# DISINI 2000-Sol00fix

Full text
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HIGH SCHOOL MATHEMATICS CONTEST

pre vious gu replus 2 square sofside1 eachm issing 1 matchstick. Thusthen gureinthese quence

: :

contains4+23n=6 n+4 matchs ticks. Thelarges tvaluen forwhich 500 matchs ticksis sucientis

:

thus82 ( which us es up682+4 =496 m atchs ticks). Now thenumb e r ofsquares in ther st gureis

3and each subsequent gurec ontains2 more s quar esthanthepreviousone . Ther efor ethenumb e r of

th

which im pliesthat `=16. BytheThe oremofPythagorasthelengthofthediago nalis

p

squareunits . Ther efor e,theprobabilityofhittingthes hade dar eais

2

6. Letus comp utethesumoftheprop e rdivisors ofeach ofthe5 p ossibleanswers inthelist:

13: 1<13

16: 1+2+4+8=15<16

30: 1+2+3+5+6+10+15=42>30

44: 1+2+4+11+22=40<44

50: 1+2+5+10+25=43<50

Theonlyone ofthesewhichqualiesasanabundantnumb e ris30. Ans we ris(C)

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8. Lett heco ordinates ofthep oint Sbe(x;y). SincePSkQRthe ymusthavethesame slop e :

y+2 0501 3

= =

x+3 109 4

or 4y03 x=1

SinceRSkQP weals ohave(bythes am eargum ent):

y01 05+2 3

= =0

x09 1+3 4

or 4y+3 x=31

Fromthe setwo equationsin2 unknowns weeasilys olveforx=5 andy=4. Thusx+y=9.

Answe ris( E)

9. Thediagram b elowshowshow gur es(a),(c),( d),and(e)canbelle dwithc op ie softhe\T"tile. No

m at terhowon e triesgur e(b)cannotb e lledwithcopie sofit.

## (e)

Answer is(B )

10. Noter stthat

5555 5 1111 1111

2 =(2 ) =32

3333 3 1111 1111

3 =(3 ) =27

2222 2 1111 1111

6 =(6 ) =36

1111 111 1 1111

Since27<32<36,wehave27 <32 <36 ,whichm eans

3333 5555 2222

3 <2 <6

Answe ris( E)

11. Letu srs tcompu tethenu mb e rofs econdsin1day,1hour,1minute ,and1s ec ond,andt henmu ltiplyby

2000. Now1dayplus1hourisc le ar ly25hours . The n1day,1hour ,plus1m inuteis25260+1=1501

m inute s. Expressed in se conds t his is 1501260 = 90060 seconds. Thus 1 day, 1 hour, 1 minute,

and 1 second is 90 ;061seconds. Th e answer to t heproblem is t his gur emultip lied by 2000; that is,

180; 122; 000, whichto thene ares tmillionis180 ;000 ;000. An sweris(D)

12. If a=b =c,t hen 100a+10b+c=100 a+10 a+a =111 a. Since a can b e any digit, in order for a

numbert obeafactorofthethree-digitnumber,itmus tb eafac torof111. Th efactorsof111 ar e1,3,

37,and111. Theonlyoneoftheseapp earing inthelistis37. Answe ris( E)

2 2 2 2 2 2

(x+y) 0(x0y ) >0 , x +2 xy+y 0x +2 xy0y >0

13.

, 4 xy>0

, xy>0

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14. Leta,b ,cb ethethreesidesoft hetriangle . Letusass umethatabc. Sinc ethep e rim et eris12,we

havea+b+c=12. Le tusnowlistallp ossibles etsofintege rs(a;b;c)s atis fyingtheab oveconditions:

(1;1 ;10); (1;2;9); (1;3 ;8); (1;4 ;7) ; (1;5 ;6); (2;2 ;8);

(2;3;7); (2;4 ;6); (2;5 ;5) ; ( 3;3 ;6); (3;4 ;5); (4;4;4)

However,itis clearthat some ofth ese\triangles"donotactu allye xis t,since inanytrianglethesum

of thelengthsof thetwo sh or ter sidesmust b e gre aterthan thelengthof thel onge st side. With this

addit ionalcondition wehaveonlythefoll owingtriangles(a;b;c) :

