2010-11 Meet 1

8/4/2019 2010-11 Meet 1

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2010-11 Meet 1, In v ua Event A

Question#1isintendedtobeaquickieandisworth1point.Eachofthenextthreequestionsisworth2points.Placeyouranswertoeachquestiononthelineprovided.Youhave12minutesforthisevent.

1. If

1

1

1

x ( )

⎛

⎝ ⎜⎜

⎞

⎠ ⎟ ⎟

=

2

3,determineexactlythevalueof x .

2. TheLCMof123,231,and312canbewrittenasapowerof2multipliedbyHiveotherprimenumbers.Doso.

3. I’moutforlunchatmyfavoritecafé,butIonlyhave$15.00.Ifthesoup-and-sandwichcomboIwanttoordercosts$13.00,andsalestaxis7%,whatistheminimumwhole-numberpercent-offdiscountcouponImustholdinmywallettoallowmetostillleavean18%tip?

(Note:taxandtipareappliedafterthecoupon,butnottoeachother.)

4. Acertainpositiveintegeristhreegreaterthanamultipleof5,Hivegreaterthanamultipleof8,andeightgreaterthanamultipleof13.Determinethevalueoftheleastsuchinteger.

=

Name:___________________________________ Team:___________________________________

NOCALCULATORSareallowedonthisevent.

Minnesota State High School Mathematics Leagu

8/4/2019 2010-11 Meet 1


SOLUTIONS

1. If

1

1

1

x ( )

⎛

⎝ ⎜⎜

⎞

⎠ ⎟ ⎟

=

2

3,determineexactlythevalueof x .

2. TheLCMof123,231,and312canbewrittenasapowerof2multipliedbyHiveotherprimenumbers.Doso.

3. I’moutforlunchatmyfavoritecafé,butIonlyhave$15.00.Ifthesoup-and-sandwichcomboIwanttoordercosts$13.00,andsalestaxis7%,whatistheminimumwhole-numberpercent-offdiscountcouponImustholdinmywallettoallowmetostillleavean18%tip?

(Note:taxandtipareappliedafterthecoupon,butnottoeachother.)

4. Acertainpositiveintegeristhreegreaterthanamultipleof5,Hivegreaterthanamultipleof8,andeightgreaterthanamultipleof13.Determinethevalueoftheleastsuchinteger.

x=

3

2

23⋅3⋅7 ⋅11⋅13⋅41

Theleftsidesimplyasksustotakethereciprocalofxthreetimes.Twoof

thosereciprocals“undo”eachother,sowe’releftwith

1

x

=

2

3 ,andx=

3

2.

Sinceallthreenumbershavedigitsthatsumtoamultipleof3,eachnumbermustbeitselfamultipleof3:

123=3⋅41 231=3⋅7⋅11 312= 23⋅3 ⋅13

SoLCM(123,231,312)= 23

⋅3 ⋅7 ⋅11⋅13 ⋅41 .

Taxandtiptogethercomprise25%ofthefoodorder,sothesubtotalbeforetax&tip

mustbenomorethan

$15.00

125%=

$15.00

1.25=$12.00 .

Amountofdiscount=$13.00–$12.00=$1.00⇒

13 1.000...

.076...

–91 ⇒

90

8%off.

8% off

(accept“8”)

Therearemanytrialanderrormethodsforsolvingthisproblem,butherewepresentaveryelegantandinstructivesolution:Letxbetheintegerinquestion.Sincexisthreegreaterthanamultipleof5,setxequal

to5k+3.Usingmodulararithmetic, x ≡5mod8

and x ≡8mod13

.Substituting,

5k +3≡5mod8 ⇒ 5k ≡2mod8 ⇒ k ≡2mod8 ;

5k +3≡8mod13 ⇒ 5k ≡5mod13 ⇒ k ≡1mod13.

