EUCLIDEAN GEOMETRY AND TRANSFORMATIONS Clayton W. Dodge University of Maine DOVER PUBLICATIONS, INC. Mineola, New York
EUCLIDEANGEOMETRYANDTRANSFORMATIONS
ClaytonW.DodgeUniversityofMaine
DOVERPUBLICATIONS,INC.Mineola,NewYork
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
Toourdaughter,KathyAlthoughwesometimesgoaroundincircles,westillreflecteachother.
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
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
CONTENTSCHAPTER1MODERNELEMENTARYGEOMETRY
1TheBeginningsofGeometry2Directedsegmentsandangles3Idealpointsandratios4ThetheoremofMenelaus5Ceva’stheorem6Somegeometryofthetriangle7Moregeometryofthetriangle8Geometricconstructions
CHAPTER2ISOMETRIESINTHEPLANE9TheAmazingGreeks10Introductiontotranslations,rotations,andreflections11Introductiontoisometries12Transformationtheory13Isometriesasproductsofreflections14Translationsandrotations15Halfturns16Productsofreflections17Productsofisometrie;asummary18Applicationsofisometriestoelementarygeometry19Furtherelementaryapplications20Advancedapplications21Analyticrepresentationsofdirectisometries22Analyticrepresentationsofoppositeisometries
CHAPTER3SIMILARITIESINTHEPLANE23Therebirthofmathematicalthinking24Introductiontosimilarities25Homothety26Similarity27Applicationsofsimilaritiestoelementarygeometry28Furtherelementaryapplications29Advancedapplications30Analyticrepresentationsofsimilarities
CHAPTER4VECTORSANDCOMPLEXNUMBERSINGEOMETRY31Thesearchforthemeaningofcomplexnumbers32Introductiontocomplexnumbers33Vectors
34Vectorsmultiplication35Vectorsandcomplexnumbers36TrianglesintheGaussplane37LinesintheGaussplane38Thecircle39IsometriesandsimilaritiesintheGaussplane
CHAPTER5INVERSION40Matchlessmodernmathematics41Inversion42Progressions,ratios,andPeaucellier’sCell43Inversionandcomplexgeometry44Applicationsofinversion
CHAPTER6ISOMETRIESINSPACE45Whatnext?46Introductiontothreedimensions47Reflectioninaplane48Basicspaceisometries49Morespaceisometries50Someapplications51AnalyticrepresentationsAppendixesA.ASummaryofBookIofEuclid’SElementsB.BasicRulerandCompassConstructionsBibliographyHintsforSelectedExercisesAnswersIndex
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
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.
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+
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
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.)
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.
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
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.
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
Exercise2.14
15.UseExercise2.14toproveHeron’sformulafortheareaKofatrianglewithsidesa,b,candsemiperimeters=(a+b+c)/2:
Thisalsoshowsthatthealtitudehtosideaisgivenby
16.UsetheaccompanyingfiguretoproveEuler’stheorem,Theorem2.16.
Exercise2.15
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
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
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.
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.
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
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
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,
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.
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.)
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
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
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.)
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,
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
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
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.
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
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
Figure5.8a
R.WeapplyCeva’stheoremtoeachofthesixtriangleslistedinFig.5.8b.Bymultiplyingthesesixequationssideforside,weobtain
from which the theorem follows, since either one or three of the collinearMenelauspointsdividethesidesexternally.
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.
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.]
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′,
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.
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.
%
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.)
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
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
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
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
fromwhichoneobtains
Figure6.17
Now
Fromthisequationweobtain
6.18 For a synthetic geometric proof of Heron’s formula, see H. Eves’ AnIntroductiontotheHistoryofMathematics,thirdedition,pages171–172.
6.19CorollaryTheareaofatriangleisgivenby
6.20TheoremTheproductoftwosidesofatriangleisequaltotheproductofthecircumdiameterandthealtitudetothethirdside.
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
Figure6.22
ButAZa=AYa,so
ThusCXa=CYa=AYa–AC=s–b.AlsoBX=BZ=AB–AZ=c–AYandBX=BC–XC=a–YC,so
whence
fromwhichthetheoremfollows.
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
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
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
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
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.
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.)
%
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
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!
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.
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
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
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
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
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.
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.
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
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
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
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.
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
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
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
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
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
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
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.
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.
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
=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.
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
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
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.
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)
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,
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
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
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
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.
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.)
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
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.
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
%
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).
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
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.
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.
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
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.
