoβα.12.7b)Letαbeinvolutoric.12.8Multiplybothsidesofeachgivenequationbyα–1.12.9Multiplybothsidesofeachgivenequationbyγ–l.12.10Multiplybothsidesofthegivenequationbyα–1ontheleftandbyβ–lonthe

right.12.12Inproperty(3),letβ=α–1.13.2ApplyTheorem13.6.13.10Seewhathappenstothesenseofagiventriangle.13.12ConsiderthecirclesP(s)andQ(t)forpart(a).13.14d)SeeTheorem13.13.14.1SeeFig.10.6.14.5UseTheorem13.13.14.10Factortheglide-reflectionintoaproductofreflectionsinthreelines,the

thirdonebeingperpendiculartotheothertwolines.Showthatiflinesmandnareperpendicular,thenσnσm=σmσn(byconsideringwhatrotationeachoftheseproductsrepresents).

14.18Findlinesa,b,csothattherotationisσbσaandthetranslationisσcσb.Theirproductisσcσa,etc.

15.2Lookatthegeometricpicture.15.4UseAnswer15.3asaguide.15.6IfABCDisaparallelogram,thenσDσc=σAσBbyTheorem15.11.

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16.2SeeHint14.10.16.4RecallExercise12.12.16.6Eliminatetranslationsandglide-reflections.16.7UseTheorem16.12andshowthatnonontrivialrotationhasmorethanone

fixedpoint.16.11ConsiderthelocationoftheimageofP.16.12FactorαAintoaproductαPαqwithpparalleltom.16.13b)RecallTheorem14.15.17.4SeeTheorem16.1.17.8SeeTheorem15.12.17.10SeeTheorem15.11.17.12SeeTheorem15.7.17.16Thisproductfactorsintosquaresofproductsoffourreflectionseach,the

firstonebeing(σaσbσcσd)2.18.1UseamapsimilartothatfortheproofofTheorem18.4.18.2Maponetriangletotheothersothatthecongruentanglescoincideandthe

triangleslieonthesamesideofthatcommonside.18.3Maponetriangletotheothersoastoformanisoscelestriangle.18.6b)ThisisTheorem2.3.18.12Reflectinthebisectoroftheapexangle.18.13SupposethatmedianAMisperpendiculartosideBC.18.14InrhombusABCD,σm(ΔABC)=ΔADCwheremislineAC.19.2Reflectinthatdiameter.19.6ThisproofisquitelikethatofTheorem19.10.19.8Considerthesumofpairsofoppositeangles.19.9ShowthatσC,σCσB,σBσA,σA= byfindingtheimageofpointAunderthis

productofhalfturnswhichisatranslation.19.10Rotatethetriangleahalfturnaboutthemidpointofitshypotenusetoforma

rectangle.19.14Placeyourfirstcoininthecenter,thenusecentralsymmetry.19.15Usesymmetryinaline.20.8PointsL,M,NarethemidpointsofthesidesoftriangleABCandthe

midpointsofthesegmentsjoiningtheverticestotheorthocenteroftriangleDEF.

20.10Sincethefigureisaparallelogram,ithasacenterofsymmetry.20.12TriangleACHisthemedialtrianglefortriangleQDR.20.14Reflecteitherthehouseorthebarnintheriverbank.Thensolvethe

problem.

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20.15Applyahalfturnaboutthemidpointofthesideofthetriangle.20.16ApplyExercise20.15foronesolution.Thereisanother(trivial?)solution.20.17Drawthediameterperpendiculartoitsbase.20.18Theiraltitudestothecommonbaselinearecongruent.20.19ShowthatσC″σA=σAσB″.20.20Showthatarotationaboutthegivenpointleavesthecirclefixed.20.21DrawlinesOAandOB.20.22a)Usea90°rotationaboutM.

b)AssumethattriangleABCiscounterclockwiseoriented.Letαandβdenote90°rotationsaboutXandY,respectively.ThenσB,αβ(C)=C,soσB,αβ= becauseitisatranslation.Hence(αβ)(α′)=B′.Letα(B′)=B″sothatα(B″)=B′.ThentrianglesB′XB″andBYB′areeachisoscelesrighttriangles,etc.

c)Draweitherdiagonalofthequadrilateralandusepart(b).d)Atrianglecanbethoughtofasadegeneratequadrilateral.

21.6Recalltheformulasforsin(θ+ϕ)andcos(θ+ϕ).21.10ProceedasinExercise21.9.21.14Findangleθsothatsinθ=b/(a2+b2)1/2andcosθ=a/(a2+b2)1/2.One

methodistosetθ=2Arctan([(a2+b2)1/2–a]/b).21.16ForthelineusetheequationsofExercise21.9.22.4ThisissimilartoExercise22.3.22.6Areflectionisinvolutoric.22.8Sinceonlythetranslationisreversed,replacerby–rinTheorem22.9.22.10SeeHint22.8.22.12StartwiththeequationsofAnswer22.9.22.14SeeExercises23to25and22.11to22.13.22.16SeeTheorems21.6and22.4andExercise22.14.22.18Fixedpointsare(0,0)and(3,3).23.1Considerthevariousplanesinwhicheachsideliesandthoseinwhicheach

pairofcorrespondingsideslie.23.2Extendeveryothersideofthehexagontoformatriangleandapply

Menelaus’theorem(Theorem4.2)severaltimes.23.4Thevaluesforπgivenby1000and1001factorsare3.1400and3.1431,

respectively.24.4SeeTheorem25.7.24.6Reflectinthebisectoroftheangleformedbyapairofcorrespondingsides

(extended).24.8SeeExercise24.7.

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24.10Itsratioisnegative.24.12Therearenonewithpositiveratios.25.6SeeTheorem25.5.25.8ShowthatQB″/B″B′=j/(1–j)andB′B/BO=k–1.UseMenelaus’theorem

ontriangleOQB’togetOP/PQ=(j–1)/(j(k–1)).25.12Tofindthevectoroftranslation,locatetheimageofpointAunderthis

productofhomotheties.25.16ByTheorem25.9.Onemustalsoshowthattheproductofatranslationand

ahomothetyisahomothety.ConsidertheimageofagivensegmentABundersuchaproduct.

25.20UseTheorem25.16.26.2SeetheproofofCorollary13.7.26.6SeeTheorem13.20.26.10LetnbetheotherbisectoroftheangleatQ.26.12ReflectfigureOA′B′CDinlinem.26.14ThisisacorollarytoExercise26.13.26.15a)Whatisometryleavesacirclefixed?

b)Whatoppositeisometrymapsacircletoitself?27.2Whatfigureisformedbytheunionofallfourimages?27.4Rememberthatareasofsimilarfiguresvaryasthesquaresofthelinear

dimensions.27.6LetthemidpointsofACandBDbeMandNinFig.27.4.Thedesiredresult

canbeobtainedbysolvingalgebraicallytheratiosresultingfromthefactsthatH(E,EA/EM)mapsMNtoABandH(E,EM/EC)mapsCDtoMN.Thealgebraissomewhatlengthy.

27.8SeeProblem27.7.27.10SeeProblem27.7.27.12Itisopposite.27.14SeeTheorem27.9.27.16SegmentsABandA′B′areparallel.28.2SeeTheorem28.1.28.3LetH(O,2)maptriangleABCtoA′B′C′.28.4SeeTheorem27.10.28.5UsetrianglesDECandFEA,andDEAandGEC.28.8ShowthatifH(M,k)mapsXYtoA′CandH(N,j)mapsXYtoBA′,thenj=k.

DeducethatMandNareequallydistantfromBC.28.9SeeTheorem28.6.28.10Useanappropriaterotationandhomothety.28.12UseProblem28.10.

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28.13Startwithasquarethatsatisfiesalmostalloftheconditions.28.14UsethemethodofAnswer28.13.28.18a)Centeranysimilarrectangleatthecenterofthecircle.

b)Restanysimilarrectanglesymmetricallyuponthediameter,d)Adiagonalisanaxisofsymmetry.

29.1UseTheorems29.2and29.4.29.2WhatdoesH(G,–2)obviouslymapintowhat?29.7Reflectoneofthecirclesinthelinetangenttoitatthegivenpointof

intersection.29.8Useahomothetyofratio–rinsteadofthereflectionrecommendedinHint

29.7.29.9DrawadiameterAOBofthesmallestcircle.Seewhatistrueifthedesired

secantterminatesatA.29.10RecallExercise2.5.29.11Firstusetheanglesizeandthenonparallelsidelengths.29.12Drawanysquarefirst.29.13Donotworryabouttheperimeteratfirst.29.14Showthattheratioofhomothetyis1/3.29.16UseExercise29.14.29.17UseTheorem7.20.29.18UsingamidpointofasideoftriangleABCasacenterofhomothetyofratio

1/3,findtheimageoftheoppositevertexandtheorthocenter.29.21UseExercise29.19.29.22ChooseanarbitrarypointononeofthelinesandapplyExercise29.21.30.4RecallExercise22.14.31.4SeeDodge,Sets,Logic#x0026;Numbers,pages249–251orDodge,

NumbersandMathematics,pages345–347.32.1a)Assume–1>0istrueandapplyproperty(3),thenproperty(1).

b)Usepart(a).c)Usei2=–1andthemethodofpart(a).

32.6SeeSections33and34.32.10b)7= .c)2= .d)Usethemethodofpart(c).f)SeeDodge,Sets,Logic&#x0026;Numbers,Exercises60-5orDodge,NumbersandMathematics,Exercises90-5.

h)Comparewitha2+b2,whichdoesnotfactoralgebraically,butwhichisequaltothesquarenumber25whena=3andb=4.

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33.4Useordered-pairnotationforthevectors.33.10ThisissimilartoExercise33.9.33.12SeeAnswer33.11.33.13UseExercise33.11.33.14SeetheproofofTheorem33.18.33.15Thevectorsummustbezero.33.16SeeExercise34.14.34.2UseFig.34.6andtrigonometry.34.6SeeTheorem34.10andAnswer34.5.34.11AssumeBEandCFmeetatO.TolocateO,equateexpressionsfor

and intermsofvectorsu= andv= ,andtheratiosBD/DC=m/n,CE/EA=n/p,andAF/FB=p/m.Thenshowthat

isascalarmultipleof .34.14RecallExercise2.5andExercise33.12.34.15UseExercise34.1434.16ReplacethevectorsuandvofExercise34.14intermsof , ,and .34.19Ifuandvarethevectorsdeterminedbythesides,then|u|=|v|.34.20LetMbethemidpointofsegmentABandletPbeanypointonthe

perpendicularbisectorofAB.Then and areorthogonal.Write||and| |intermsof and .

34.21UseExercise33.11.IfthetwocongruentmediansareBB′andCC′,let=uand =vandset .

35.4Multiplyzbyz–1toget1.35.6SeeExercise21.6.35.7UsemathematicalinductionandExercise35.6.35.8UseExercise35.7andDefinition35.17.35.9UseExercise35.6.35.10Giventhatnisnegative,useExercise35.9.35.14UseDefinition35.17.35.16SeeAnswer35.15.35.18ApplyDeMoivre’stheorem(Exercise35.7).35.19through35.23UseExercise35.18.35.24d)Thisisacorollarytopart(c).35.25Foradifferentproof,seeExercises36.28through36.30.35.26Eliminate fromthetwoequations2z+ =5+iand2 +z=5–i.35.27through35.29UsethemethodofExercise35.26.35.30UsingthemethodofExercise35.26,fromthegivenequationandits

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conjugate,obtain ,whichhasnosolutiononlywhenthecoefficientofziszero.Observethatifa +biszeroalso,thenallcomplexnumberszsatisfythegivenequation.

36.2ApplytheisometriesofSections21and22.36.4Usesimilartriangles.36.6Itsufficestoprovethetheoremforasecond-orderdeterminant,soadda

multipleofoneroworcolumntotheotherandevaluatetheresultingdeterminant.

36.7SeeHint36.6.36.10Parts(a)and(b)aretruebyTheorem36.17,butnotethatthisexerciseis

somewhatstrongerinstatingtheproportionalityconditionasbothnecessaryandsufficient.

36.12TrianglesDEFand areoppositelycongruent.36.14UseTheorem36.19andaddappropriatemultiplesoftwooftherowstothe

thirdrowsoastoreducetheelementzb+(1–z)atozero.ThenapplyCorollary36.16.

36.15UseExercise36.5asastartingpoint.36.18ShowthatthedeterminantofCorollary36.20isequalto

Adding and subtracting rows so these factors appear as elements of thedeterminantsimplifiesthealgebra.

36.19ShowthatthisexpressionisequivalenttothatofExercise36.18.36.20Howcanatrianglebebothclockwiseandcounterclockwise?36.21Assuminga=0(forconvenience),deducethat eisreal,sothatc=kbfor

somescalark.Nowseewhatmustbetrueofb.36.22SeeHint36.6.36.23SeeHint36.6.36.25Textsoncollegealgebra,linearalgebra,ormatrixtheorygenerally

investigatedeterminantsindetail.UseExercise36.22.36.27f)Thealgebraicexpressionforeitherproductissymmetricinbandc.36.28SeeAnswer36.29.36.30UseExercises36.28and36.29.SeealsoAnswer35.25.37.2Showthattwopointsonthatlinesatisfythatlinearequation.37.6UseTheorem37.4.37.10StartwithTheorem37.10,addthesecondcolumntothefirst,andfactor2

fromthefirstcolumn.Thensubtractthefirstcolumnfromthesecondandfactorifromthesecondcolumn.

37.12Showthatanylineconcurrentwiththefirsttwolineshasanequationofthe

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form

forsomerealconstantshandk.Thedeterminantformedfromthefirsttwoequationsandthislinearcombinationequationiszero.

37.14UseExample34.17.37.16MultiplythethirdrowbyiandapplyExercise37.2.?37.18b)UseExercise36.14.37.20ApplyExercises37.18and37.19.38.4TakeabsolutevaluesintheequationofTheorem38.9.38.5UseTheorem38.8.Thenletk=1.38.6SolveforzandapplyTheorem38.3.38.13Showthata–b=a(1– b/a ).38.14Thesolutionsarequitesimple.38.16Ifrissucharoot,then–ristheother.Nowz2–r2isafactorofthe

equation.Theotherrootsare(1±i )/2.38.17LetQdenotethecircumcenteranduse|a–q|=|b–q|=|c–q|=R,the

circumradius.39.14ThisissimilartoTheorem39.6.40.2UseCeva’stheorem.40.3UseCeva’stheorem.40.8SeeAppendixA.41.6IfthereisnorayCPotherthanCAtocurvec,thenccoincideswithlineCA.

Thiscaseisnotdifficult.WhentheraysCPandCQlieonoppositesidesofrayCA,thenconsideranglesPAA′andP′A′A(solongasA≠A′)andanglesQAA′andQ′A′A.

41.8SeeTheorem42.41.15UseExercise6.18.41.16ThisisaconversetoExercise45.41.18a)LocatetheinverseS″ofthecenterSofcircles.ApplyExercise47.

b)UseExercise47.42.3Drawtheirlineofcentersandtheradiitooneoftheirpointsofintersection.42.4SeeExercise45.42.6SeeTheorems40,43,and44.42.8Usesimilartriangles.42.10InFig.42.9,letlineCPP′meetABatT.Thenuserighttriangles.42.11UseTheorem42.8.42.12Modifythegivenprooftoshowthatthelistedratiosarestillequal.42.13FortheconstructionuseTheorem42.3andExercise42.11.