( 2;5;5); (3;4 ;5); (4;4;4)

Wecannowexam inethe4s tatementsandconc lude that(i),(ii)and(iv)arec le arlyt rue. Asfor (iii),

wes eethattr iangle(3;4 ;5) ab oveisright-angled; hence(iii)isfals e. An sweris(D)

1

15. The ar ea of t he origi nal triangle is bh. The new tr iangle has alt itude h+m and base b0x . We

2

1

ne ed to nd x sucht hat th e ar ea of the ne w triangle is bh . Clearly the area of the new triangle is

4

1

(h+m )(b0x). Thus

2

1

1

bh= (h+m)( b0x )

4

2

bh

=b0x

2(h+m )

bh 2 bh+2 bm0bh b (2m+h)

x=b0 = =

2(h+m ) 2(h+m ) 2(h+m)

Answe ris( E)

Senior Prelim ina ry

1. Antoninoaver age s15km/hfortherst20km .Thism eansittake shim20 = 15=4= 3hourstocoverth e

rs t20 km. Inor dertoaver age20 km/hfor a 40km dis tance ,he mustc over th edis tance in 2hours.

Heonlyhas2 = 3hr emaininginwhichtoc ove rth elast20km . H isspeedove rth islas t20kmthe nmus t

b e(on average)20 = (2= 3)=30 km/hr. Answer is(B )

2. LetC b ethec entreofthecircle. Sincet hep oint sOandB aree quidistantfromthecentreofthec ir cle

and also e quidistant from the p oint P, and s inc ebot h O and B lie ont he x-axis ,wese e that P has

co ordinates( 3;y)with y<0. Th e slop e ofOC is2 = 3. Sinc e PO isth e tan ge ntlin etothec ir cle atO

weknowt hatPO?OC. Ther eforethes lop eofPO is0 3= 2. H oweve r,t heslop eofPO iscom putedto

b e(y00)= (300)=y=3. Toge therth eseim plythat y=0 9= 2. Ans we ris(C)

3. First of all the numb e r of p os sible ways to cho ose a pair of dis tinc t students from a se t of ve is

0 1

5

5!

= =10. From this we needonly e liminatet hos ewhos e age die rence is 1. Clearl y the re ar e

2 2! 3!

exac tly 4 s uch, nam ely (6;7), (7 ;8), (8;9),and ( 9;10). S oourprobabilityo f successis 6 ou t of10, or

3 = 5. Ans we ris(C)

4. Thes traightsectionsoft heb eltar etange nttoall3pulle ysandthusp e rp e ndic ulartot heradiusofe ach

pulle yatthep ointofcontact . Thust hestraightse ctionsoftheb e ltarethes am ele ngthsasthedis tance s

p

2 2

b e tweenthec entre softhepulleys ,whichare5,12,and 5 +12 =13;sothest raightse ctionsofb elt

addupt o30units. Thecurve ds ectionsofbelt,whentakentogethe r,m akeuponec ompletepulley,or

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5. Letnb ethenumb erofquiz zesMarkhasalr eadytake n. Le txb ehistotals coreonallnqu iz zes. The n

orientations ,on ewithasingleve rte xabovethehorizontalbas eandonewith

asingle ve rte x b elowth ehorizontalbase . Forthersttyp eweh ave6 s uch:

totalof13e quilat eraltr iangle sso far. Howe ve r,there ar etwo other swhich

ares keweds om e whatt oth eedge soft heoutertriangle: (2;6;8)and(3;4 ;9),

which givesu sat otal of15e quilat eralt riangle s. A nswer is(A)

7. Thec ritical ideahere is tor ecognize that whenthe s qu ar e cove rsas mucho f thet riangleas p os sible,

the triangle will also cove r as mu ch of the square as pos sible , and that at this p oint the am ount of

trianglecovered isthesameas theamountof s quar ecove red . Le tA b etheareaof thetr iangle . The n