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
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
%
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.
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.
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°
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.
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.”
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
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
%
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.
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.
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
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
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
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.
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.
%
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.
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
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.
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
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.
Figure19.9
19.10TheoremIfABQPandBCRQareparallelograms,thensoalsoisACRPaparallelogram.
Wehave(seeFig.19.10)
soACRPisaparallelogram.
Figure19.10
19.11TheoremIfABPQandBCQRareparallelograms,thensoalsoisACPRaparallelogram(Fig.19.11).
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.
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.
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
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.
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.
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
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
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
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.
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.
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.
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)
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
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
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
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:
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
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
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
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
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.
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
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
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
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.
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?
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
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.
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′.
%
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.
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.
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.
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
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
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.
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.
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.
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
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.
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.
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,
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
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
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.
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.
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
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
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,
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.
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
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.
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
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
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
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)
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.
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,
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
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+
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
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
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
–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?
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
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
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
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
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.
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.
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
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.
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
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.
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).
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
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
Hence
and
Sinceu andv havedistinctdirections, it follows that eachcoefficientmustbezero;thatis,
Thus
sowehave
Nowtherearealsorealnumberspandq,suchthat
Thenwehave
Bycollectingterms,obtain
andagainthecoefficientsofuandvmustbothbezero,yielding
andhence
Finally
34.25Theexamplesgivenintheselasttwosectionsindicatethatvectormethods
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
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.
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
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).
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
(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
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.
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.
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
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
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).
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.
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
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
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
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.
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
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
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,
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
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
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.
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
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
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
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.
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.
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,
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
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.
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.
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
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
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.
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
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.
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.
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
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,
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
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,
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.
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.
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:
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
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′.
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.
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
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.
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
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,
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.
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
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.
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
|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′.
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
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
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
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
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
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.
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
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.
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
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
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
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.
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
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
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
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)
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.
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
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
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
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.
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.
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
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
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
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).
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).
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
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
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.
49.6DefinitionAglide-reflectionistheproductofareflectioninaplaneandatranslationthroughavectorparalleltothemirrorofthereflection(Fig.49.6).
Figure49.6
49.7DefinitionArotatoryreflectionistheproductofareflectioninaplaneanda rotation about an axism perpendicular to the mirror of the reflection (Fig.49.7).
%
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.
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
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
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
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
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).
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.
%
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).
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
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
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.
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
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
51.6Similarequationsdeterminereflectionsintheothercoordinateplanes.ThusweobtainthenexttwoimmediatecorollariesfromTheorems48.15,48.19,and51.5.
51.7CorollaryAhalfturnaboutthez-axis(seeFig.51.7)isgivenby
51.8CorollaryAcentralinversionintheorigin(seeFig.51.8)isgivenby
51.9TodeveloptheequationsforareflectioninthegeneralplaneII,letuswritetheequationofIIinthestandardform
Figure51.7
Figure51.8
LettingσIImapP(x,y,z) toP′(x′,y′,z′),wehavethelinePP′noemal toII,sothatthecoordinatesofPandP′satisfytheequation
or
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
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.
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
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.
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′.
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.
FigureB.7
FigureB.8a
FigureB.8bB.9ConstructionDrawalinenparalleltoagivenlinemandpassingthroughagivenpointP.Drawanyline throughP tocut linematA.UsingConstructionB.4,drawanangleatPcongruenttotheangleatAandontheoppositesideoflinePA.Thenlinenistheterminalsideofthisconstructedangle.(SeeFig.B.9.)
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).
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
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,
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
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
Wylie,C.R.,Jr.,IntroductiontoProjectiveGeometry.NewYork:McGraw-Hill,1970Yaglom,I.M.,ComplexNumbersinGeometry,translatedbyE.J.F.Primrose.NewYork:AcademicPress,1968
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.
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.
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
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.
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.
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.
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.
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&Numbers,Exercises60-5orDodge,NumbersandMathematics,Exercises90-5.
h)Comparewitha2+b2,whichdoesnotfactoralgebraically,butwhichisequaltothesquarenumber25whena=3andb=4.
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
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
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.
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.
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.
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
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
=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.
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
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
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
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
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
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.
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.
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
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.
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).
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
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.
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
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.
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
=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.
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.
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
–(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
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
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
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
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.
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.
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.
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
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
,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.
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
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
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
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
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.
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.
%
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′.
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,
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
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).
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
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.
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.
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
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.
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Wessel,C,31.10–31.11
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