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42.14SeeTheorems42.3and42.4.42.15DrawtherayfromthecenterofcirclestopointP.42.16Considerwhathappensifthecurvesaretangentatthecenterofinversion.43.6Ifαandβareinverse,thenβα= .43.8UseTheorem43.9.43.12Findthebilineartransformationthatmaps0,1,and–1tothedistinctreal

imagesr,s,andt.Considerthetwocasesr=∞andr≠∞.44.2b)Thisisaconversetopart(a).

d)UseExercise6.18.h)LetthecircleshavecentersSandT,andletQbethefootoftheperpendiculartoSTfromanypointPontheradicalaxis.Showthat(SQ)2–(TQ)2isequaltothedifferenceofthesquaresofthecircles′radii.HencethelocationofQisindependentofthelocationofP.

44.3SeeTheorem44.1andExercise44.2.44.4IfthecircleonB,C,DdoesnotpassthroughA,then(inTheorem44.3)B′,

C′,D′lieonacircle,sothatB′C′+C′D′>B′D′.44.5UseTheorem42.3.44.7UseExercise47.44.8Thecenteristhecenterofsimilarity.44.10Replacerrby1intheequationsofExercise44.9,thensolveforxandyin

termsofx′andy′.Byrecallingthatinversionisinvolutoric,muchlaborissaved.

44.11UseExercises44.5and44.6.44.12Invertinthepointofintersectionofthetwocirclesorthogonaltothegiven

circleandeachpassingthroughtwooppositepoints.44.15InvertinthecirclecenteredatOandorthogonaltocirclesn.44.16Invertthecirclesintheirpointofconcurrenceandlookattheresulting

triangle.44.17DrawacircleonABasdiameter.46.2SeeSection14.46.8SeeDefinition47.10.46.10SeeTheorems49.10to49.13.46.12SeeTheorem47.9.46.14SeeTheorem49.9.47.8SeeTheorem13.19.47.12SeeExercise47.11.48.8SeeTheorem49.2.48.12SeeTheorem15.11.

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48.16SeeTheorem15.12.49.8Considerthetracesofthethreeplanesinaplaneperpendiculartolinemand

seeTheorem16.4.49.10SeeTheorem47.9.50.2SeeTheorem14.9.50.10SeeTheorem17.5.50.11Consideratrianglecutfromthesidesofthetrihedralanglebyaplanenot

throughitsvertex.50.12SeeTheorems49.11and49.12.50.17SeeExercise50.2.50.20Factorthehalfturnsintoreflectionsinappropriateplanes.51.4SeeAnswer51.5.51.6Wehaveh,k,lproportionaltoA,B,C.51.8SeeTheorem23.51.10UseExercise51.8andTheorem51.10.

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ANSWERSAnswersare includedhere toalternatepartsof allquestionshavingmore thanone part and to all other odd-numbered questions. In the double-numberingsystem,thefirstnumberreferstothesectioninwhichtheexerciseappears,andthesecondnumberreferstotheexercisenumber.ThusExercise1.3isExercise3ofSection1.1.18 .1.3LetA,B,C,Ddenotetheverticesbetweensidesdanda,aandb,bandc,

candd,respectively.SincetheareaoftriangleABC,forexample,isgivenby absinB,thentheareaKofquadrilateralABCDishalfthesumofthefourtrianglescutfromthequadrilateraltwoatatimebyadiagonal.Thatis,

sincesinθ≤1forallθ.Furthermore,equalityholdsiff

1.5Eachrighttrianglewithlegs3and4hasarea6,sothetotalareaofthissquareis6·4+1=25,makingitssideoflength5.Thatis,therighttrianglehassides3,4,and5.

1.7a)11c)6

2.1ByTheorems2.10and2.11,d(AB)+d(BC)+d(CA)=d(AC)+d(CA)=0.2.3SupposethatBisbetweenAandC.Thend(AB)+d(BC)=d(AC),so

d(AB)=–d(BC)+d(AC)=d(CB)–d(CA).Theothercasesaresimilar.2.5UsingthenotationofTheorem2.19,letALbethebisectorofangleA.

Thensin BAL=sin LAC,so

2.7ByTheorem2.12,ifOiscollinearwithA,B,C,D,then20N=OC+ODand20M=OA+OB.Then

etc.2.9Thegivenexpressionisequalto

whichreducestozerowhenmultipliedout.2.11 ThisistheconversetoExercise2.8.LetMandNbethemidpointsofAB

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andCD,respectively.ThenAM=MBandCN=ND.Now

andalso

HenceMN=0,soMandNcoincide.2.13 LetALdenotethebisectorofangleAintriangleABC.SinceBLjLC=

AB/CAbyExercise2.5andBL+LC=BC,then

ByStewart′stheorem,

fromwhichweobtain

andfinally

2.15 TheareaK=ah/2,soK2=(ah/2)2.UsingExercise2.14toreplaceh2,theresultingexpressionfactorsintos(s–a)(s–b)(s–c).

3.1a)False;anideallinehasmorethanoneidealpoint.c)TrueFalse;thereisjustone.

3.2a)False;theideallinehasinfinitelymanyidealpoints.c)Truee)Falsewhennistheidealline.g)False;thereisjustoneideallineintheextendedplane.

3.3a)Anordinarytrianglehavingthreeordinaryvertices.c)DrawtwolinesthatintersectatordinarypointA.Theothertwoverticesareonthelineatinfinityinthatplane.

3.4a)( ,0)c)(0,0)e)Thepointatinfinityg)( ,0)i)( ,0)3.5OneneedonlyconsiderthevariouspossiblepositionsforpointP.

3.7a)LetMandNbethepoints.ThenAM/MB=AN/NC,soAM/AB=AN/AC.Since BAC= MAN,thentrianglesABCandAMNaresimilarbySAS.

3.9For(AB,CD)=(AC/CB)(DB/AD)=(CA/AD)(BD/CB)=(CD,AB),etc.3.11 a)If(AB,CD)=(AB,CE),thenDandEdivideABinthesameratio,soD

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=EbyTheorem3.5.3.13 Sincethemidpointofasegmentdividesthesegmentintheratio1,its

harmonicconjugatemustdividethesegmentintheratio–1.Thepointatinfinitydoesso.

3.15 a)Forexample,

c)Forexample,

4.1SupposeMenelauspointL=B,forexample,andL,M,Narecollinear.TheneitherN=BandMisarbitraryonlineAC,orN≠BandM=A.InthefirstcaseBL=0andNB=0,soby2.18theequationforMenelaus’theoremholds.Othercasesaresimilar.

Conversely,ifanyfactorinthenumeratoroftheMenelausformulaiszero, thena factor in thedenominatormust alsobe zero.All suchcasesyieldthreecollinearMenelauspoints.

4.3Notwithoutsomesortofagreementastohowtheratioinwhichanidealpointdividestwoidealpointsshallbedefined.Thereappearstobenoadvantageinattemptingtodothis.

4.5ApplyMenelaus’theoremwherePQcutssideBCatpointL.Then,ifPQisparalleltoBC,wehaveBL/LC=—1.

4.7ByapplyingExercise4.2,theproofgiveninthetextholdsforthiscasealso.

4.9SeeHint4.8.

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4.11 Wearegivensin BAL=sin L′AC,sin LAC=sin BAL′,etc.ThenapplythetrigonometricformofMenelaus’theorem.Menelaus’formulaforL,M,NisequaltothereciprocalofhisformulaforL′,M′,N′,soifeitheroneisequalto—1,thentheotheris—1also.

4.13 ByMenelaus’theorem,(AZ/ZB)(BK/KC)(CY/YA)=—1.Also,BZ≅CX,CX≅CY,andAY≅AZ,sincetangentsfromanexternalpointtoacirclearecongruent.Nowsubstitutionyieldsthedesiredresult.

4.15 LetLandNbethepointsasinFig.4.7.DrawlinesNAandNA′throughNtowardL.FromanotherpointO,notoneitherline,drawtwolinesOAA′andOBB′cuttingNAinAandBandNA′inA′andB′.DrawBLandB′L.DrawathirdlinethroughOtocutBLinCandB′LinC′.DrawACandA′C′tomeetatM.NowtrianglesABCandA′B′C′arecopolaratO,sotheyarecoaxial.Thatis,L,M,Narecollinear.

4.17 LetmandnbethelinesNAandNA′ofAnswer4.15,andletthegivenpointPbepointL.

5.1Assume(BL/LC)(CM/MA)(AN/NB)=1andletBLandCMmeetatQ.LetAQcutBCatL′.TherestoftheproofissimilartothatfortheconverseofTheorem4.2.

Onestudentgavethefollowingproofofthisconverse.Assumingthegivenequation,letALmeetBMatOandCNatQ.Then,usingMenelaus′theoremon trianglesALB andALC cutbyCQN andBOM, respectively,obtain

Bymultiplyingthesetwoequationssideforsideandsimplifying,obtainAQ/QL=AO/OL,soO=Q.Thetheoremfollows.

5.3StartingwiththeequationofTheorem5.2,wemayuseTheorem2.19tomakethereplacements

TheequationofTheorem5.3results.5.5When B=90°,thenB=D=F,sothealtitudesconcuratB.5.7ByTheorem5.9,BL/LC=—BL′/L′C,CM/MA=—CM′/M′A,and

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so

5.9IfBL/LC=CM/MA=AN/NB=randtheceviansconcur,thenr3=1,sor=1,sincerisreal.

5.11 UseTheorem5.3,eachfractioninwhichequals1inthiscase.5.13 SeeAnswer4.11.UsethissametechniquealongwithTheorem5.3.6.3Themedialtriangleisjusthalfaslarge(inlineardimensions)asthegiven

triangle,sotheircircumcirclesalsohaveradiiofratio1/2.Buttheninepointcircleisthecircumcircleofthemedialtriangle.

6.5InFigure6.7,letBCcutB′B″atX.ThenXbisectsCA′soBX= BC.SincethealtitudeoftriangleBXB′tosideBXishalfthatoftriangleABC,thentrianglesABCandBB′B″havethesamealtitudes.HenceKBB′B″= KABC.

6.7ByTheorem6.6,sinceBB′≅CC′then∆BGC′≅CGB′bySAS.ThusBC′≅CB′fromwhichAB≅AC.

6.9LetAXandAYbetheinternalandexternalbisectorsofangleBACasshownintheaccompanyingfigure.Letαandβbethemeasuresoftheanglesthusformed.Then2α+2β=180°,soα+β=90°.Thetheoremfollows.

Answer6.96.11 ThecircumdiameterSTofTheorem6.12bisectssideBC,sinceitis

perpendiculartochordBC.Thetheoremfollows.6.13 FromTheorems6.15,6.16,and6.17,wehave

fromwhichthetheoremfollows.6.15 InthefigureforExercise6.15,OC=(a+b)/2,OQ=(a+b–d)/2,QD=

d/2,andOD=(b–a)/2.SinceOQDisarighttriangle,thenapplythePythagoreantheoremtoobtainthedesiredresult.

6.16 a)LettheinscribedanglebeABC,anddrawdiameterBODandassumeit

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doesnotcoincidewithBA.ThentriangleOABisisosceles,soexternalangleAODhastwicethemeasureofeitheroppositeinteriorangleandspecificallyangleABO.Hencem( ABO)=m( ABD)= m( AOD)=m(arcAD).Similarlym( OBC)= m(arcDC),whetherornotdiameterBODcoincideswithsideBC.NowtheanglesmaybeaddedorsubtractedtogivethedesiredresultaccordingtowhetherOliesinteriororexteriortoangleABC.

c)LetchordsABandCDmeetatE.SinceAECisanexterioranglefortriangleECB,then AEC= ECB+ EBC=(arcDB+arcAC)/2.

6.17 LetthetangentsfrompointPbePTandPU,andletObethecenterofthecircle.InrighttrianglesPOTandPOU,PO=PO,OT≅OU,and T≅U=90°.Hence∆POT≅∆POUbyHL(hypotenuseandlegofone

righttrianglecongruenttothecorrespondingpartsofasecondrighttriangle).

6.19 LetthechordsACandBDmeetatE.Since ADC≅ ABCand DAB≅ DCBbyExercise6.16a,then∆ADE~∆CBE,soAE/DE=CE/BE,andthetheoremfollows.

6.21 ByExercise6.20,Theorem6.15,andCorollary6.19,

6.22 a)LetBPbethecircumdiameterfromB.Then BPC≅ A,sosin=sin BPC=a/2R.

6.23 ThisisacorollarytoTheorem6.20.7.1ByTheorem6.11,eachpairofexcenterssubtendsarightangleateach

vertexofthetrianglenotcollinearwiththeseexcenters.Thetheoremfollows.

7.3ThemidpointofIbIcisTbythelastparagraphintheproofofTheorem7.2.NowAImeetsthecircumdiameterSOTthatistheperpendicularbisectorofsideBCatpointSonthecircumcircle.ThenA=TiffAliesontheperpendicularbisectorofBC;thatis,iffAB≅AC.

7.5Sinceeachside(suchasBC)interceptsarightangleateachofthetwovertices(EandF)oftheorthictrianglenotlyingonthatside,thetheoremfollows.

7.7LetMbethemidpointofthehypotenuseABinrighttriangleABC,and

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dropperpendicularMNtolegBC.SinceACisparalleltoMN,thenBN≅CN.AlsoMN=MN,so∆BMN≅∆CMNbySAS.ThusCM≅BM≅AM.

7.9ReferringtoFig.7.6,sinceCA,CB,andCFareperpendicular,respectively,toBH,AH,andAB,thenCA,CB,andCFcontainthealtitudes(AE,BD,andHF)oftriangleHAB.HenceCisitsorthocenter.

7.11 PointPliesonthecircumcirclebyTheorem6.12andnotinsidetriangleABC.Now,whenAB<AC,thenthepointRliesoutsidesegmentABandQliesinsidesegmentAC.Thestatedcongruencesarealltrue,butinthelastequation,AB=AR–RBandAC=AQ+QC,soitisnottruethatAB≅AC.

7.13 SinceANaisparalleltoOA′,andsinceAHistwiceOA′becausetheyarecorrespondingaltitudesegmentsforthegiventriangleanditsmedialtriangle,thenANa=OA′.ThusANaA′OisaparallelogramanditsdiagonalsAA′andONabisecteachother.

7.15 InFig.7.16,OA′isparalleltoandhalfthelengthofAHbyExercise7.13.HenceAOandHA′meetatapointPsothatAP=2OPbythesimilartrianglesAHPandOA′P.ThusAPisacircumdiameterandPliesonthecircumcircle.

7.17 FortheorthocentricquadrangleABCH,finditscentroidsGa,Gb,Gc,G.Thedesiredquadrangle,sayLaLbLcL,istoABCHasABCHistoGaGbGcG.Hence,sinceGaGbGcGisone-thirdthatofABCHinlinearsize,extendeachofAGa,BGb,CGc,andHGtwiceitsownlengthtoLa,Lb,Lc,andL,theverticesofthedesiredorthocentricquadrangle.

7.19 ThisisacorollarytoTheorem7.9.8.1a)SinceAXBisarightangle,asareAPXandBPX,since XAB≅

PAXand XBA≅ PBXwehave∆ABX~∆AXP~XBPbyAA(twoanglesofonetrianglecongruentrespectivelytotwoanglesoftheother).Fromthelattertwotriangles,AP/PX=PX/BP,andthetheoremfollows,c)Since∆EHJ~∆EFG,thenEJ=2HJ,so5HJ2=HJ2+(2HJ)2=EH2=r2.ThusHJ=r/ ,soHJ=s/2.

e)Theconstructionclearlysatisfiesthegivenconditions,g)SincePCBAisaparallelogram,PC≅AB.TrianglePCQisisoscelessincethebisectorofanglePisalsoanaltitudeofthattriangle.

8.3LetthecircleA(r)cutthesidesoftheangleBACatpointsBandC.DrawcirclesB(r)andC(r)tocutcircleA(r)withinangleBACatpointsDandE.Then DAE= BAC–120°,andthebisectorofangleDAEisthebisector

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ofangleBACalso.An alternative procedure, suggested by a student, is to bisect the

supplement of the given angle and then erect a perpendicular to thatbisector.