BytheZirstresult,k ∈ 2,10,18,26,...{ } Bythesecondresult,

k ∈ 1,14, 27,40, ...{ }

TheZirstksatisfyingbothconditionsisk=66,makingx=5k+3=5(66)+3=333.

(Formoreinformationonproblemslikethese,see:

http://en.wikipedia.org/wiki/Modular_arithmetic

http://en.wikipedia.org/wiki/Chinese_remainder_theorem)

333

Minnesota State High School Mathematics Leagu2010-11 Meet 1, In v ua Event A

http://en.wikipedia.org/wiki/Chinese_remainder_theorem







8/4/2019 2010-11 Meet 1


1. Figure1showstheobtuseanglebetweenlinesl1 andl2 as133°,andtheobtuse

anglebetweenlinesl1 andl3 as144°.

Determinethemeasure,indegrees,ofthe

obtuseanglebetweenlinesl2 andl3 .

2. Inrighttriangle ABC(Figure2),pointDison

side AC andE ison AB .

Givenm∠ ACB =m∠ ABD =m∠ ADE = 37° ,

determinethedegree-measureof∠BDE .

3. InpentagonPENTA,thedistancesPE ,PN ,

PT ,PA,ET andNAareallequal(Figure3).

Ifm∠EPA=76° ,determinethedegree-

measureof∠PEN .

4. Someanglesinaconvex21-gonarerightangles,andtherestarecongruentobtuse

angles.Whatisthelargestpossibledegree-measureforanangleinthispolygon?


Name:___________________________________ Team:___________________________________

E

D

B

A C

Figure2

Figure3


Figure1

m∠BDE =

m∠ ABC =

2010-11 Meet 1, In v ua Event B

8/4/2019 2010-11 Meet 1


1. Figure1showstheobtuseanglebetweenlinesl1 andl2 as133°,andtheobtuse

anglebetweenlinesl1 andl3 as144°.

Determinethemeasure,indegrees,ofthe

obtuseanglebetweenlinesl2 andl3 .

2. Inrighttriangle ABC(Figure2),pointDison side AB andEison AC .

Givenm∠ ACB =m∠ ABD =m∠ ADE = 37° ,

computethedegree-measureof∠BDE .

3. InpentagonPENTA,thedistancesPE ,PN ,

PT ,PA,ET andNAareallequal(Figure3).

Ifm∠EPA=76° ,determinethedegree-

measureof∠PEN .

4. Someanglesinaconvex21-gonarerightangles,andtherestarecongruentobtuseangles.Whatisthelargestpossibledegree-measureforanangleinthispolygon?

E

D

B

A C

Figure2


Figure1Theinterioranglesshownarethesupplementsofthegivenobtuseangles.Usingthetriangleenclosedbythethreelines,47°+36°+ x =180°

impliesthat x =97° .

97°

Within ABC ,∠ ABC and∠C arecomplements,so

m∠ ABC = 90°−37° = 53° .∠CBD and∠BDE are

alternateinteriorangles,andsince BC DE ,∠CBD

and∠BDE arecongruent.m∠BDE = 53°−37° = 16° .

16°

Figure3

APN isequilateral,som∠ APN =60° .Thus

m∠NPE =76°−60° =16° .Since PEN is

isosceles,

m∠PEN =180°−16°

2=82° .

82°

m∠BDE =

m∠ ABC =

Letr=thenumberofrightangles.Thenthereare(21–r)obtuseangles,ofameasurewe’ll

callm.Thesumoftheanglesinthe21-gonisthen180° 21−2( )= 90°r +m 21− r ( ) .

Rewritingtheleftsideas90°⋅38 ,wehave

21− r ( )m=90° 38− r ( ) ⇒ m=90°38− r

21− r

⎛

⎝ ⎜⎞

⎠ ⎟ .

Wewant

38−r

21−r

<2 ,occurringwhenr<4.Thenearestvalueisr=3,leadingtom=175° .