FromtheTheoremofP yt hagoraswehave

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Fore xac tly thesam ereasonas ab ovewes ee that B mustb e theintege rpart of5 =4,i.e .B =1. The n

n. Thus theprobab ilitythatoneball ofeachc olourischose nis:

mn mn mn 2 mn

14. Addingt hetwogive nequationstoge therwege t:

2 2

15. Alltheun shadedtriangle sinthediagramb e lowar eright-angle dand

thusarecongruent. B ytheThe or emofPythagor aswehave

2 2 2

2. Eachoft he10s traightline sinte rsectseachoftheothe rsexact lyonc e. Thism ake sfor90int ersections;

however,each oftheinte rsections isc ount edtwic ein thisapproach, de p en ding up onwhich ofthetwo

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3. Letus trysuccessivelyt om akeu pe achofthegivenamountsus ing6 coins :

inga right -angle dtrianglewith hyp otenus e40 m and one side of

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Whenweadd all20oftheabovee quat ionstogethe rweget

1

: : :

b =1+3+5+111+39= 20(21+192)=400

20

2

whe re we have used t he well-known formula for the sum of an arithm etic progre ssionwith n ter ms,

1

havingrstt ermaandc om m ondie renced : n(2a+(n01)d ). Answe ris( E)

2

9. Ac cordingtothedenitionofth eop eration3wehave

01 2 1 1

23( 01)=2 0(0 1) = 01=0

2 2

Ans we ris(C)

10. Leta b e thelengthof PQ , QR , andRS. The n th e radiiof th e3 circ le s ar e a,2 a,and 3a. The area

2 2 2

b e tweentheinne randmiddlec irc le sisthe n(2a) 0a =3 a ,andtheare ab etwe enthemiddleand

3

2 2 2 2 2

outercirclesis(3a) 0(2a) =5 a . Thust heratiowewantis3 a :5a = . An sweris(D)

5

Part B

1. (a) Fort his part of theques tion, t he simples t me tho d is s im ply to list all the pos sible numb er s. In

incre asin gorde rtheyare :

111; 112;113 ;122;123 ;133 ;222; 223 ;233 ;and333

foratotalof10 numbers.

(b) Again,m ostjuniorstudentswills im plytr ytolistallt hep os sib leinte ge rs . Inincre asingor dert hey

are :

1111;1112 ;1113;1114;1122 ;1123; 1124 ;1133 ;1134; 1144 ;1222;1223 ; 1224 ;1233;

1234;1244 ;1333;1334;1344 ;1444; 2222 ;2223 ;2224; 2233 ;2234;2244 ; 2333 ;2334;

2344;2444 ;3333;3334;3344 ;3444; and4444

foratotalof35 numbers.

Amores ophisticate dapproach(whichc anb egene ralize d)follows: Wer stdenen(k;d)tob eth e

numberof k-digitinte gerse ndingwiththedigitdands atis fyingthetwo conditions(i)and (ii)in

theproblem state me nt. Sinc e a k-digitnumb erending withthed igit dc onsists ofapp e ndingth e

digitdto all(k01)-digi tnumb ersendingwitha digitles sthanor equaltod ,wehave

n(k;d)=n(k01;1)+n(k01 ;2)+111+n(k01 ;d ) (3 )

Furtherm or e, we als o have n( 1;d) = 1 for all digits d an d n(k;1) = 1 for all integer s k. Th e

re lation ship( 3)all owsustocr eatethefoll owingtable ofvaluesfor n(k;d):

kn d 1 2 3 4

1 1 1 1 1

2 1 2 3 4

3 1 3 6 10

4 1 4 10 20

Eache nt ryinthetableisthesumofth eentriesinthepre viousrowupt oandinc ludingthecolum n

containingthegivene ntry(note thepr esence ofPas cal'sTrianglein thetab le ). Fromtheanswe r

toparts(a)and (b)are:

(a):n(3;1)+n(3;2)+n(3;3)=1+3+6=10

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2. SinceABCD is a s quarethe line s AC and BD are p e rp e ndic ular. Sinc e thecirc le had radius1 unit,

Recognizingthatmultipli c ationisc ommutativeforre alnumb er swecanreorganiz ethepro duc tsin

eachof theab oveine qualitie sandsumthethr ee inequali tiest ogetth edesir edre sult.