8.5OnalinemarkAP=aandPB=1,where isdesired.OnalineparalleltoABmarkA′B′=2rwhereristhecompassopening.LetAA′andBB′meetatO.LetOPcutA′B′atP′.ConstructP′X′=((A′P′)(P′B′))1/2byProblem8.10.SinceP′X′isperpendiculartoA′B′,lettheperpendiculartoABatPcutlineOX′atX.Seetheaccompanyingfigure.Bysimilartriangles,PX= .

8.7Compute ,andmarkpointsB′andD′onsidesABandADoftrapezoidABCD,sothatAB′= AB,andAD′= AD.DrawparallelsthroughB′tosideBCandthroughD′tosideCDtomeetatC′.ThenAB′C′D′isthedesiredtrapezoid.

Answer8.58.9Letbandhbeabaseandcorrespondingaltitudeofthegiventriangle,and

letthegivensegmenthavelengthc.Constructxsothatxc=bh.Ontheperpendicularbisectorofthegivensegment,markasegmentoflengthxfromthebaseline.Thepointsoconstructedistheapexofthedesired

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triangle.8.11 a)AtapointEonabaselinemerectaperpendicularBEoflengthhb.See

theaccompanyingfigure.SwingarcB(tb)tocutmatV.ConstructatB,oneithersideofBV,raysmakinganglesequaltoB/2andcuttingmatAandC.

Answer8.11ac)AtpointEonabaselinemerectaperpendicularBEoflengthhb,asintheaccompanyingfigure.DrawcircleB(c)tocutmatA.LetcircleA(b)cutmatC1andC2.ThentrianglesABC1andABC2aresolutions.

Answer8.11ce)AtapointConalinemconstructsegmentCAoflengthbandmakingthegivenangleCwithm.DrawcircleA(ma)tocutmat and .LocateB1andB2onm,sothat bisectsB1Cand bisectsB2C,asseenintheaccompanyingfigure.BothtrianglesAB1CandAB2Caresolutions.

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Answer8.11e8.13 SinceOCAisanexterioranglefortriangleCOD,then OCA= COD+

ODC=2 ODC.Similarly, AOB= AOD+ ADO= OCA+ODC=2 ODC+ ODC=3 ODC.Sinceitrequiresuseofamarkedstraightedge,itdoesnotsatisfytheEuclideanrestrictions.

8.15 40°7′8.16 c)Thelinethrough(a,b)and(c,d)hastheequationy–b=(x–a)(d–b)/

(c–a).e)Byparts(a)and(d),sinceonlylinesandpointscanbedrawn,thenonlyrationalnumberscanbeconstructed.

g)Circlex2+y2+ax+by+c=0andliney=vx+umeetatthepoints(p,q)where

andq=vp+u.Twocirclesx2+y2+ax+by+c=0andx2+y2+dx+ey+f=0meetontheline(a–d)x+(b–e)y+(c–f)=0,iftheymeetatall,sothiscasereducestothatofacircleandaline.i)Noneofthesequantitiesinvolvesonlysquarerootsandrationaloperationsappliedtorationalnumbers.

9.1Measuringtheshadowofthepyramidrequireslocatingthepointdirectlybelowtheapexofthepyramid,whichpointisinaccessible.Fortwoshadowobservations,thedistancebetweenthetipsofthetwoshadowsofthepyramidistotheheightofthepyramidasthecorrespondingmeasurementsforthestickaretooneanother.

9.3a)LetABbeasideofthegivenpolygoninthecircleofcenterO.LetMbethemidpointofarcABandNthemidpointofchordAB.Theny=m(AM)andx=m(AB).NowusethePythagoreantheoremontrianglesOANandAMN.

10.1 SeeTheorem15.1.10.3 ThroughthevectorfromBtoA,thenegativeofthegivenvector.

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10.5 SeeTheorem14.11.10.7 180°10.9 SeeTheorem16.1.11.1 a)(5,0),(6,0),(5,2)c)(3,–3),(3,–2),(1,–3)

e)(5,–1),(5,–2),(7,–1)g)(0,0),(0,1),(2,0)i)(6,0),(5,0),(6,–2)k)(5,4),(6,4),(5,2)m)(0,0),(1,0),(0,2)

11.2 a)x=0,x= .c)y=0,y=x–3e)y=x–5,y=–3g)y=xi)x=3,y=0k)x=0,x= ,y=2m)x=0,x=0

11.3 Ahalfturnaboutitscenter,andareflectionineitherofthetwolinesthroughitscenterandparalleltoapairofsides,and,ofcourse,theidentitymapι.

11.4 a)SameasAnswer11.3.Alsoa90°ora270°rotationaboutitscenter.Alsoareflectionineitherdiagonal.

c)Theidentity,ahalfturnaboutitscenter,andreflectionsinthediagonals.e)Rotationsthroughmultiplesof60°,reflectionsinperpendicularbisectorsofitssides,andreflectionsinitsdiagonalsthatpassthroughitscenter.

11.5 a)Atranslationof5unitsinthenegativex-direction.c)Arotationof270°aboutthepoint(3,0).e)Arotationof90°aboutQ.g)Thisisself-inverse,i)Thisisself-inverse.k)Aglide-reflectionof5unitsinthenegativex-directionwithmirrory=2.

m)Theidentity.11.6 Theisometriesof(g)and(i)areinvolutoric,thoseof(a),(c),(e),(k),and

(m)arenot.11.7 a)Thereisnone.

c)4e)4g)2

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i)2k)Thereisnone.m)1

11.8 a)y=x/2c)y=–x/2

11.9 a)y=1c)y=–1

11.11TherighttriangleABCwithrightangleatCandhypotenusemidpointManditsimageunderthehalfturnaboutMformarectangleACBC′whosediagonalsarecongruentandbisecteachotheratM,establishingthetheorem.

12.1 ForanypointAintheplane,letα(A)=Bandβ(B)=C.Then(βα)(A)=C.Sinceβisone-to-one,nopointotherthanBismappedtoCbyβ.Similarly,AistheonlypointmappedtoBbyα.Henceβαisone-to-one.

ForanypointX in theplane,sinceβ isonto, there isapointYsuchthatβ(Y)=X.Similarly,thereisapointZsuchthatα(Z)=Y.Then(βα)(Z)=X,soβαisonto.

12.3 Letαbeatranslationof1unitinthepositivex-direction,andβareflectioninthey-axis.Thenβα.mapstheoriginto(–1,0)andαβmapstheoriginto(1,0).Henceβα≠αβ.

12.5 Sinceαand(α–1)–1arebothinversetoα–1byTheorem12.11,thanα=(α–1)–1byTheorem12.12.

12.7 a)Wehaveα=αι=α(αα–1)=(α2α–1=αα–1=ι.12.9 Supposeαγ=βγ.Thenα=αι=α(γγ–1)=(αγ)γ–1=β(γγ)–1=β(γγ–1)=βι=

β.Theothercaseissimilar.12.11a)Letα,β,γbeareflectioninthey-axis,areflectioninthelinex=1,and

atranslationof1unitinthepositivex-direction.c)Yes,butα=βifαandβcommute.

12.13By(1),takea∈S.By(2*),sincea∈Sanda∈S,thenι=a–1a∈S.By(2*),usingaandι,a–1=a–1ι∈S.Hencecondition(2)issatisfied.Now,forgivenaandbinS,thena–1andbareinSby(2),soab=(a–1)–1bisinSby(2*).Hence(3)issatisfied.

13.1Theorem13.3showsthatanisometryαmapsanythreecollinearpointsintothreecollinearpoints.Hencelinesmapintolines.IfPisanypointonacircleofradiusrandcenterO,letP′andO′betheimagesofPandO.Sinceαisanisometry,m(OP)=m(O′P′),soP′liesonthecircleofcenterO′andradiusr.Similarly,allsuchpointsP′atdistancerfromO′areimagesofpointsonthegivencirclecenteredatO.

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13.3 IfA,B,andA′arecollinear,thenB′isalsocollinearwiththem.LetlineABA′B′cutthemirroratF.Bydefinitionofareflection,AF=FA′andBF=FB′.ThenAB=AF+FB=FA′+B′F=B′A′.

13.5 ForanypointA,σm(A)=A′iffA=A′ormistheperpendicularbisectorofAA′.ButthenA′=A∈mormistheperpendicularbisectorofA′A,soσm(A′)=A.Thus =σm.

13.7 InFig.13.16c,segments(suchasAB)generallyparalleltothemirrormhavegenerallythesamesensesastheirimages.Thatis,ABandA′B′arebothsensedgenerally“downward”alongm.Segments(suchasCA)roughlyperpendiculartomhavetheirsensesapproximatelyreversed.Thatis,CAandC′A′aredirectedroughlytowardthemirror,butonoppositesidesofit,sowhenCAisdirectedrightward,C′A′isdirectedleftward.ThusthesenseofangleBACisoppositethatofangleB′A′C′.Itfollowsthatthesenseofanytriangleisreversedbyareflection.

13.9 Eachreflectionreversesthesenseofatriangle,soaproductoftworeflectionspreservessense.Itfollowsthatanydirectisometry,beingaproductofanevennumberofisometries,preservesthesenseofatriangle.

13.11Corollaries13.17and13.18establishthistheoremforatriangle(apolygonof3sides).Supposethetheoremistrueforanypolygonofksides,k≥3.LetA1A2…Ak+1beapolygonofk+1sidesandletabeanisometry.Thenα(A1A2…Ak)≅A1A2…Ak,andα(ΔAkAk+1A1)≅ΔAkAk+1A1byhypothesis.TheremustbeatleastonevertexAi,i<k,suchthatA1,Ai,andAkarenotcollinear.Sinceα(AiAk+1)≅AiAk+1,itfollowsthatα(Ak+1)liesonthesamesideofα(A1Ak),relativetotherestoftheimagepolygon,asdoesAk+1,relativetoA1Akandthegivenpolygon.Nowwehaveα(A1A2…Ak+1)≅A1A2…Ak+1byadditionofregions.

13.13a)A′(0,0),B′(1,0),C′(0,–3),D′(1,–1),E′(2,–3),F′(15,–12),G′(–3,5)c)A′(0,4),B′(1,4),C′(0,1),D′(1,3),E′(2,1),F′(15,–8),G′(–3,9)e)A′(0,0),B′(0,1),C′(3,0),D′(1,1),E′(3,2),F′(12,15),G′(–5,–3)14.1 Theproofgivenin10.6isquitesufficient.

14.3 ThisisacorollarytoTheorems14.2and14.4.14.5 ByTheorem13.13,anyisometryαisaproductofnotmorethanthree

reflections.Sinceitisdirect,itisaproductofexactlytworeflectionsinmirrorsmandn,arotationifmandnintersectoratranslationifmandnareparallel,theidentitymapbeingaspecialcaseofeitherarotationoratranslation.

14.7 For(βα)(A)=β(A)=(1,–2)and(αβ)(A)=α(1,–2)=(1,2).

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14.9 Itisalsoatranslation,butnotnecessarilytheinverseofthetranslationβα.14.11Areflectionisaglide-reflectionwhoseglideiszero.No,atranslationis

directandaglide-reflectionisopposite.14.13Lettheglide-reflectionbeα=σgσbσa,inwhichgisperpendiculartoboth

aandb.Thenα–1=σaσbσg=σgσaσbbyHint14.10,soα–1istheinverseofthegivenglide,followedbythegivenreflection.

14.15No,suchaproductisadirectisometry.14.16a)(0,0),(0,1),(–3,0),(–1,1),(–3,2),(–12,15),(5,–3)c)(0,0),(0,–1),

(3,0),(1,–1),(3,–2),(12,–15),(–5,3)e)(2,0),(2,1),(–1,0),(1,1),(–1,2),(–10,15),(7,–3)g)(2,–4),(2,–5),(5,–4),(3,–5),(5,–6),(14,–19),(–3,–1)14.17a)(5,0),(6,0),(5,3),(6,1),(7,3),(20,12),(2,–5)c)(–1,2),(0,2),(–1,5),(0,3),(1,5),(14,14),(–4,–3)14.19ThecenteroftherotationαβisthepointsymmetrictothecenteroftherotationβαwiththemidpointofABascenterofsymmetry.Whennandmareparallel,thatis,whentheanglesoftherotationsareopposites,theproductβαisatranslationandαβisitsinversetranslation.

15.1 Since,forgivenangleABC,anyreflectionmapstriangleABCtoacongruenttriangle,itmapsangleABCtoacongruentangle.Buteachisometryisaproductofreflections.

15.3 ByTheorem13.12, =ι,butthereexistpointsAandBsuchthatσm(A)=BandA≠B,soσm≠ι.Ofcourse,Acanbetakenasanypointnotonm.

15.5 Sincebandaareperpendicular,thenbothσbσaandσaσbrepresent180°rotationsabouttheirpointPofintersectionbyTheorem14.9.

15.7 LetmbethelineonAandB,andletaandbbethelinesthroughAandBandperpendiculartom.LetlinescanddbedrawnparalleltoaandbandspacedsothatthedirecteddistancefromatobisequaltothatfromdtocandwithpointConlinec.Finally,letlinenpassthroughCperpendiculartoc,andletncutdinD.ThenσCσBσA=σDσDσCσBσA=σDσdσnσnσcσbσmσmσa=σDσdσcσbσa=σD,sinceσdσc=σaσb.

15.8 a)(1,0),(2,0),(1,–3),(2,–1),(3,–3),(16,–12),(–2,5)c)(–8,–6),(–9,–6),(–8,–3),(–9,–5),(–10,–3),(–23,6),(–5,–11)15.9 a)(0,0),(–1,0),(0,–3),(–1,–1),(–2,–3),(–15,–12),(3,5)c)(4,6),(3,6),(4,3),(3,5),(2,3),(–11,–6),(7,11)15.11Ifαandβaretranslations,thentherearepointsA,B,Csothatα=σBσAandβ=σCσB.Thenβα=σCσBσBσA=σCσA.Now,takingDsothatABCDisaparallelogram,thenα=σBσA=σCσDandβ=σCσB=σDσA,so

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16.1 Ifa,b,careallparallel,thentakedsothatthedirecteddistancefromctodisequaltothatfrombtoa.Ifthegivenlinesareconcurrent,thenreplace“distance”by“angle”inthefirstsentenceofthisanswer.

Conversely, if σcσbσa = σd, then σbσa = σcσd, so σcσd is the sametranslation or rotation asσbσa.Hence the four lines all pass through thecenter of the rotation, or all are perpendicular to the direction oftranslation. In either case, they form a pencil and the directed angles ordistancesareasstatedinthefirstparagraphofthisanswer.

16.3 Eachdirectisometryisaproductoftworeflections.Eachoppositeisometryisaglide-reflectionbyTheorem16.4,soitcanbewrittenasaproductσgσbσawheregisperpendiculartobothaandb.NowletBbethepointofintersectionoflinesbandg,sothatσB=σgσb.

16.5 Thetransformationsofeachsetformasubsetofthoseaboveand,ineachcase,theconditionsofExercise12.12aresatisfied.

16.7 Anontrivialrotationhasjustitscenterfixed,forifarotationαhasXasafixedpoint,thenletα=σbσawherelineapassesthroughX.Sinceσa(X)=X,wehaveX=α(X)=(σbσa)(X)=σb(X),andXisafixedpointunderσb,too.ThusXliesonlineb,also.ThisispossibleonlywhenXisthecenterofrotationorwhenlinesaandbcoincidesothattherotationistheidentitymap.