175°

2010-11 Meet 1, In v ua Event B

SOLUTIONS

8/4/2019 2010-11 Meet 1


1. UsingFigure1,showingcertainsegmentslabeledaslength1,determineexactlythevalueofcosθ.

2. Giventhattheratioofcos x tosin x is3:2,determineexactlytheratio(tan x ):(cot x ).

3. If

cos α =

1

3andα andβ areacutecomplements,determineexactly

tan 180°−β ( ) .

4. Determineexactlythecoordinates(bothofthem)ofthehighestpointonthegraphof

f x ( )= 2 cos 3 x −π

4

⎛

⎝ ⎜⎞

⎠ ⎟ −5

⎡

⎣⎢⎢

⎤

⎦⎥⎥

2

ontheinterval0≤ x ≤2.


Name:___________________________________ Team:___________________________________


Figure1

osθ =

an 180°−β ( )=

2010-11 Meet 1, In v ua Event C

1

1

1

!

an x

cot x

=


8/4/2019 2010-11 Meet 1


1. UsingFigure1,showingcertainsegmentslabeledaslength1,

determineexactlythevalueofcosθ.

2. Giventhattheratioofcos x tosin x is3:2,determineexactlytheratio(tan x ):(cot x ).

3. If

cosα =

1

3and

α +β = 90° ,determineexactly

tan 180°−β ( ) .


f x ( )= 2 cos 3 x −π

4

⎛

⎝ ⎜⎞

⎠ ⎟ −5

⎡

⎣⎢⎢

⎤

⎦⎥⎥

2

ontheinterval0≤ x ≤2.


Figure1

cosθ =

an 180°−β ( )=

2010-11 Meet 1, In v ua Event C

tan x

cot x

=

SOLUTIONS

3

3

accept

1

3

hereas

kindofgrace

eriod,butwarn

tudentsaboutthe

equirementsof

determineexactly”)

3

2

1

1

1

!

Thebottom-righttriangleisrightisosceles,soits

hypotenusehaslength 2 .UsingthePythagorean

Theoremontheuppertriangleyieldsahypotenuseof

length 3 ,sothedesiredcosineis

adj

hyp=

1

3

=

3

3.

cos x

sin x

= cot x =3

2 ,so

tan x =2

3 ,and

tan x

cot x

=

2

3

( )3

2( )

=2

3⋅2

3=4

9.

4

9

Sinceαandβarecomplements,theircofunctionsareequal,so

sin β =1

3.Thecirclediagramshowntotherightrevealsthat

tanβ =1

2 2 ,andviareZlectionsandnegative-angleidentities,

weseethat

tan 180°−β ( )= tan −β ( )= −tanβ =−1

2 2

=− 2

4.

− 2

4

Asin#1,accept

quivalentssuchas

−1

2 2

,butbe

nstructiveto

tudentsaboutthe

ewgrading

uidelines)

Thinkingtransformationally,therangeofy=cosxis–1≤cosx≤1,sowe:• Multiplyallsidesby2:–2≤2cosx≤2

• Subtract5fromallsides:–7≤2cosx–5≤–3• Squareallsides,whichreversestheinequality:9≤(2cosx–5) 2≤49

Sothemaximumpossiblevalueoccursat49-butisthereanx-valuebetween0and2

thatpairswith49?

2 cos 3 x −π

4

⎛

⎝ ⎜⎞

⎠ ⎟ −5

⎡

⎣⎢⎢

⎤

⎦⎥⎥

2

= 49 ⇒ 2 cos 3 x −π

4

⎛

⎝ ⎜⎞

⎠ ⎟ −5= ±7 ⇒ cos 3 x −

π

4

⎛

⎝ ⎜⎞

⎠ ⎟ = −1or 6

⇒ 3 x −

π

4=π +2k π ⇒ 3 x =

5π

4+2k π ⇒ x =

5π

12+

8π

12k ,andtheonlyxthatworksis

5π

12.