5. (a) Let us place 2 ( oute r) c oins next to th e original c oin s o t hat they touch each othe r. Then th e

centre s of t he3 coins form anequilate raltriangle with s ide le ngth e qualto twic ethe radiu s of a

singlec oin. Thereforeth eangle betweenthec entre soft he2 (outer)coinsme as ure dat thec entr e

ofther stc oinis60 . Since6suchan glesmakeupafullr evolutionar oun dtheinnercoin,wecan

have e xac tly 6 out ercoins e achtouching theoriginal (inne r)c oin and alsotouch ing itsothe r two

ne ighb ours.

(b) The reare6 non-over lappings pace swhoseareaswemus tadd;eachisfoundbetween3coinswhich

simultaneous lytouchother ,andwh os ec entr esformthee qu ilate raltriang lem ent ionedinp ar t(a)

ab ove . Thisequilate raltrianglehassidelength2,s inc ewearegive ntheradiiofthecoinsas1. O ur

st rategyto c ompute theare aof one such sp ac e is to n dt he area of theequilater al triangle and

sub tracttheare asofthe3circularsectorsfoundwithint hetrian gle. Thea ltitudeoftheequilater al

p

theequilate ral triangleare each one-sixth of the area of thecoin; there are 3 suchse ctors which

give s us a total area of one-half the area of a s ingle coin to b e subtrac te d fro m the are a of th e

p

equilate raltr iangle . Thus thear ea ofa singles paceis 30(= 2) . Sinc ethe re are6 s uch spac es,

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Senior Final

2. LetP beth eintersectionofAC andBDas inthediagrambelow. The

sumoft heinte rioranglesofa(r egular)p e ntagonis(502)180 =540 .

Soeachinte riorangleinaregularp e nt agonhasm eas ure540 = 5=108 .

Since 4ABC is isosc eles with vertex angle equal to 108 . The base

ages. Thuswearefor cedtoconclude thatt hesummustb e16,sincethe rearetwo distinctse tsofage s

which su mto 16in theab ovetable. Theonly age sfound in thesetwo se ts ar e2,3 ,5,9, and10. We

noticethat1 and6donotappear. An sweris(D)

4. Let V b e the volu me of a full tub (in litr es , say) . The n therate at which thehot water c an ll th e

tub isV= 10 litre s per minute. Similarly t he r at eatwhichthe c oldwate rc anll thetub is V= 8 litre s

p e rm inute. O ntheothe rh anda full tube mpt ie sat therateofV= 5litr esp e rm inut e. Ifall threear e

happ e ningatthesam etimethentherateatwhichthetub llsis:

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Since 19 is pr im e, in orde r for 19( k+10) to b e a p er fe ct s quare, k+10 mus t contain 19 as a factor.

Thesmalles ts uchvalue o c curs whe nk+10=19,i.e .whenk=9,andweindee dgeta p e rfec ts qu ar e

2

int hisc ase,name ly19 . Answer is(B )

5

6. Letnb et henumberin que stion. The nncanb e writtenas 10 +awher eaisanumberwithatm os t

5digit s. Moving theleft-mostdigit(thedigit1)to thee xtre mer ightpro duc esa numb e r10a+1. Th e

5

inform ationintheproble mnowte llsusthat10 a+1=3(10 +a)=300000+3a,or7a=299999. This

yie ld sa=42857 . Son=142857(andtheot hernumb e rwecr eatedis428571) ,thesumofwhos edigit s

is1+4+2+8+5+7=27. An sweris(D)

7. Letusorganiz et hissolut ionbycons ide ringt hes iz eofthelarges tcu b ei nthesubdivisionoftheoriginal

cu b e. Thelar ge stcould havea sideofs iz e4c m, 3c m, 2c m, or1 c m. Ineachofthes e4 c as eswewill

de terminetheminimumnumb er ofc ub esp os sible. Inther stcase,whe n the reis acubeofside4 cm

pre se nt,wec anonlyinc ludec ub esofside1cmtoc om pletethesubdivision,andwewouldne ed61s uch