16.9 Ifarotationthroughangleθisinvolutoric,thenitssquare,arotationthroughangle2θ,istheidentitymap.Thisimpliesthat2θisamultipleof360°,soθisamultipleof180°.Sincetheidentityisnotcalledinvolutoric,thenθisanoddmultipleof180°,andtherotationisahalfturn.

16.11Sincetheisometryαmapsmtomandnton,thenα(P)mustlieonm,sincePliesonmandmisafixedline.Similarly,α(P)mustlieonn,sincePliesonnandnisafixedline.Henceα(P)=P,theonlypointonbothmandn.

16.13a)Yesc)Yese)No;theproductoftwohalfturnsisatranslation.17.1 Otherdualpairsare17.4and17.8,17.5and17.9,and17.11and17.12.17.3 Ifa=borifaisperpendiculartob,thenclearlytheotherconditions

follow,mainlybyCorollary15.9.Ifσbσa =σaσb, then (σbσa)2 = , so applyExercise 16.11.The other

casesaresimilarorfollowfromTheorem16.14.17.5 Conditions(1),(2),and(3)areequivalentbyTheorems14.4and14.10.

Conditions(2)and(4)arealgebraicallyequivalent;simplymultiplyontherightbothsidesofeitherequationbyσb.

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17.7 TheseresultsareanalagoustoTheorem17.5.17.9 ThistheoremisdualtoTheorem17.5,soitsproofissimilar.17.11SinceσCσbσAisaproductofanoddnumberofreflections,itisaglide-

reflection.ThenapplyTheorem15.7.17.13ApplyTheorem15.12.17.15Thisidentityisequivalentto

whichistrue,sincethefirstandfourthfactorsareinversesofoneanother,asarethesecondandlast,andthethirdandfifth.

17.17InExercise17.16,eachproduct(σAσBσC)2isthesquareofahalfturn,henceitistheidentity.InThomsen'srelation,eachsuchproductisatranslation.

18.1 Letm(AB)=m(A′B′),m( A)=m( A′),andm( B)=m( B′)intrianglesABCandA′B′C′.LetanisometryαmapΔA′B′C′toΔABC″sothatCandC″areonoppositesidesofAB,asinFig.18.4.LettingmdenotelineAB,letσm(C)=X.SinceABisthebisectorofangleCAC″,XliesonrayAC″.Similarly,XliesonrayBC″.HenceX=C″,thepointofintersectionofthetworays.Now(α–1σm)(ΔABC)=α–l(ΔABC″)=ΔA′B′C′,andthetheoremfollows.

18.3 LetrighttrianglesABCandA′B′C′havem( C)=m( C′)=90°,m(AC)=m(A′C′)andm(AB)=m(A′B′).LetanisometryαmaptriangleA′B′C′totriangleA″B″C,withB″onrayCBandAandA″onoppositesidesofBC.LettingmdenotelineBC,thenσm(A)=A″.Thusm(BA)=m(BA″)=m(B″A″),soB=B″.Nowσm(ΔABC)=ΔA″B″C≅ΔA′B′C′.Seetheaccompanyingfigure.

Answer18.3

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18.5 a)IfAMisthemedian,thentrianglesABMandACMarecongruentbySSS.

c)IfAUistheanglebisector,thentrianglesABUandACUarecongruentbySAS.

18.6 a)LetthebisectorofapexangleAmeetthebaseBCatU,andletmdenotelineAU.NowσmmapslineABtolineACand,sincem(AB)=m(AC),σm(B)=C.Soσm( ABC)= ACB.

18.7 ByTheorem17.8,σD=σCσBσA,soσDσN=σCσBσAσN=σCσBσNσC=σNσBσCσC=σNσB,thenexttothelastequalityusingTheorem17.13.

18.9 ByTheorem18.9,σN(ΔABN)=ΔCDNandσN(ΔBCN)=ΔDAN.18.11IntrapezoidABCD,letm( A)=m( B)andletmbetheperpendicular

bisectorofAB.NowrayBC=σm(rayAD).Letσm(D)=X.ThenJHiesonrayBCand,sinceABisparalleltoCD,thenXalsoliesonCD.HenceX=C.Thatis,σm(AD)=BC,sothetrapezoidisisosceles.

18.13IfmedianAMisperpendiculartosideBC,thenσm(B)=C,wheremislineAM.ButthentriangleABCisisosceles.

18.15LettheperpendiculardiagonalsACandBDmeetatN.ByTheorem18.9,m(BN)=m(DN).LettingmdenotelineAC,thenσm(B)=D,som(AB)=m(AD)andm(CB)=m(CD).Thetheoremfollows.

19.1 LetPbeanypointonacircleofcenterO,andletmbeanydiameter.Ifσm(P)=P′,thenσm(OP)=OP′,soOP≅OP′andP′liesonthecirclewheneverPdoes.

19.3 Fortheconverse,supposechordsABandCDarecongruent.LetαbearotationaboutthecenterofthecirclecarryingABtoCD.Thatis,assumeα(A)=Candα(B)isonthesamesideofCasisD.Sinceα(B)isattheintersectionofthegivencircleandcircleC(AB),thenα(B)=D.Sincethedistancefromthecenterofthecircletoachordispreservedbyα,thetheoremfollows.

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Answer19.1319.5 Rotatethetriangleahalfturnaboutthemidpointofanyside,forminga

parallelogram.NowapplyTheorem19.6.19.7 For(σbσaσcσbσa)(A)=(σbσaσcσb)(B)=(σbσaσc)(C)=(σbσa)(D)=σb(E)=F.

Butσbσaσcσbσa=σc,soσc(A)=F.19.9 SinceσC′σCσB′σBσA′σAisatranslation,itistheidentityiffithasafixed

point.Now(σC′σCσB′σBσA′σA)(A)=(σC′σCσB′σBσA′)(A)=(σC′σCσA′σBσB′)(A)=(σC′σCσA′σB)(C)=(σC′σBσA′σC)(C)=(σC′σBσA′)(C)=(σC′σB)(B)=σC′(B)=A.Thetheoremfollows.

19.11Onetangentmapsintotheotherbyareflectioninthatdiameterthatpassesthroughthegivenpoint.

19.13LineDFisfixedundertheglide-reflectionσcσbσa(seeTheorem7.6),soDFisthemirror.PointDisfixedunderσaandmapsintoapointD′onrayFEunderσbsothatm(ED′)=m(DE).Finally,σccarriesD′topointD″onrayDFsothatFD″=FD′=FE+ED′=FE+ED.NowDD″=DF+FD″=DF+FE+ED.Seethefigureonpage262.

19.15Drawthatdiametermofthecircumcircletothepolygonthatpassesthroughtheonecoinorbisectsthesidejoiningthetwocoinsyouropponenttakesonhisfirstplay.Taketheoneortwocoinsattheoppositeendofdiameterm.Fromthenon,taketheimageinmirrormofthecoinorcoinsyouropponenttakes.

20.1 Sinceσdσgσf=σe,thenσdσg=σeσf.20.3 Since,forexample,CABUisaparallelogram,thenσCσAσBσU= ,soσU=

σCσAσB.SincethenσUσC=σCσV,thenCisthemidpointofUV,etc.20.5 TheconditionσC(U)=VstatesthatCisthemidpointofUV,etc.NowσU

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=σBσAσCimpliesσBσU=σAσC,som(BU)=m(AC),butBU= WU,etc.20.7 InTheorem6.20,sincetrianglesABDandAPCaresimilar,thenADand

APareisogonalconjugates.Thatis,thealtitudeandthecircumdiameterissuingfromavertexofatriangleareisogonalconjugates.Thetheoremfollows.

20.9 TheproofofTheorem20.15showsthatStrisectsmedianAA′.SimilarlyStrisectsmediansBB′andCC′.

20.11Asstated,(σfσeσd)(Y)=(σfσe)(Z)=σf(X)=Y,sincetangentsfromapointtoacirclearecongruent.Thusσfσeσdhasafixedpoint,soitisnotaglide-reflection,butjustareflectioninalinethroughthatfixedpointandalsothroughthepointofintersectionofthesethreemirrorsbyTheorem16.1.Thatis,d,e,andfconcur.

20.13Supposethemidpointsarenotalldistinct.ThentherearedistinctpointsPandSonmwithimagesα(P)=Qandα(S)=Tonn,suchthatsomepointRisthecommonmidpointofPQandST.NowσR(PS)=QT.Sinceσnleavesthepointsofnfixed,then(σnσR)(PS)=QT,too.ByTheorem13.20,then,α=σRorα=σnσR.IneithercasethedesiredmidpointsallcoincideatR.

20.15LetperpendicularsBB1andCClbedroppedfromtheverticesBandContomedianAA′.ThehalfturnσA′carriesA′BintoA′CandrayA′B1intorayA′C1.SincethereisjustoneperpendicularfromCtolineAA′,itfollowsthatσA′(B1)=C1.HencetrianglesBA′BlandCA′C1arecongruent,andthetheoremfollows.

20.17LetABCDbeacyclictrapezoidwithsidesABandCDparallel.ReflectinthatdiameterperpendiculartoABandCD.ThenAandDmaptopointsonlinesABandCD,respectively.Buttheyalsomaptopointsonthecircle.HencetheymaptoBandC.Thusm(AD)=m(BC).

20.19WehaveσC″=σC′σCσC′andσB″=σB′σB′byTheorem17.9.Also,

Now

usingalsoTheorem17.13.HenceσC″σA=σAσB″,soAisthemidpointofB″C″byTheorem17.9.

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20.21ThebisectorsaandboftheinterioranglesatAandBpassthroughO.Sinceσbσacarriesminton,thenσbσaisahalfturn(sincemandnareparallel).ThusaandbintersectatOinarightangle.ThusthecircleonABasdiameterpassesthroughthevertexOofthatrightangle.

20.22a)A90°rotationaboutMcarriesoneofthetrianglesAMCandBMDintotheother.

c)LetMbethemidpointofeitherdiagonalofthequadrilateral.Thenapplypart(b).Finally,applypart(a).NotethatWYandXZdonot,ingeneral,meetatM.

e)Theproofsarethesame.21.1 Theequationsforthehalfturnsare:forσO,x′=–xandy′=–y;forσA,x′=

–x+2handy′=–y+2k;forσC,x′=–x+2aandy′=–y+2b;andforσD,x′=–x+2a+2handy′=–y+2b+2k.ThenforσCσDσAσO,x′=–(–[–x)+2h]=2a=2h)+2a=x,andy′=–(–[–(–y)+2k]+2b+2k)+2b=y,soσCσDσAσO= .

21.3 SinceCisthemidpointofthesegmentjoiningP(x,y)andP′(x′,y′),thenh=(x+x′)/2andk=(y+y′)/2,fromwhichthedesiredequationsfollow.

21.5 Letαhavetheequationsx′=x+handy′=y+k,andβtheequationsx′=x+mandy′=y+n.Thentheequationsforβαarex′=(x+h)+m=x+(h+m)andy′=(y+k)+n=y+(k+n),equationsforatranslation.

21.7 Letαandβbethegivenrotations.Thenβαhasequations

andinasimilarmanner,

Fortheseequationstorepresentatranslation,wemusthavesin(θ+ϕ)=0andcos(θ+ϕ)=+1.Thisoccursonlywhenθ+ϕisamultipleof360°.

21.9 Replaceθby–θtogetx′=xcosθ+ysinθandy′=–xsinθ+ycosθ.21.11Sinceahalfturnisinvolutoric,usethesameequations.21.13Thetranslationisx″=x′+randy″=y′+s.21.15Ifa=l,thenb=0,sox′=x′+candy′=y+d.22.1 ThemirrorimageofpointP(a,b)inthex-axisisthepointP′(a,–b).Thus

x′=xandy′=–ydefinethereflectioninthex-axis.Theequationsforthereflectioninthey-axisareobtainedinasimilarmanner.

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22.3 LettingαandβdenotethesereflectionsandusingtheequationsasgiveninTheorem22.4,wehave,forβα,

andinasimilarmannerweobtain

whichareequationsforthedesiredrotation.22.5 Thisisstraightforwardsubstitution.22.7 Sameasforthegivenreflection.22.9 Theyarex′=(x–h)cos2θ+(y–k)sin2θ+h+rcosθand

22.11IntheequationsofTheorem22.6,takea=cos2θ,b=sin2θ,c=h–hcos2θ–ksin2θ,andd=k–hsin2θ+kcos2θ.

22.13Thereisanangleθsuchthata=cos2θandb=sin2θ.Thentheresultfollowsreadily.

22.15Letustakeαwithequationsx′=ax–by+candy′=e(bx+ay)+d,andβwithx′=fx–gy+handy′=j(gx+fy)+k,wheree=±1,j=±1,a2+b2=1,andf2+g2=1.Thenβαhasequations

and

Theseequationshavetheproperform.Allthatremainsistoshowthatthesumofthesquaresofthecoefficientsofxandyis1.Tothatend,notingthate2=1,wehave

22.17a)Wehavex=x′andy=–y′,andx=–x′andy=y′.c)Wehavex=(x′–h)cos2θ+(y′–k)sin2θ+handy=(x′–h)sin2θ–(y′–k)cos2θ+k

e)Nowx=(x′–h)cos2θ+(y′–k)sin2θ+h–rcos2θandy=(x′–h)sin2θ

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–(y′–k)cos2θ+k–rsinθ.22.19Thisistrue,sincesin(θ+180°)=–sinθandcos(θ+180°)=–cosθ.22.21Letthemirrorsm,n,ppassthroughthepoints(a,b),(c,d),(e,f),

respectively,eachwithinclinationθ.Then,usingequationsasinTheorem22.6,wefindthat(σpσnhastheequations

and

Itfollowsthatσpσnσmhastheequations

and

Nowtranslatethepoint(a,b)bytheproductσpσnto

=(g,h).Thenabitofalgebrashowsthatσpσhσmisareflectioninalineparalleltothegivenlines,andhastheequationsx′=(x–g)cos2θ+(y–h)sin2θ+gandy′=(x–g)sin2θ–(y–h)cos2θ+h.

23.1 LetthetrianglesABCandA′B′C′becopolaratO.Lettheplanesofthetwotrianglesintersectonlinem.ThenABandA′B′bothlieinplaneABO,anditfollowsthattheymeetatapointthatliesinallthreeoftheplanesABO,ABC,andA′B′C′.Hencetheselinesmeetatapointonlinem.Similarly,BCandB′C′meetonm,andCAandC′A′meetonm.Hencethetrianglesarecoaxialinlinem.TheconverseisestablishedasinTheorem4.7.

23.3 LetthefivepointsbeA,B,C,D,E.LetABandDEmeetatL.LetanylinethroughLcutBCatMandCDatN.NowMEandNAmeetatF,thesixth

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vertexoftheinscribedhexagonABCDEFbytheconversetoPascal′smystichexagramtheorem.

24.1Itsufficestoshowthatahomothetypreservesangles.UsingFig.24.1,letthehomothetymapangleBACtoB′A′C′.SinceB′O/BO=A′O/AOandAOB= A′OB′,thentrianglesAOBandA′OB′aresimilar.Similarly,trianglesAOCandA′OC′aresimilar.Bysubtractingcongruentangles,wehavem( BAC)=m( B′A′C′).

24.3Theratioistheproductofthetworatiosandthecenteristhecommoncenterforthetwohomotheties.

24.5Usinganycenter,rotateonetrianglesoitssidesareparalleltotheother,thenapplyExercise24.4.

24.7Lettingthecirclesofradii3and5becenteredat(0,0)and(a,0)witha>8,thecentersofhomothetyareat(3a/8,0)and(–3a/2,0).Theratiosare±3/5.

24.9BysimilartrianglesanysuchpointsOandO′dividethelineofcentersinternallyandexternallyintheratiooftheirradii.HenceallsuchpointsOandO′coincide.Thusthesepointsarethecentersofsimilitude.