5π

12, 49

⎛

⎝ ⎜⎞

⎠ ⎟

(Award1pointforeachcorrect

coordinate)

8/4/2019 2010-11 Meet 1


1. Determineexactlytheleastvalueof x thatsatisHiestheequation x −4( ) x +4( )= 9 .


y + x

2+6 x = 4 .

3. Forwhatvalueofk willthegraphsof y=k and

y =−

1

4 x

2

+

3

2 x −

1

2intersectatonly

onepoint?

4. Determineexactlyeachofthefourcomplexrootsof

x 2+25

x 2+2 x +

10

x

+2=0 .


Name:___________________________________ Team:___________________________________

Minnesota State High School Mathematics Leagu2010-11 Meet 1, In v ua Event D


k=

8/4/2019 2010-11 Meet 1


1. Determineexactlytheleastvalueof x thatsatisHiestheequation x −4( ) x +4( )= 9 .


y + x

2+6 x = 4 .

3. Forwhatvalueofk willthegraphsof y=k and

y = −1

4 x

2+3

2 x −

1

2intersectatonly

onepoint?

4. Determineexactlyeachofthefourcomplexrootsof

x 2+25

x 2+2 x +

10

x

+2=0 .

Minnesota State High School Mathematics Leagu2010-11 Meet 1, In v ua Event D

SOLUTIONS

TheleftsideoftheequationsimpliZiesintothedifferenceoftwosquares:

x

2−16= 9 ⇒ x

2=25 ⇒ x = ±5 .Thelesserofthosetwosolutionsis–5.

Completingthesquarewithrespecttoxontheleftsideoftheequationyields

y + x +3( )

2

−9= 4 ,andisolatingy,wehave y = − x +3( )

2

+13 ,whichhasitsvertex

(maximumpoint)at(–3,13).

−5

−3, 13( )

(Award1pointforeachcorrectcoordinate)

Thisproblemisreallyanexerciseinrecognition.Wearegivenahorizontallineandavertically-orientedparabola,whichwillhaveasingleintersectiononlyatthevertex.They-coordinateofthevertexcanbefound,amongothermethods,bycalculating:

−b2

4a+c

⎛

⎝ ⎜

⎞

⎠ ⎟ = −

3

2( )

2

4 − 1

4( )

+1

2

⎛

⎝

⎜⎜

⎞

⎠

⎟ ⎟ = − −

9

4+2

4

⎛

⎝ ⎜⎞

⎠ ⎟ =7

4.

7

4

Thetwofractionaltermsarebothrelatedto

5

x

,sorewritetheZirsttwotermsas

x +5

x

⎛

⎝ ⎜⎞

⎠ ⎟

2

−10 .Thentheentireleftsidecanbeexpressedas

x +5

x

⎛

⎝ ⎜⎞

⎠ ⎟

2

+2 x +5

x

⎛

⎝ ⎜⎞

⎠ ⎟ −8 ,

andsetting

y = x +5

x ,wehave

y

2+2 y −8=0⇒ y +4( ) y −2( )=0⇒ y = −4or 2 .

So

x +5

x = −4⇒ x

2+4 x +5=0⇒ x =

−4± 16−20

2= −2± i ,

or

x +5

x =2⇒ x

2−2 x +5=0⇒ x =

2± 4−20

2=1±2i .

−2± i ,1±2i

(Award1point foreachpairof

correctroots)

k=

(Alsoaccept

1.75or

13

4)

8/4/2019 2010-11 Meet 1


1. Convexpolygon A

1 A

2... A

2n+1hasncongruentangles

(∠ A

2n+1 A

1 A

2≅ ∠ A

1 A

2 A

3≅ ...

≅∠ A

n−1 A

n A

n+1) ,andtheothern+1anglesarecongruenttoeachother.

If A

2n+1 A

1isparallelto

A

n A

n+1and

m∠ A

1 A

2 A

3=m∠ A

2n A

2n+1 A

1−2° ,Hindthevalueofn.