24.11a)(0,0),(1,0),(0,2),c)(0,0),(–1,0),(0,–2),e)(–2,0),(0,0),(–2,4),g)(–4,0),(–2,0),(–4,4).

24.13Ahomothetywiththesamecenterandwhoseratioisthereciprocaloftheratioofthegivenhomothety.

24.14a)Obvious.c)Sincetheratioofaproductofhomothetiesistheproductoftheratios,andsincer2=1iffr=±1,andforn>2,rn=1onlywhenr=1orperhaps–1(forrealr),andthecasesforr=1andr=–1arealreadycovered,thenthesmallestsuchnisnevergreaterthan2.

25.1TheresultfortrianglesfollowsfromTheorem25.2.Forthepolygon,usethemethodofAnswer13.11.

25.3Thecenteristhemidpointofthelineofcentersandtheratiois–1.25.5ThisisadirectcorollarytoTheorem25.5.25.7ThecenterliesbetweenAandA′ifftheratioofhomothetyisnegative.

SimilarlyforBandB′.25.9Ifj≠1andk≠1,thentheproductofthetwohomothetiesinreverseorder

isahomothetyofratio(k–1)/k(j–1).25.11ByDefinition25.1,thecenterisafixedpoint.ForanyotherpointP,letits

imagebeP′.ThenOP′=koOP.ThusP=P′onlyifk=1.Likewise,ifalineisatdistancedfrom0,thenitsimageisatdistancekdfromO.HencetheonlyfixedlinesarethosethroughO.

25.13LetthehomothetyH(0,k)mapangleABCtoA′B′C′.Iftheratiokis

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positive,thenasapointP,interiorto ABC,movesawayfromthecenterOofhomothety,itsimageP′movesinthesamedirectioninside A′B′C′.ItfollowsthatanglesABCandA′B′C′aresimilarlyoriented,andhence,thattrianglesABCandA′B′C′aredirectlysimilar.

25.15ApplyTheorems25.10and25.14.25.17ByTheorem25.10.25.19ApplyTheorem25.9andHint25.16.26.1ByDefinition13.2.26.3ThisresultissimilartothatofTheorem13.4.26.5LetsimilaritiesαandβeachmaptriangleABCtoA′B′C′.Sinceαandβare

transformations(byTheorem26.4),thenβ1αisatransformationand,infact,asimilarity.Sinceα–1αmapstriangleABCtoitself,itistheidentitymap.Henceα=β.

26.7a)ThenABandA′B′areparallelsopointQisthecenterofhomothety,andtherotationreducestotheidentity.

26.9PointQwillliebetweenAandA′andbetweenBandB′.26.11Followthehintgiveninthetext.26.13Letαandβbesimilaritiesofratiosjandk.Ifα(AB)=A′B′andβ(A′B′)=

A″B″then(βα)(AB)=A″B″,andA″B″=k·A′B′=jk·AB.Thusβαisasimilarityofratiojk.Thetheoremfollows.

26.15a)Acirclecisnotasegment,andanyrotationαaboutitscentercarriescirclectoitself.So,ifβisasimilarity,thensoalsoisβαasimilarity,andβ(c)=(βα)(c).Thusthereareinfinitelymanysimilaritiescarryingonegivencircletoanother,c)Yes,infinitelymanysimilarities,e)Threeofeach,g)Twoofeach.

27.1DrawanotherparallelthroughthethirdvertexandthenapplyTheorem27.2.Alternatively,ifaparalleltosideBCoftriangleABCcutssidesABandACatMandN,thenH(A,AN/AC)mapsBCtoMN,andH(A,AM/AB)alsomapsBCtoMN,soAN/AC=AM/AB,andthetheoremfollows.

27.3AlinesegmentparalleltoBCandoflengthr/(r+1)timesthelengthofBC.

27.5ThisisacorollarytoTheorem27.4andtheratioisDC/AB.27.7AreflectioninthebisectorofangleAEDfollowedbythehomothetyH(E,

ED/EA).27.9ThisissimilartoTheorem27.5.27.11 .27.13SincethesidesoftrianglesEFAandADBaredirectlyparallel,thereisa

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similaritymappingonetriangletotheother.ThustriangleEFAisisoscelessincetriangleADBisisosceles.

27.15ApplyExercise27.14twice.28.1One-fourththeareaoftriangleOAM,or ,whereristheradiusof

thecircle.28.3Hint28.3establishesthetheorem.28.5TrianglesDECandFEA,andalsotrianglesDEAandGECaresimilar

byAA.ThenDE/EF=EC/AEandEC/AE=EG/DE,fromwhichthedesiredresultfollows.

28.7A90°rotationfollowedbyahomothetycarriesonetriangletotheother.28.9LetAUbeanexternalbisectorofangleAoftriangleABC,andusethe

proofofTheorem28.6.28.11AreflectioninthebisectorofangleBfollowedbythehomothetyH(B,

BA/BU)mapstriangleBAUtotriangleBCA.Thetheoremfollows.28.13FromanypointP′onrayCAbeyondpointAdropaperpendicularP′Q′to

BC,andconstructsquareP′Q′R′S′containingpointAinitsinterior.Seetheaccompanyingfigure.LetCS′cutBAatS.NowSisavertexofthedesiredsquare,homothetictoP′Q′R′S′withpointCascenter.

Answer28.1328.15Ingeneral,onlythreeverticesofthedesiredparallelogramwilltouchthe

giventriangle,soseeAnswer28.13aswellasProblem28.10.28.17UsethemethodofAnswer28.13andProblem28.10.29.1ByTheorem29.4,HdividesNOintheratio–1/2.ByTheorem29.2,

HG/GO=2.Then

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29.3TheyconcuratN.29.5FirstapplyTheorem29.5,thenapplyahomothetyH(O,OP/OP′).29.7DrawatangentmtooneofthecirclesatthegivenpointP.Reflectthat

circleinm,anddrawthecommonchordoftheimagecircleandtheothergivencircle.NotethatthisreflectioncanalsobeeffectedbyH(P,–1).

29.9DrawdiameterAOBofthesmallestcircleandletcircleB(2s),wheresistheradiusofthesecondcircle,cutthethirdcircleatC.ThenACisthedesiredsecant.Toshowthisfact,letACcutthesecondcircleatD.ThenOD=s,soH(O,AC/AD)=H(O,AB/AO)=H(O,2).

29.11AtangleP,ofthegivensize,constructPQandPRonitssidesoflengthsequaltothenonparallelsides.Seetheaccompanyingfigure.Thendraw,joiningthesidesofanglePandparalleltoQR,segmentsA′B′andD′C′oflengthsmandn,wherem/nisthegivenratio.TrapezoidA′B′C′D′issimilartothedesiredtrapezoid,souseahomothetytogettrapezoidABCDwhosenonparallelsideshavetheproperlengths.

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Answer29.1129.13Drawarighttrianglehavingtheproperratiooflegs,thenuseahomothety.29.15SinceOistheorthocenterofthemedialtriangleA′B′C′ofatriangleABC

underthehomothetyH(G,— ),thentheninepointcenterN′ofthemedialtriangleliesthree-fourthsofthewayfromHtoO.ByExercise29.14,H(H, )mapstriangleA′B′C′andN′intoGaGbGcanditsninepointcenterN″,soN″liestwo-thirdsofthewayfromHtoN′;thatis,N″lies( )( )=wayfromHtoO.HenceN′=N.

29.17ByTheorem7.20,theorthocentricquadrangleformedbythecircumcentershasthesameninepointcenterNas,andiscongruentto,thegivenorthocentricquadrangleABCH.Hencetheratiois1/3andthecenterisNforthehomothety.

29.19Asimilarityconsistingofahomothetyandarotationcenteredatthegivenpointmapsthegivenlinetothelocusofthethirdvertex.

29.21LetthegiventrianglebeABCandthegivenpointP.ConstructanytwotrianglesPQRandPSTsimilartotriangleABC,withQandSlyingonthefirstline.ThenRTcutstheotherlineinthedesiredvertex.

29.23ThehomothetyH(A,2)readilyestablishestheresult.29.25ReferringtoFig.6.20,rotatetriangleABDaboutAthrough90°– C=

DAC,andapplythehomothetyH(A,AC/AD).ThenADmapstoACandABmapstoAP,sincem( BAD)=m( PAC)andm( ADB)=m( ACP)=90°.ThetheoremfollowsfromthesimilartrianglesADBandACP.

30.1Sincex′=kxandy′=ky,then,ifP(x,y),wehaveOP′=(x′2+y′2)1/2=(k2x2+k2y2)1/2=k(x2+y2)1/2=k·OP.

30.3LetPbeanypoint,letαbethetranslationthroughvector(a/k–a,b/k–b),letα(P)=Q,andletH(0,k)mapPandQtoP′andQ′.Thenthevector

=k(a/k–a,b/k–b)=(a–ak,b–bk)=(a(1–k),b(1–k)),thevectorofTheorem30.3Thetheoremfollows.

30.5Letpk=aandqk=b.Thenx′=k(px–qy+c/k),andy=±k(px+qy+d/k),anisometryfollowedbyahomothetycenteredattheorigin,intowhichanysimilaritycanbefactored.

30.7Arccos inthefirstandfourthquadrants,respectively;thatis,53°8′and306°52′.

30.9SeeAnswer22.15.30.10a)c)Wehavex′=(–17x–y+7)/10andy′=(x–17y+19)/10.30.11a)Thereflectioninthex-axis:x′=xandy′=–y.

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c)Wehavex′=(13x–lly–23)/10andy′=(–11x–13y+31)/10.30.12a)Wehavex′=2y+3andy′=2x.c)Wehavex′=3xandy′=3y.e)Wehavex′=–x+ y–5andy′=– x–y+5.31.1a)Multiplythetwogivenequationssideforside.31.364days31.5e)Theinterpretationisi2=–1.32.1a)If–1>0,thenby(3),1=(–1)(–1)>0.Now(1)isviolatedsinceboth–

1>0and+1>0.c)Ifi>0,thenby(3),–1=i2>0.Butthisviolatespart(a).e)Since02≠–1,wecannothavei=0.g)Bypart(f)32.2a)5+ic)–5–ie)12+5ig) +3i/2i)–64k)Yes;therealnumbersareasubsetofthecomplexnumbers.32.3Rememberthat,whena,b,c,darerealnumbersanda+bi=c+di,then

a=candb=d.a)x= andy=–3c)x=5andy=10e)x=y=±1/g)x=–y=±1/i)x=3sandy=4s,wheres=±132.4a)Bydefinitionc)i4=(i2)2=(–1)2=+1e)1/i=i3/i4=i3=–i32.5a)ic)–ie)–1g)–1i)1k)–1m)l0)–i32.7Twounitseastandthreeunitssouth.32.9Yes,butnotarealnumber.Itisacomplexnumber.32.10a)± iItcannotbedoneintherealnumbersystem.

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c)(x+ i)(x– i).Itcannotbedonewithrealfactors,e)Thisprocedureworkswith26/65,19/95,16/64,and49/98only.

g)Twoskewlines.Itcannotbedoneinaplane.i)Threemutuallyperpendicularlines.Theycannotbefoundinaplane.33.1LettingthefourthvertexoftheparallelogramofFig.33.7bedenotedby

D,thenfromtrianglesABCandADC,weobtainv+w= =w+v.InFig.33.8,usingtrianglesACDandABD,obtain(u+v)+w=

=u+(v+w).33.3InFig.33.7,letv=(a,b)andw=(c,d).TranslatingAtotheorigin,then

wehaveA(0,0),B(a,b),andbyDefinition33.4,C(a+c,b+d)since=(c,d).Nowv+w= =(a+c,b+d)byDefinition33.4.

33.5Ifu=(a,b)andv=(c,d),thenu–v=u+(–l)v=(a,b)+(–c,–d)=(a–c,b–d),avector.Hencetheclosure.Nowu–v=(a–c,b–d)≠(c–a,d–b)=v–u.Theassociativityissimilarlydisproved.

33.7Theorem19.10establishesthisresult.33.9LetABCDbethequadrilateralhavingmidgointsM,N,0,PforitssidesAB,

BC,CD,DA.Let =2u, =2v,and =2w.Then =( +)/2=u+vand =( + )/2=(( + + – )/2=

u+v.Since = ,thenMNOPisaparallelogram.Seetheaccompanyingfigure.

Answer33.933.1133.13ByExercise33.11, ,etc.33.15Usingtheusualnotation,let =uand =v.Then =(u+v)/2,

=–u+v/2,and =–v+u/2.Now + + =0,so

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thesethreevectorsdoindeedformatriangle.33.17LetABCDbethetrapezoidwith =u, =ku,and =v.LetM

andNbethemidpointsofthenonparallelsides and .Then ==

=,establishingthetheorem.

34.1Since|v|=(a2+b2)1/2,then|v|≥0.Hence(1).For(2),clearly|0|=0.If0=|v|=(a2+b2)1/2,thena2+b2=0,soa=b=0,andv=0.For(3),|cv|=|c(a,b)|=|(ca,cb)|=((ca)2+(cb)2)1/2=|c|(a2+b2)1/2=|c||v|.For(4),from0≤(bc+ad)2obtain2abcd≤b2c2+a2d2,so

andfinally,|u+v|≤|u|+|v|.Geometrically,thispartistrivial;itsimplystatesthatonesideofatriangleisalwayslessthanorequaltothesumofthe other two sides.Hence it is often called the triangle inequality. For(5),replaceuin(4)byu–v,obtaining

so|u|–|v|≤|u–v|34.3|–v|=|–(a,b)|=|(–a,–b)|=((–a)2+(–b)2)1/2=(a2+b2)1/2=|v|.34.5Forthedistributivity,letu=(a,b),v=(c,d),andv=(e,f).Then

34.7Ifu=(a,b)andv=(c,d),then|uv|=|(ac–bd,ad+be)|=((ac–bd)2+(ad+bc)2)1/2=((a2+b2)(c2+d2))1/2=|u||v|.

34.9Ifv=(a,b),then|v |=|(a,b)(a,–b)|=|(a2+b2,0)|=a2+b2=|v|2.34.11UsingFig.34.24,letCeviansAD,BE,CFfortriangleABCsatisfy

(BD/DC)(CE/EA)(AF/FB)=+1,andletBD/DC=m/nandCE/EA=n/p,sothatAF/FB=p/m.Letu= andv= .IfBEandCFmeetatO,thentherearescalarssandtsuchthat =s(p(v–u)+n(–u)),and=t(p(u–v)+m(–v)).Equate and

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,andsetthecoefficientsofuequaltoeachother.Dothesameforthecoefficientsofv,andsolveforsandt,obtainings=m/(np+pm+mn)andt=n/(np+pm+mn).Now

,amultipleofnu+mv.Since =(nu+mpv)/(m+n),alsoamultipleof +mv,thendoespassthroughO.

34.13LetMbethemidpointofthehypotenuseABoftherighttriangleABCwithu= andv= .Then =(u+v)/2byExercise33.11,so

byTheorems34.21and34.22,Example34.17, and thePythagoreantheorem,so =| |/2.

34.15EitherapplyCeva’stheorem(Example34.24),orthemethodusedinthatexampletoestablishthistheorem.

34.17Let =uand =v.Then

byTheorems34.20and34.22.34.19Letuandvbevectorsrepresentingadjacentsidesoftherhombus.Then|u|

=|v|andthediagonalsareu+vandu–v.FromExample34.17,(u+v)(– )+( + )(u–v)=2u –2v =0.