2. Determinethesmallestbaseb>10inwhich2010bisequivalenttoabase-10multiple

of2010.

3. Calculatethevaluesofk forwhichthegraphsof y=x and

y = −1

4 x

2+kx −

1

2intersect

atonlyonepoint.

4. exagonHEXGON canbedissectedintorighttriangles,

asshowninFigure4.Theninelinesegmentsin

thediagramallhavedifferentinteger

lengths,andGO hasasshorta

lengthaspossible.

Calculatethesmallest

possiblevalueforthe

perimeterofHEXGON .

5. DeterminethesmallestpossibleLCMofabc,bca,andcabifa,b,andcare

distinctdigits(witha<b<c)andarealsotherootsof x 3+18 x

2+mx +n=0 .

6. Thegraphof y = f(x),x≤9isthelefthalfoftheparabolawhichintersectsthe y -axis

at–4andthe x -axisat–2.Writeaformulafor f(x).

Eachquestionisworth4points.Teammembersmaycooperateinanyway,butattheendof20minutes,submitonlyonesetofanswers.Placeyouranswertoeachquestiononthelineprovided.

Team:___________________________________

Minnesota State High School Mathematics Leagu2010-11 Meet 1, Team Event

=

=

E H

X

G

N

O

Figure4

(x)=

=

8/4/2019 2010-11 Meet 1


1. Convexpolygon A

1 A

2... A

2n+1hasncongruentangles

(∠ A

2n+1 A

1 A

2≅ ∠ A

1 A

2 A

3≅ ...

≅∠ A

n−1 A

n A

n+1) ,andtheothern+1anglesarecongruenttoeachother.

If A

2n+1 A

1isparallelto

A

n A

n+1and

m∠ A

1 A

2 A

3=m∠ A

2n A

2n+1 A

1−2° ,Hindthevalueofn.

2. Determinethesmallestbaseb>10inwhich2010bisequivalenttoabase-10multiple

of2010.

3. Calculatethevaluesofk forwhichthegraphsof y=x and

y = −1

4 x

2+kx −

1

2intersect

atonlyonepoint.

4. exagonHEXGON canbedissectedintorighttriangles,

asshowninFigure4.Theninelinesegmentsin

thediagramallhavedifferentinteger

lengths,andGO hasthe

shortestlengthofall

thosesegments.

Calculatethesmallest

possiblevalueforthe

perimeterofHEXGON .

5. DeterminethesmallestpossibleLCMofabc,bca,andcabifa,b,andcare

distinctdigits(witha<b<c)andarealsotherootsof x 3+18 x

2+mx +n=0 .

6. Thegraphof y = f(x),x≤9isthelefthalfoftheparabolawhichintersectsthe y -axis

at–4andthe x -axisat–2.Writeaformulafor f(x).


b=

n=

E H

X

G

N

O

Figure4

f(x)=

k =

9

670

2± 2

2

(Award2pointsforeachcorrectvalue;accept

1±2

2

,decimalequivalents,etc.)

94

56700

1

10 x −9( )

2

−

121

10

or...

1

10 x

2−

9

5 x −4

SOLUTIONS (page 1)

Notethattheword

“calculate”willometimesbeused

ncaseswherea3-

placedecimal

approximationis

notnecessary,in

ordertomaskthe

natureofa

solution.)

8/4/2019 2010-11 Meet 1


1. Imaginealinethatdividesthe(2n+1)-gonin“half”,withthencongruentanglesofonemeasureononesideoftheline,andthe(n+1)congruentanglesofanothermeasureontheotherside.Forthetwoindicatedsidestobeparallel,theexterioranglesoneachsideofthedividinglinemustsumto180°.Furthermore,ifthetwotypesofinterioranglesdifferby2°,thensodotheirexteriorangles.Setupthe

equation

180°

n

=180°

n+1

+2° andsolveforntoobtainn=9 .

2.