34.21Let =uand =vintriangleABC.Thenmedians

Assumem(BB′)=m(CC′),sothat|v–2u|2=|u–2v|2.Then

fromwhichweobtain3u =3v ;thatis,|u|=|v|.35.1Eitherproductgives(–7,22)=–7+22i.35.5Use35.16andExercise35.4.35.7Bydefinition,(cisθ)1=cis1θ.Nowsuppose(cisθ)n=cisnθ.Then(cis

θ)n+i=(cisθ)ncisθ=cisnθcisθ=cis(nθ+θ)=cis((n+1)θ)byExercise35.6.Thetheoremfollowsbymathematicalinduction.

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35.9Sincecisαcisβ=cis(α+β)byExercise35.6,letα=ϕandβ=θ–ϕ.Thenα+β=θ,andthetheoremfollows.

35.11Wehaveeiθ/eiϕ=ei(θ–ϕ

35.13 =(cis60°)27=cis1620°=-–1.35.15a+bi=a—biiff2bi=0iffb=0.35.17Forz+ =2a,z =a2+b2,and(z– )/i=2bwhenz=a+bi.35.192,2cis120°,2cis240°.__35.21Wehavecis45°=( /2)(l+i)andcis225°=–( /2)(l+i).35.23cis45°,cisl35°,cis225°,cis315°;thatis,± /2± i/235.24a)|z|=(a2+b2)1/2=(a2+(–b)2)1/2=| |.

c) =ac–bd–(ad+bc)i=(a–bi)(c–di)– .

e) =a+bi.35.25|z+w|2+|z–w|2=(z+w)( + )+(z–w)( – )=2z +2w =2(|z]2+

|w|2).35.27From(5–2i)z+ =6–16iand(5+2i) +z=6+16i,eliminate to

obtainz=2–3i.35.29Since|z|isreal,thenIm(|z|+z)=Im(z)=Im(1+5i)=5.Then|z|=–z+1

+5i=–Re(z)+1.Squaringthisequation,get(Re(z))2+25=(Re(z))2+(Im(z))2=|z|2=(Re(z))2–2Re(z)+1,so25=–Re(z)+1,andRe(z)=–12.Hencez=–12+5i.

36.1ForRe(z)=x=((x+yi)+(x–yi))/2=(z+ )/2.Also(z– )/2i=((x+yi)–(x–yi))/2i=2yi/2i=y=Im(z).

36.3 Forthend+z=w,sod=w–z.36.5 Bydefinition,eiθ=cosθ+isinθ,soz=rcosθ+irsinθ.Thencosθ=

(Re(z))/r,etc.36.7 Itsufficestoprovethetheoremforasecond-orderdeterminant.Thus

and

36.9 Thenaddinganappropriatemultipleofoneoftherowsorcolumnstotheotherwillproducearoworcolumnofzeros,andCorollary36.16applies.

36.10a)Theorem36.17proveshalfofparts(a)and(b).Ifthedeterminantis

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zero,thenad–bc=0,soad=bcanda/c=b/d.Asimilarresultholdsevenwhenc=0ord=0.

c)Thedeterminantofthematrixwhoserowsare1,2,3,and1,1,1,and2,3,4iszero,butnoroworcolumnisamultipleofanyotherroworcolumn.

36.11ByTheorem36.19,thedeterminantequationofCorollary36.20holdsifftrianglesABCandBCAaredirectlycongruent,whichimpliesm( A)=m(B)andm( B)=m( C)andm( C)=m( A),andtheseequationsare

trueifftriangleABCisequilateral.Ofcourse,iftriangleABCisequilateral,thentrianglesABCandBCAaredirectlycongruent.

36.13ByCorollary36.21,thegivenconditiondetermineswhethertrianglesABCandACBareoppositelycongruent;thatis,whetherm( B)=m( C).(SeeTheorem2.3.)36.15Letb–a=zandc–a=w.Thencos BAC=cosZOW.Furthermore,iftheaffixofUis1andifv=z/w,then

36.16a)Wehave

36.17Ifa–c=(c–b)cis60°,thenangleABCis120°,etc.36.19Since–(c–b)=(b–a)+(a–c),then1/(c–b)+1/(b–a)+1/(a–c)=0

iff,clearingoffractions,

andthetheoremfollowsfromExercise36.18.36.21Wemayassume,withoutlossofgenerality,thata=0.Butifnot,then

subtractatimesthelastcolumnfromthefirstcolumnandatimesthelastcolumnfromthesecond.Thusweassumethat

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Sincebbandccarereal,thenbcisreal,sothereisarealnumberk,suchthatbc=kbb(sincebbisreal),whencec=kb.Then0=bc–cc–bb=bkb–k2bb–bb,sok2–k+1=0ifb≠0.Nowkisreal,butk2–k+1hasnorealroots.Thus6=0,soc=0,too.Thatis,thegivendeterminantconditionrequiresthata=b=c,sotriangleABCdegeneratestoasinglepoint.

36.23SeethecommentinAnswer36.7.Then

36.25Denotebyaijtheelementinthezthrowandjthcolumnofthedeterminant.DenotebyAijthedeterminantformedfromthegivendeterminantAbydeletingthezthrowandalsothe7’thcolumn.Then

wherethedeterminantisofordern.36.27a)ByExercise36.14,a*b=zb+(1–z)a.

c)For(0*1)*1=z*1=z+(1–z)z,and0*(1*1)=0*1=z.e)Forc=zb+(1–z)acanbesolvedforaorb,solongasz≠0andz≠1;thatis,when0,1,zformanondegeneratetriangle,g)A*(B*C)=(A*A)*(B*C)=(A*B)*(A*C)byparts(d)and(f).

36.29|z–w|2= == .

37.1ByTheorem36.21,triangleABCisoppositelycongruenttoitself.TheonlywaytriangleABCcanbebothclockwiseandcounterclockwiseorientedisforthattriangletodegenerateintothreecollinearpoints.

37.3ThiscorollaryfollowsimmediatelyfromthesecondproofofTheorem37.2.

37.5SinceA2liesonlineOA1iffa1/a2isreal,thetheoremfollowsfromTheorem37.4.

37.7Clearly,lineOBhastheparametricformz=tb,wheretisreal,sincethenOZ=t·OB.Thenz=a+tbisobtainedfromz=tbbythetranslationz′=z+a.

37.9LinesOBandODareperpendiculariffb=ridforsomerealnumberr;thatis,iffb/dispureimaginary.ThenapplyAnswer37.7.

37.11IntheproofofTheorem37.10,iftriangleABCisclockwiseoriented,thentheequation“ ”becomes ,in

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whichCisthenondirectedmeasureofangleBCA.ThenthesignofsinCischanged,andthetheoremfollows.

37.13 .37.15ApplyExercise37.14.37.17a)1

c)337.18a)FromTheorem33.18,g=a+((c–a)+(b–a))/3=(a+b+c)/3.

c)ByCorollary36.20,sincetriangleABCisequilateral,wehavethedeterminantequationshownattheleftbelow.ThecorrespondingdeterminantfortriangleA′B′C′isshownattheright.Wehave

WecanshowthisseconddeterminanttobezerobyapplyingExercise36.23severaltimes.

37.19Wehave(c–a)/(b–c)=r.38.1Forr2=|z–c|2=|(z–c)(z–c)|=zz–cz–cz+cc.38.3Letz1,z2,z3bethevaluesfort1,t2,t3,Thenwehavethreeequationszk=

(atk+b)/(ctk+d),whichreducetotka+b–tkzkc–zkd=0fork=1,2,3.Thesethreeequationsinthefourunknownsa,b,c,dmaybesolvedtoyieldvaluesofthreeoftheunknownsintermsofthefourthunknown.Thealgebraforthegeneralcaseissomewhatmessy.

38.5Theorem38.8takescareofthecasewhenk≠1.Fork=1,wehave|z–a|=|z–b|;thatis,m(AZ)=m(BZ),soZliesontheperpendicularbisectorofAB.

38.7(ad–bc)(cd–cd)and|(ad–bc)/(cd–cd)|.38.9ForthecirclewithcenterEandradiusr,rreal,useTheorem38.5,letting

z0=e+r,z∞=e–r,andz1=e+ir.Thenletd=z.Nowc=1,b=i(e+r),anda=e–r.

38.10a)z=tc)z=(4+2i)t/((1+i)t+2–2i)e)z=((2+4i)t+5–3i)/(2it+1–4i)38.11Whend≡0,theequationbecomes(ct+d)z=(b/d)(ct+d),sincea=bc/d.Ifalsoc≠0,thenallpointsintheplanesatisfywhent=–d/c.Ifc=0,thenz=b/distheonlypointthatsatisfiestheequation.Asimilarsituationexistswhend=0.

38.13Ifa≠0,thena–b=a(1–b/a)=a(1–ab/aa).So,when|a|=1,then|a–b|=|a||1–ab|aa|=|1–ab|andthetheoremfollows.Thesituationis

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similarwhen|b|=1.38.15c=(1+i)a–biandd=(1–i)b+ia,c=(1–i)a+biandd=(1+i)b–ia,

orc=(a+b+bi–ai)/2andd=(a+b–bi+ai)/2.38.17ThecircumcenterQandthecircumradiusaregivenby

38.19z=(ca–cb)/(a–bb).39.1ByDefinition33.6.39.3ByTheorem39.1andthediscussionin34.12.39.5Becauseeachdirectsimilaritycanbefactoredintoaproductofoneor

moreofthethreeformslistedintheproofofTheorem39.4.39.7Wehavez′=(z–a)e–iθ+a,z″=z′,andz″′=(z″–a)eiθ+a.Hence,since

aisreal,z″′=((z–a)eiθ+a–a)eiθ+a=(z–a)e2iθ+a.39.9Ifz′=a(az+b)+b,thenab+b=0anda2=1.Sincez′≠z,thena=–1,

whencebisarbitrary.Thuswehavez′=–z+b,anequationforahalfturn.39.11Thisequationrepresentsareflectionintherealaxisfollowedbyadirect

similarity,aformintowhichanyoppositesimilaritycanbefactored.39.13z′=((a–b)/(a–b))z+(ab–ab)/(a–b).39.15Ifz′=z+bandz′=z′+c,thenz′=z+(b+c),atranslation.39.17z′=–z.39.19z′=–z.40.1a)DrawcircleB(A)(withcenterBandpassingthroughpointA).Draw

circleA(B)tocutcircleB(A)atP.DrawcircleP(B)tocutcircleB(A)atQ,anddrawcircleQ(B)tocutcircleB(A)atC.Seetheaccompanyingfigure.c)DrawC′thereflectionofCinlineAB(bypart(b)).ThencirclesC(D)andC′(CD)meetatthedesiredpoints.

Answer40.1a40.3ChoosepointCsothatsidesACandBCoftriangleABCarecutbythe

otherparallelinpointsMandL,andletBMandALmeetatOasintheaccompanyingfigure.ThenCObisectsABbyCevA′stheorem.

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Answer40.340.5Startingfromanyoneofthesixpoints,thereare5choicesforthesecond

vertex,4forthethird,3forthefourth,2forthefifth,and1choiceleftforthelastvertex.These120=5!possibilitiesappearinduplicatepairssinceABCDEFandAFEDCB,forexample,giverisetothesamehexagon.Thenumberofhexagonsisthus5!/2=60.

40.6a)Constructanequilateraltriangle.40.9Propertiesofcirclesandlines,anglesbetweencurves(andorthogonality),

andcrossratiosaresometopicsstudiedininversivegeometry.41.1a)(6,8)c)(10,0)e)(100,0)g)(5,5)i)(200/53,–700/53)k)(0,0)41.2

a)(70/13,40/13)c)(10,0)e)( ,0)g)(6,2)i)(215/58,–175/58)k)(5,0)41.3Letαdenotetheinversionandletα(P)=P′.ByDefinition41.4,ifP≡CandP≡∞,thenCP·CP′r2.ThenCP′·CP=r2andPliesonrayCP′.Henceα(P′)=P.Alsoα(C)=∞andα(∞)=C.Thusαisnottheidentity,andawillbeinvolutoricwhenwehaveprovedittobeatransformation.Tothatend,letPbeanypointintheinversiveplane,andletα(P)=P′usingDefinition41.4.SincethereisjustonepointP′onrayCPsuchthatCP·CP′=r2,thenP′isuniqueandαisone-to-one.Toshowthatαisonto,notethatα(P′)=P,soP′isthepre-imageofP.Thusαisatransformationoftheinversiveplane.

41.5SeeTheoremsandCorollaries13.17,13.18,15.1,and26.5,41.7ByDefinition41.4,eachpointonsuchalinemapsintoanotherpointonthatline,andbyTheorem41.5,thislinemapsontoitself.

41.9Seetheaccompanyingfigure.Theminussignswillchangetoplussigns.41.11Seethefigure.41.13Seethefigure.

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%

Answer41.9

Answer41.1141.15LetcirclescutthecirclecofinversionorthogonallyatpointsTandU,

andletCbethecenterofc.LetPbeanypointotherthanTandUoncirclesandletCPcutsagainatP′(Fig.42.3).NowCP·CP′=CT2byExercise6.18.HencePandP′areinversepoints.

41.17ByDefinition41.4,CQ′=r2/CQ.ByTheorem41.9,trianglesCPQandCQ′P′aresimilar,soPQ/Q′P′=CP/CQ′.HencewehaveQ′P′=PQ·CQ′/CP=(PQ/CP)(r2/CQ)(SeeFig.41.9.)41.18a)LetSbethecenterofthecirclesofFig.41.14,andletS″beitsimage.ByExercise41.17,A′S″=(r2·AS)/(CA·CS)andS″B′=(r2·CB)/(CS·CB).ThenA′S″/S″B′=CB/CA,sinceAS≅SB.SinceCB CA,thenA′S S″B′,soS″isnotthecenterofcircles′.

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Answer41.1342.1LetcirclescandshavecentersCandSandletthemmeetatpointsPand

Q.ThecirclesareorthogonalifftheirtangentsatPareperpendicular(bydefinition),ifftheirradiiCPandSPareperpendicular(sinceeachradiusisperpendiculartothetangenttothatcircle),iffeitherradiuscoincideswiththeothertangent.

42.3LetthetwocirclesmeetatPandQ.TheyareorthogonaliffradiiCPandDPareperpendicular.ButthisistrueiffCDPisarighttrianglewithrightangleatP;thatis,iffr2+s2=CD2.

42.5For1/AB–1/AP=1/AP′–1/ABiff2/AB=1/AP+1/AP′iff2/AB=(AP+AP′)/(AP·AP′)iffAB=2·AP·AP′/(AP+AP′).Also,CP/CB=CB/CP′iffCP·CP′=CB2iffCB=(CP·CP′)1/2.

42.7a)TakeoneadditionallinkQPsuchthatm(CA)–m(AP)<2m(PQ),andfixpointQsothatm(CQ)=m(QP)(seeFig.42.9).NowPtracesacircle(circleQ(P))thatpassesthroughC,soP′,itsinverse,tracesastraightline.

42.9IfAandBseparateCandD,thenoneofCandD(sayC)isbetweenAandB,andtheother(inthiscase,D)isoutsidesegmentAB.Thenoneratio(AC/CB)ispositiveandtheotherratio(AD/DB)isnegative.Nowtheirquotient,whichis(AB,CD),isnegative.Thisargumentisreversible.

42.11UseTheorem42.8andconsidertwocases:Pinsidethecircle,andPoutsidethecircle.

42.13IfthepointsAandBarecollinearwiththecenterSofthecircles,thendiameterSABissuchadesiredline.Inthiscase,nootherlinepassesthroughAandB,andanycircleorthogonaltosandpassingthroughAmustalsopassthroughtheinverseA′ofpointAincirclesbyTheorem42.4.ButnocirclecanpassthroughthethreedistinctcollinearpointsA,B,

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andA′.HencelineOABistheuniquesolutioninthiscase.If S, A, and B are not collinear as in the accompanying figure,

constructtheinverseA′ofpointAincirclesbyExercise42.11.Thenthecirclec throughA,A′, andB is orthogonal to circle s byTheorem42.3.ThiscirclemaybeconstructedbytakingitscenterCastheintersectionoftheperpendicularbisectorsofAA′andAB.Inthissecondcase,lineABisnot a diameter of circle s, and any circle throughA and orthogonal to smustalsopassthroughA′byTheorem42.4.Hencecirclec is theuniquesolutioninthiscase.

Answer42.1342.15Letthetwocirclesbec1andc2andletthecenterofcirclesbeS.By

Theorem42.4,theinverseP′ofPincirclesliesonbothc,andc2.ThusP′=PorP′=Q.SincethecirclescannotbetangentatP(Why?),thenPdoesnotlieoncircles.HenceP′≠P,soP′=Q.

43.1Lettheinversionsbez′=c+r2/(z–c)andz″=c+s2/(z′–c),whererandsarerealandnonzero.Thenwehave

ahomothetywithcenterCandratios2/r2.43.3Theycommutewhentheircirclesareorthogonal.43.5Seethediscussionin38.6.ThenuseAnswer38.10,replacingzbyz′andt

byz.43.7ByTheorem43.7,thebilineartransformationanditsinversewillbeequal

ifa=–d.43.9Theidentitymapisthebilineartransformationz′=(1z+0)/(0z+1).The

inverseofabilineartransformationisthebilineartransformationgivenbyTheorem43.7.Also,theproductofthetwobilineartransformationsz′=(az+b)/(cz+d)andz″=(ez+f)/(gz′+h)isgivenby

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abilineartransformation.ThetheoremfollowsbyExercise12.12.43.10a)z′=(1–i)z/(z–i).Fortheotherpartsofthisanswer,followthismapof

part(a)byeachofthemapsofparts(c)and(e)inAnswer43.5toobtainthecorrespondinganswershere.

43.11a)–1,(1+3i)/2,∞,2c)∞,1,–i,0

43.13a)No44.1Thecirclethroughthecentersisnotorthogonaltoanyofthethreecircles,andthecenterofacircledoesnotinvertintothecenteroftheinversecirclebyExercise41.18.

44.2a)LetthetwocirclescanddmeetatPandQ,letthecenterofthecirclesorthogonaltocanddbeO,letcirclescutcirclecatSandS′anddatTandT′andletOPcutcagainatQ′anddagainatQ″.Now,seetheaccompanyingfigure,m(OS)=m(OT)andOP·OQ′=OS2=OT2=OP·OQ″byExercise6.18.HenceOQ′=OQ″,soQ′=Q″,andthiscommonpointisQ.HenceO,P,andQarecollinear.

Answer44.2ac)Theproofofpart(a)readilyacceptsthiscondition,g)SinceeachsuchpointPliesontheradicalaxisofcirclessandu,andcirclestandu,thenthetangentsfromPtosanduarecongruent,andthosetotanduarecongruent.Hencethosetosandtarecongruent.ThusPliesontheradicalaxisofsandt.

i)Seetheanswertopart(a).

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k)Seetheanswertopart(a).44.3Findthecirclesorthogonaltothethreegivencircles(byExercise44.2)

andinvertincircles.44.5DrawanytwocirclescanddthroughbothAandB.ByTheorem42.3,

theyeachareorthogonaltos.Nows,c,anddinvertintothreecircles(orlines)s′,c′,andd′witheachofc′andd′orthogonaltocircles′.HencetheirintersectionsA′andB′areinverseincircles′byExercise42.15.

44.7LetthecircleofinversionhavecenterCandradiusr,andletthegivencirclehavecenterTandradiust.LetAandBbetheendsofthatdiameterofcircleT(t)thatiscollinearwithC,andletA′andB′betheirimages.Seetheaccompanyingfigure.ByExercise41.17,A′B′=(r2·AB)/(CA·CB)=(r2·2t)/((CT–t)(CT+t)=(r2·2t)/(CT2–t2).SinceA′B′isadiameteroftheimagecircle,thentheradiusoftheimagecircleisr2t/(CT2–t2).

Answer44.744.9Since ,thenx′=Re(z′)=k·Re(z)/(x2+

y2)=kx/(x2+y2),andy′Im(z′)=ky/(x2+y2).44.10a)x′=x′2+y′2

c)2x′+1=0e)b2x′2+a2y′2=a2b2(x′2+y′2)2;no.g)x′2y′2=(x′2+y′2)2

44.11ByExercises44.5and44.6,theinversesoftanduareinverseins′,theinverseofs.Buts′isastraightline,sot′andu′arereflectionsofoneanotherins′,andhencetheyarecongruent.

44.13a)LetPbeanypointonasemicirclewhosediameterisAOB.ThenthecirclesonA,O,PandonB,O,Pareorthogonal.c)Anangleinscribedinasemicircleisarightangle.

44.15InvertinthecirclecenteredatOandorthogonaltocirclesn.ThenlineOABandcirclesnareself-inverse.CirclesOAandOBmapintolines

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perpendiculartolineOAB,henceparalleltoeachother,andalsotangenttocirclesn.Circless0,s1,…,sn–1mapintocirclesnestedbetweent′andu′asshownintheaccompanyingfigure,hencecongruenttooneanotherandtocirclesn.Thetheoremfollowsreadilyfromthisfigure.

Answer44.1544.17DrawthecircleonABasdiameter.UnderinversionincenterA,thiscircle

mapsintoadiameteroftheinverses′ofcircles.AlsodiametrallineABofcirclesisinvariantandmapstoadiameterofs′.HencetheinverseofB,lyingonthesetwodiametersofs′,isthecenterofs′.

45.1No45.3

46.1SeeTheorem14.4andExercise14.2.46.3SeeCorollary15.9.46.5a)Areflectioninaplane.

c)Acentralinversionoraglide-reflectionorarotatoryreflection.46.6a)1

c)2e)2g)2i)3

46.7Isometries(a)and(i)areopposite;(c),(e),and(g)aredirect.

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46.9SeeTheorem49.3.46.10SeeTheorems49.10and49.13.

a)Theplaneofthereflectionandalllinesandpointsinthatplane;alllinesandplanesperpendiculartothemirrorplane.

c)Nopoints,butalllinesandplanesparalleltothevectorofthetranslation.

e)Theaxisofrotationandallpointsontheaxis;allplanesperpendiculartotheaxis.

g)Inadditiontothoseofpart(e),allplanescontainingtheaxis;andalllinesperpendiculartotheaxis.

i)Theplaneofreflection,andinitanylinethatisparalleltothevectoroftranslation;andanyplanethatisbothparalleltothevectoroftranslationandperpendiculartothemirror.

46.11a)SeeTheorem13.3.c)Thisisacorollarytopart(a).e)SeeTheorem15.1.

46.13SeeTheorem15.12.47.3SeeTheorem13.12.47.5SeeTheorem13.16.47.7ByTheorem47.8andthefactthatareflectionisnottheidentitymap.47.9Infinitelymany.(Considerhowmanyisometriesoftheplanemapagiven

pointAtoanotherpointA′.)47.10a)Consider,forexample,changingonlyDtoD′(0,0,–1).c)Applyparts(a)and(b)(perhaps).

47.11LettheplanethroughAperpendiculartoDD′cutDD′atpointP.ThentrianglesADPandAD′ParecongruentbyHL.Sincenowm(DP)=m(D′P),thenthisplaneisthedesiredperpendicularbisectorofDD′.SimilarlythisplanepassesthroughBandC.ButA,B,Cdetermineaplane.Hencetheydeterminetheperpendicularbisectorplane.

48.1Foranyrealconstantk,x+y+z=kandx+y+z=k+ .48.3SeeTheorem14.5.48.5SeeTheorem14.12.48.7SeeTheorem14.15.48.9SeeTheorems14.9and14.10.48.11SeeTheorem14.14.48.13SeeTheorem15.8.48.15ByTheorem48.15andCorollary48.16.48.17SeeTheorem15.8.49.1SeeTheorem16.1.

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49.3a)SeeTheorem14.14.c)LetΠbetheplanecontainingthetwoaxes.Factortherotationsinto

forappropriateplanesΔandΓ.Theirproductis.

49.5ByTheorem47.9,eachdirectisometryisaproductoftwoorfourreflectionsinplanes.Intheformercaseitisarotationoratranslation(ortheidentitymap).Thelattercaseyieldsascrewdisplacementbyanargumentsimilartothatforpart(d)ofTheorem49.3.

49.7Arotationaboutthesameaxisthroughtwicetheangle.49.9ByTheorem47.9itiseitherareflectionoraproductofthreereflections.

Butaproductofthreereflectionsinplanesformingapencilisareflection,andareflectionhasfixedpoints.Also,ifthethreeplanesconcurinapoint,thenthatpointisfixed.

49.11SeeTheorem16.4.49.12a)Ifthemirrorsformapencil,thentheyrepresentasinglereflection.So

supposejusttwomirrorsmeetalongalinem.WithoutlossofgeneralitywemayassumethatplanesΔandΓmeetinmwheretheisometryis

.NowapointPmapsintoapointP′bythemap ifftheperpendicularbisectorofPP′passesthroughlinem.ButthenPisafixedpointinαiffΓisthatperpendicularbisector,andthethreeplanesformapencil.Ontheotherhand,ifPisfixedunder ,thenPliesonm,soΓmustcontainP,andthethreeplanesmeetatleastinpointP.

49.13Arotationhasitsaxisasasetoffixedpoints,sothisisometrymustbeeitherofthetworemainingdirectisometries:atranslationorascrewdisplacement.

49.15Lettherotatoryreflectionbe withΔperpendiculartoΓandtoΓ.Then andthetheoremfollows.(Thetheoremisobviousgeometrically.)49.17ForareflectionσΓ,letmbeanylineperpendiculartoplaneΓandletΔbeanyplanecontaininglinem.Now ,arotationofzerodegreesaboutlinemfollowedbyareflectioninplaneΓ,hencearotatoryreflection.Similarlyacentralinversionisarotation(ahalfturn)aboutthelineofintersectionofitsfirsttwomirrors,followedbyareflectioninitsthirdmirror,henceitisarotatoryreflection.

50.1SeeTheorem18.2.50.3Applyatranslationcarryingonefacetotheother.50.5InFig.50.4,ABCDisarectangle.LetEbeitscenter.Thenareflectionin

theplanethroughEandparalleltothebaseplane(throughAandB)ofthe

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parallelepipedcarriestherectangleintoitself,andspecifically,diagonalACtodiagonalBD.

50.7ByTheorem50.6.50.9LettriangleABCbetheisoscelestrianglewithapexA.Thetriangleand

theplanecontainingitsbaseareinvariantinareflectionintheperpendicularbisectorplaneofthebaseBC.

50.11InthetrianglementionedinHint50.11,thetracesofthebisectorplanesaretheanglebisectors,whichconcuratapointI.NowthelinethroughIandthevertexofthetrihedralangleisthelineofconcurrenceofthebisectorplanes.

50.13SeeTheorem19.6.50.15ThisisacorollarytoCorollary50.14.50.17Letthelinesintersectathalftheangleoftherotationinaplane

perpendiculartotheaxisofrotation.SeeTheorem14.9.50.19Justasacircirculardiscisgeneratedbyrotatingasegmentaboutoneofits

endpoints.50.21a)Areflection(inthexy-plane).

c)Atranslation(of1unitinthez-direction).e)Aglide-reflection(of1unitinthez-directionandwithmirrortheyz-plane).

51.1SeeTheorem21.3.51.3SeeTheorem22.2.51.5Factorthecentralinversionintoaproductofreflectionsincoordinate

planes.51.7Weassumethatthez-axisliesineachofthetwodistinctplanesΔandΓ,

whoseequationsareAx+By=0andEx+Fy=0,withA2+B2=1andE2+F2=1.NowequationsforσΔarex′=x–2A(Ax+By),y′=y–2B(Ax+By),andz′=z,andthoseforσΓarex′=x–2E(Ex+Fy),y′=y–2F(Ex+Fy),andz′=z.HencetheirproductσΓσΔisgivenbyx′=Gx–Hy,y′=Hx+Gy,andz′=z,whereG=(1–2A2)(1–2E2)+4ABEFandH=2AB(1–2E2)+2EF(1–2B2).VerificationthatG2+H2=1isstraightforwardbuttedious.HenceGandHarecosθandsinθforsomeangleθ.

51.9SetA=B=D=0andC=1.

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INDEXUnlessstatedtrisectionotherwise,numberrefertosectionnumbersinthetext.AAS,18.6Abeliangroup,12.16Absolutegeometry,40.16Absolutevalue,34.2,35.17Addition,vector,33.6Affix,35.12Ahmes,1.3Alexandria,9.9,9.13,23.1,23.2Altitudes,concurrence,7.11Amasis,KingofEgypt,9.4Amplitude,35.17Angle,35.17dihedral,50.1rotation,10.3,14.8trisection,8.2,Exercises8.13–8.16

Apex,Exercise18.5Apollonius,9.11,23.9problem,40.11

Arabs,9.13,23.2Archimedes,9.10–9.11,23.1Area,19.6–19.8,37.10,Exercise37.10triangle,6.15–6.21

Argand,J.,31.11Argument,35.17Aristotle,9.7Arithmeticmean,42.6Arithmeticprogression,42.6ArithmeticTeacher,45.12ASA,18.5Associative,12.6,33.8,34.10,34.14Axiomaticsformal,40.21material,9.1,40.21

Axisimaginary,32.7radical,Exercise44.2real,32.6,32.7rotation,48.8

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Babylonians,1.2,1.7–1.9Base,2.6Basis,35.1Bell,E.T.,1.4Between,2.4Bible,1.7Bilineartransformation,43.6BlackDeath,23.4Boldface,2.15Bolyai,J.,40.17Bombelli,R.,31.5Brianchon,C.J.,Exercise23.5theorem,Exercise23.5

Brocard,H.,40.14Cavalieri,23.9Cayley,A.,40.14line,40.9

Centercentralinversion,48.18homothety,24.1,25.1inversion,41.4ninepoint,7.16radical,Exercise44.2rotation,10.3,14.8similitude,24.2,25.6

Centralinversion,46.2,48.18center,48.18equation,51.8

Centroid,Exercise5.8,6.2,Exercise37.18,50.21Ceva,G.,40.2theorem,5.2,5.3,5.10,5.11,Exercise5.4,34.24,Exercise34.11,40.2–40.3

Cevian,5.1CharlesII,KingofEngland,23.5Chasles,M.,40.14China,1.6Christina,QueenofSweden,23.10Circle,19.1–19.5,38.1–38.5crossratio,42.11inscribed,6.2inversion,seeInversionninepoint,6.1,6.5,7.13–7.16

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notation,8.4reflection,41.4,Exercise44.6squaring,8.2,Exercise8.16

Circular,42.1Circumcircle,6.2Cis,35.17Clairaut,A.,40.4Clockwise,13.15Closure,Exercise32.1Coaxial,4.6Coingame,Exercise19.14Collinear,2.6Commutative,14.6,33.8,34.10,34.14Compass,8.7equivalence,8.8rusty,8.16–8.19

Complexnumbers,31.1–39.14conjugate,35.16order,Exercise32.1polarform,35.17polynomialform,32.5pureimaginary,35.18vectors,35.9–35.11

Component,vector,21.4Composition,12.3Concurrent,2.6Conformal,41.7Congruent,2.8Conjugatecomplex,35.16harmonic,3.11isogonal,Exercise4.11,20.12,20.13isotomic,Exercise4.11,5.6vector,34.17,34.19

Constructible,8.2,Exercise8.16Construction,8.1–8.19,AppendixBclassical,8.2,Exercise8.16impossible,8.2,Exercise8.16meanproportional,8.10squareroot,8.11