2010b= 2

b3

0

b2

1

b

0

1

= b 2b2

+1( ) ,whichwearetoldequals2010k =2⋅3⋅5⋅67⋅k for

someintegerk.Ashortinvestigationof0,1,2,...,9revealsthatifthelastdigit

ofb 2b

2

+1( ) is0,thenthelastdigitofbmustalsobe0.Setb=10nforsome

integern.Thenb 2b

2

+1

( )= 10n 200n

2

+1

( )= 2 ⋅3⋅ 5 ⋅67⋅k

.So67mustdivideeithernor200n

2

+1.Wewantthe

smallestbaseb>10,soletn=67,whichmakesb=670 .

3. Substitutexforyandrearrange:−

1

4x

2

+ k −1( ) x − 1

2=0 .Bythequadraticformula,

x =

− k −1( )± k −1( )2

−4 −

1

4( ) −

1

2( )

2 −

1

4( )

Adiscriminantof0leadstoasinglex,sok −1( )

2

−4 −1

4( ) − 1

2( ) = 0⇒ k −1( )

2

=1

2⇒ k −1= ±

1

2⇒ k =

2± 2

2.

4. ThesmallestPythagoreantriplesare(3,4,5),(5,12,13),(6,8,10),(7,24,25),(8,15,17),...Beginby

choosingthesmallestvalueineachtripleasthelengthofGO .GO=3failsbecauseGH=4,and XGH

cannotbeanother3-4-5triangle.GO=5works,butthenHO=13,forcingON=84andHN=85,which

willmakeforaverylargehexagonperimeter.ThekeywillbetomakehypotenuseHOassmallas possiblesothat HON usesa“small”Pythagoreantriple.ThisisaccomplishedbysettingGO=6

andlabelingthediagramasshown.Theperimeteroftheresultinghexagonis94 .

5. Thegivencubicpolynomialtellsusthatthesumoftheroots,a+b+c,equals18.Therefore,

abc =100a+10b+ c = 99a+9b+ a+b+ c( )= 99a+9b+18= 9 11a+b+2( ) .Similarly,

bc a = 9 11b+ c +2( ) ,and

cab = 9 11c +a+2( ) .SotheLCMcontainsa9.Buildatable:

a b c 11a+b+2 11b+ c +2 11c +a+2 LCM

5 6 7 63= 32

⋅7 75=3⋅52

84= 22

⋅3⋅7 22

⋅34

⋅52

⋅7=56700

4 6 8 52= 22

⋅13 76 =22

⋅19 94 = 2⋅47 22

⋅32

⋅13⋅19⋅47= 417924

3 6 9 41 77 =7⋅11 104 = 2

3

⋅13 2

3

⋅3

2

⋅7

⋅11

⋅13

⋅41= 2954952

4 5 9 51= 3⋅17 66 = 2⋅3⋅11 105= 3⋅5⋅7 2⋅33

⋅5⋅7⋅11⋅17=353430

2 7 9 31 88 =23

⋅11 103 23

⋅32

⋅11⋅31⋅103= 2528856

1 8 9 21=3⋅7 99=32

⋅11 102= 2⋅3⋅17 2⋅34

⋅7⋅11⋅17=212058

ThesmallestpossibleLCMisclearly56700 .

6. Usingvertexformforaquadraticfunction, f x ( )= a x − h( )

2

+k ⇒ −4 = a 0− 9( )2

+k ⇒ k = −4−81a .

Rewritingusingthex-intercept, f x ( )= a x −h( )

2

+k ⇒ 0= a −2−9( )2

+ −4−81a( ) ⇒ 4 =121a−81a ⇒ a = 1

10.

Therefore,k = −4−81a = −4−

81

10=

−121

10,andonepossibleequationis

f x ( )= 1

10x −9( )

2

−121

10.


SOLUTIONS (page 2)

Figure1

9

1 2

1 7

1 5

2 4

1 0

8

6

E

H

X

G

O

2010-11 Meet 1

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