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Coplanar,3.1Copolar,4.6Cotes,R.,31.7Counterclockwise,13.15Crossratio,3.6,3.7,42.11Cube,duplication,8.2,Exercise8.16Cyclicorder,10.4Cyclicquadrilateral,7.6DarkAges,23.1,23.3Dedekind,R.,9.7Delineisrectisseinvicemsecantibus,40.2Demaximusetminimis,23.7DeMoivre,A.,31.7–31.8,Exercise31.3theorem,31.7–31.9,Exercise31.2,Exercise35.7

DeMorgan,A.,23.5,23.8Desargues,G.,23.11two-triangletheorem,4.7,4.9,4.10,Exercise23.1

Descartes,R.,23.9–23.11,31.6Determinant,36.9–36.17Dihedralangle,50.1Direct,10.4,13.14,13.15,25.12,Exercise25.18,41.7,Exercise46.8,47.10Directeddistance,33.3Directedlength,2.9,33.3Directedmeasure,2.13Directedsegment,33.3Discoverymethod,45.10,45.11Displacement,screw,46.3,49.1Distancedirected,33.3inversesofpoints,Exercise41.17

Distributivelaw,Exercise36.27Divide,3.2externally,3.2harmonically,3.11improperly,3.2internally,3.2

Dodge,C.,Exercise45.4Doublepoint,12.7Dual,4.9,4.10,5.10,5.11,Exercise17.1Pappus’theorem,4.10–4.12

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Duplicationofcube,8.2,Exercise8.16e,35.17Egyptians,1.2–1.6Elements,2.1,8.6,9.2,9.8,AppendixAEpsilon-ruler,Exercise4.16Equalvectors,33.2Equationcentralinversion,51.8glide-reflection,22.9halfturn,21.11,Exercise39.9,Exercise51.7homothety,30.1–30.5inversion,43.1–43.3reflection,22.2–22.6,39.8–39.10,39.12,51.5,51.10rotation,21.6,21.9,39.2,51.3,51.4similarity,30.6,39.5–39.13translation,21.3,39.1,51.2

Equicircle,6.2Equivalentcompasses,8.8Eratosthenes,9.11ErlangerProgramm,12.22Euclid,9.2,9.8–9.9,12.23,40.9Elements,2.1,8.6,9.2,9.8,AppendixAfifthorparallelpostulate,40.15–40.16,Exercise40.8,AppendixA

Euclideancompass,8.7Euclideanconstruction,8.1–8.19Euclideangeometry,12.19,12.21Euler,L.,31.9line,29.2,Exercise29.3theorem,2.16,Exercise2.16

Eudoxus,9.7Eves,H.,8.17,40.4,Exercise40.1,Exercise40.4,Exercise40.7,43.10Excircle,6.2Exponentialform,35.17Extendedplane,3.2Extendedspace,3.2Externalcenterofsimilitude,25.6Externaldivision,3.2Fagnano’sproblem,20.17Falsetheorem,7.23,Exercise7.12Fermat,P.,23.7,23.9,23.10

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Feuerbach,K.,7.21,40.12theorem,7.21,44.6

Fifthpostulate,40.15–40.16,Exercise40.8,AppendixAFixedpoint,10.3,12.7Formalaxiomatics,40.21Fréchet,M.,40.19Frustum,1.4,Exercise1.4Fundamentaltheoremofalgebra,31.13Galileo,23.6Game,Exercises19.14–19.15Gauss,C.F.,31.13–31.15,Exercise31.4,40.17plane,35.12

Geometricmean,42.6Geometricprogression,42.6Geometryabsolute,40.16Euclidean,12.19,12.21Klein’sdefinition,40.20non-Euclidean,40.16–40.17planeequiform,12.20–12.21projective,4.10,12.21

Gergonne,J.,40.11point,5.4

Glide-reflection,11.9,14.16,46.3,46.9equation,22.9

GreatestEgyptianpyramid,1.4Greeks,23.2Group,12.15,Exercise12.12,Exercise12.13,16.6–16.10,25.16,25.17,26.11,

26.12abelian,12.16

Halfturn,11.6,15.5,46.1,48.14equation,21.11,Exercise39.9,Exercise51.7

Halmossymbol,2.3footnoteHamilton,W.,31.15Harmonicconjugate,3.11Harmonicdivision,3.11theorem,42.5

Harmonicmean,Exercise6.15,42.6Harmonicprogression,42.6Heron,9.11formula,Exercise2.15,6.17

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Hexagramtheorem,23.12,Exercises23.2–23.3,40.9Hilbert,D.,40.14Hipparchus,9.11Homography,43.6Homothety,24.1,25.1center,24.1,25.1equation,30.1–30.5,39.3ratio,24.1,25.1

HorblitandNielsen,45.12Horner’smethod,1.6HundredYears’War,23.4Hypatia,9.11–9.12i,imaginaryunit,32.5,35.10–35.18i,vector,34.23Idealline,3.2Idealplane,3.2Idealpoint,3.2,38.4,41.3Idempotent,Exercise12.7Identity,11.3,12.8Image,10.1,35.12Imaginaryaxis,32.7Imaginarycoefficient,35.17Imaginaryunit,35.17Impossibleconstruction,8.2,Exercise8.16Improperdivision,3.2Incircle,6.2Indivisibles,method,23.14Inequality,32.3,Exercise32.1triangle,Answer34.1

Infinity,23.13,38.4,41.3point,3.1,3.2

Inscribedcircle,6.2Internalcenterofsimilitude,25.6Internaldivision,3.2Invariantpoint,10.3,12.7InverseseealsoInversionreflection,13.12rotation,14.13similarity,24.5transformation,11.4,12.10

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translation,14.5vector,34.14

Inversion,41.1–44.10,41.4center,41.4circleorline,41.11–41.14equation,43.1–43.3line,Exercise44.6plane,41.3power,41.4radius,41.4rotatory,46.3,49.9

Involutoric,11.4,15.3Iota,11.3,12.8Irrationalityofpi,40.13Isogonalconjugate,Exercise4.11,20.12,20.13Isometry,10.1–22.10,46.1–51.11,13.2,47.2order,Exercise11.7

Isomorphic,35.6Isotomicconjugate,Exercise4.10,5.4Jones,W.,40.5Kepler,J.,23.6Klein,F.,12.22,40.20,Exercise40.9geometry,12.22,40.20

Kirkmanpoint,40.9Lambert,J.,40.16LarousseDictionary,31.16Lawofsines,Exercise6.22Legendre,A.,40.16Leibniz,G.,23.14Lemoine,E.,40.14Length,2.7directed,2.9

Limits,theoryof,23.15Line,2.4,37.2–37.8ideal,3.2ordinary,3.2

Linearcombination,35.1Lobachevsky,N.,40.17geometry,40.17,40.19

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Magnitudeconstructible,8.3,Exercise8.16vector,34.1,34.2

Mahavira,31.3Malfatti’sproblem,40.10MapseealsoTransformationconformai,41.7identity,11.3,12.8involutoric,11.4,15.3

Mascheroni,L.,40.7–40.8Materialaxiomatics,9.1,40.21MathematicsMagazine,45.12MathematicsTeacher,45.12Matrix,36.8Mean,42.6Meanproportional,8.10construction,8.10

Measure,2.7directed,2.13

MechanixIllustrated,Exercise8.14Medialproperty,Exercise36.27Medialtriangle,6.2Median,6.2,50.20theorem,6.6,6.7,50.21

Menelaus,9.11,40.2point,4.1theorem,4.2,4.3,5.8,5.10,5.11

Methodofindivisibles,23.14Mirror,10.4,13.9Moderncompass,8.7Monge,G.,40.6Multiplicationscalar,33.10vector,34.8–34.14

Mystichexagramtheorem,23.12,Exercises23.2–23.3,40.9Newton,I.,23.14,31.7Ninepointcenter,6.2,7.16Ninepointcircle,6.1,6.2,6.5,7.13–7.16,29.4Ninepointradiustheorem,6.5Noncommutativityofrotations,14.15Non-Euclideangeometry,40.16–40.17One,1,34.11

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Opposite,10.4,13.14,13.15,25.12,Exercise25.18,Exercise46.8,47.10.Ordercomplexnumbers,Exercise32.1cyclic,10.4isometry,Exercise11.7

Orderedpair,31.14–31.16,Exercises31.4–31.5Ordinaryline,3.2Ordinaryplane,3.2Ordinarypoint,3.2Oresme,23.9Orthictriangle,6.2Orthocenter,6.2Orthocentricquadrangle,7.18Orthogonal,34.17,Exercise41.15,42.3Oughtred,W.,23.5Overbar,2.15footnotePacioli,L.,31.4Pappus,9.11ancienttheorem,Exercise44.15theorem,4.8–4.12

Papyrus,Rhind,1.3Parallel,Exercise3.2Parallelpostulate,40.15–40.16,Exercise40.8,AppendixAParallelepiped,50.3–50.13Parallelogram,18.9–18.11law,33.7

Parent,A.,40.4Pascal,B.,23.12line,40.9theorem,23.12,Exercises23.2–23.3,40.9

Peacock,G.,31.2Peaucelliercell,42.9Peaucellierimage,42.9Pencil,2.6Pi,1.3,1.7,1.8,Exercise1.2,Exercise1.6,23.5,Exercise23.4,23.13,40.5irrationality,40.13

PiMuEpsilonJournal,Exercise6.15Planeextended,3.2Gauss,35.12ideal,3.2inversive,41.3

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ordinary,3.2Planeequiformgeometry,12.20–12.21Plato,9.7Plimpton322,1.8Plücker,J.,40.14line,40.9

Poincaré,H.,40.19Pointdouble,12.7fixed,10.3,12.7Gergonne,5.4ideal,3.2,38.4,41.3infinity,3.1,3.2,41.3invariant,10.3,12.7Menelaus,4.1ordinary,3.2

Polarform,35.17Polygongame,Exercise19.15Poncelet,J.,40.8,40.11Positive,2.9Postulate,fifthorparallel,40.15–40.16,Exercise40.8,AppendixAPowerofinversion,41.4Preimage,10.1Principleofpermanenceofforms,31.2Problem,epsilon-ruler,Exercise4.16Productisometry,10.5scalar,33.10transformation,12.3vector,34.8–34.14

Progression,42.6Projectivegeometry,4.10,12.21Proportional,mean,8.10Ptolemy,Claudius,9.11theorem,44.3

Ptolemy,KingofEgypt,9.9,9.13Pureimaginary,35.18Pyramid,greatestEgyptian,1.4Pythagoras,9.5–9.6Society,9.5–9.6theorem,1.5,9.6,Exercise9.2,34.12,AppendixA(47–48)triple,Exercise

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45.3Quadrangle,orthocentric,7.18Quadrilateral,cyclic,7.6Radicalaxis,Exercise44.2Radicalcenter,Exercise44.2Radiusofinversion,41.4Range,2.6Ratio,3.3cross,3.6,3.7homothety,24.1,25.1similarity,26.1

Real,35.18Realaxis,32.6,32.7Realpart,35.17Rectangularform,35.17Reflectioncircle,41.4,Exercise44.6line,10.4,13.9,22.2–22.6,39.8–39.10,39.12plane,46.1,47.4,51.5,51.10point,11.6,15.5rotatory,46.3,49.7

Rhindpapyrus,1.3Rich,B.,45.12Riemann,B.,40.18Ringisomorphic,35.6Roberval,23.9Rotationangle,10.3,14.8center,10.3,14.8equation,21.6,21.9,39.2,51.3–51.4inverse,14.13noncommutativity,14.15plane,10.3,14.8space,46.1,48.8

Rotatoryinversion,46.3,49.9Rotatoryreflection,46.3,49.7Rustycompass,8.16–8.19Saccheri,G.,40.16Salmonpoint,40.9SASpostulate,13.1,18.1Scalar,33.11

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Scalarmultiplication,33.10Screwdisplacement,46.3,49.1Segment,2.4directed,33.3

Sense,13.15,47.10direct,10.4left,47.10opposite,10.4right,47.10

Side,triangle,6.2Sigma,15.10,47.4Similarity,24.1,26.1equation,30.6,39.5–39.13inverse,24.5ratio,26.1

Similitude,centerof,24.2,25.6Sines,lawof,Exercise6.22Space,extended,3.2Squareroot,construction,8.11Squaringacircle,8.2,Exercise8.16SSS,18.4Staudt,K.von,40.14Steiner,J.,40.9,40.10point,40.9

Stewart,M.,40.3theorem,Exercise2.10

Subtraction,vector,33.13Superposition,12.23,45.6Thales,9.3–9.4,Exercise9.1TheonofAlexandria,9.11Theoremaltitudes,5.5,7.11,28.7anglebisectors,6.10,20.4area,19.6–19.8,37.10,Exercise37.10areaoftriangle,6.15–6.21associativity,12.6,33.8,34.10,34.14Brianchon’s,Exercise23.5Ceva’s,5.2,5.3,Exercise5.4,34.24,Exercise34.11,40.2,40.3commutativity,14.6,33.8,34.10,34.14DeMoivre’s,31.7–31.9,Exercise31.2,Exercise35.7

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Desargues’,4.7,Exercise23.1distancebetweeninverses,Exercise41.17dualofPappus’,4.11,4.12duality,4.10equivalentcompasses,8.8Euler’s,2.16,Exercise2.16false,7.23,Exercise7.12Feuerbach’s,44.6harmonicdivision,42.5Heron’sformula,Exercise2.15,6.17Horner’smethod,1.6isometry,13.13,16.14,17.2–17.17medians,6.6,6.7,20.11,29.1,50.21Menelaus’,4.2,4.3,5.8mystichexagram,23.12,Exercises23.2–23.3,40.9ninepointcenter,7.16ninepointcircle,7.13–7.14,29.4ninepointradius,6.5Pappus’,4.8Pappus’ancient,Exercise44.15parallelepiped,50.3–50.13parallelogram,18.9–18.11perpendicularbisectors,7.1,20.1–20.3Ptolemy’s,44.3Pythagorean,1.5,9.6,Exercise9.2,34.12,AppendixA(47–48)Stewart’s,

Exercise2.10Thomsen’srelation,17.17,Exercise17.17two-triangle,4.7

Theoryoflimits,23.15Thomsen’srelation,17.17,Exercise17.17Transform-solve-transform,18.1Transformation,12.2seealsoMapbilinear,43.6circular,42.1identity,11.3,12.8inverse,11.4,12.20

Translation,10.2,14.1,46.1,48.1equation,21.3,39.1,51.2inverse,14.5plane,10.2,14.1

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space,46.1,48.1vector,14.1

Trapezoid,18.12–18.14Trianglearea,6.15–6.21inequality,Answer34.1medial,6.2orthic,6.2

Trichotomy,Exercise32.1TrigonometricformCeva’stheorem,5.3complexnumber,35.17Menelaus’theorem,4.3

Trilateral,5.10Trisectinganangle,8.2,Exercise8.13–8.16True,2.18“Truemetaphysicsof ”31.13–31.14Two-triangletheorem,4.7,4.9,4.10,Exercise23.1Two-YearCollegeMathematicsJournal,45.12Unitvector,34.5UniversityofAlexandria,9.9,9.13Vector,10.2,33.1addition,33.6complexnumbers,35.9–35.11component,21.4conjugate,34.17,34.19i,34.23inverse,34.14magnitude,34.1,34.2multiplication,34.8–34.14one,1,34.11orthogonal,34.17product,34.8–34.14subtraction,33.13translation,14.1unit,34.5

Vertex,2.6Vieta,23.9Volume,50.12Wallis,J.,23.13,Exercise23.4Well-defined,Exercise33.7

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Wessel,C,31.10–31.11

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