3 3

sincethecub e ofs ide 4c m us esup64 c m ofthe125 c m in t heoriginal c ub e. Thus, in thiscasewe

have62cub e s in thesub divis ion. Ifwelo ok at theot here xtre me case,name lywhe n thelarges tcube

in thes ub division hassidelengthof1c m, weclearlyneed125 cu b esforthesub d ivision. Weals onot e

he rethatwewillcertainlyd ecreaset henumb e rofcub e sinasub divis ionifwet rytore places etsof121

cu b es bylarger cub e swhe neve rp oss ible. Nowconside r thecasewhen t hereis a cubeof sidelength3

cm pre sent. If we place itanywhe rebut in a corner, thes ub division c anonlybecom ple ted by c ub e s

ofs ide le ngth1c m, whichgive sus1+98=99c ub es intotal. Ifweplac eit ina corne r, wec an the n

plac e4c ub esofs idelength2cmonones ideoft helargerc ub e ,2m ores uchc ub e onas econdsideand

a thir d suchc ub e onth e thirdside; th is give sus1 largecubeand7 me dium cub e sfor a tot al volum e

3

of27+7(8)=83c m which me an swest ill 42 small cubes,for agrand totalof50 cubes. Itis e as yto

se e that ifthe large st cubehasside length2 cmwecan plac e at m os t 8 of themin the original cube

and ther emainde r ofthe volume mus t b e m ade upof c ub e s ofs ide le ngt h 1 c m; thi s givesa total of

8+61=69 c ub es . Thus thesmalle stnumb erofcubespos sibleis50andinthisc as ethe reare7 c ub e s

ofs idele ngth2 cm. An sweris(D)

8. Wen eedtondas eque nceofall9counc illorsbeginningwithAandendingwi thE suchthate achpair

of consecut ive councillors ar e`on speakingte rm s' with each other . When on e rs t looks at the table

provide d,itlo oksalit tledaunting. However ,arstobs ervationist hatam ongthec ouncillor sothe rthan

AandE (whoneedto app e arat th ee ndsofthesequence)c ou ncillor sF andH are only`on speaking

te rms ' with 2 others, one ofwhich is counc illor B. Thusc ounc illo r F mus t re ce ivethe rumor from B

andpass it t oI, or vic ever sa. Sim ilarly, councillorH must he ar therumor fromB andp as sit toC,

or vic e ve rs a. Thus we mus t haveeithe r I0F 0B0H 0C or C0H0B0F0I as consecutive

counc illorsinthese quence. Sin ceA andE lieonth eends,an dne ithe rofthemar e`onsp e akingte rm s

with' e ithe r counc illors C or I, we see that councillors D and Gmus t b e place done on e ith ere nd of

the ab oves ubsequence of 5 c ouncillors. This leaves us with eithe r D0I0F 0B0H 0C0G or

G0C0H 0B 0F 0I0D . Counc illors A and E can b e place d on the front and r ear of e ithe r

of the se se quences to give s t he nal s eque nce as e it her A0D0I0F 0B 0H 0C 0G0E o r

A0G0C0H0B0F0I0D0E. Ine ithe rcasethefourth p e rsonafte rcouncillor A(whostarted

therumor)t ohe artherumorwas councillorB. Answe ris(A)

9. Letuse xaminet het emp er atur edie rencesonthere sp e ctivepairsofthe rm om et ers. Adie re nceof24

onAcorr esp on dstoa die re nceof16 onB;t husthe yar ein theratioof3:2. Adi e renceof12 on

B corresp ondsto adi erenceof72 onC;thusthe yareintheratioof1:6. Nowa te mp e raturedrop

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10. Thenumb e r of p ositiveinte gers le ssthan ore qualto n wh ich ar emult iplesof kis theintege rpart of

n=k (that is ,p er formthe divisionand disc ar d thedecimal fraction ,ifany). This inte gerisc omm on ly

de note dbn=kc. Thust henumberofpos itiveint egersb etwe en200 and2000whicharemultiplesof6 is

2000 200

0 =333033=300

6 6

Similarly,thenumb erof p osit iveintegersb e twe en200and2000whichar emultipl e sof7 is

2000 200

0 =285028=257

7 7

Inord er to countthe numb e r ofpos itive inte ge rsb etwe en200and 2000 which a remultiples of 6 or 7

wec ouldaddtheab ovenumb er s. This ,howeve r,wouldc ountthemult iplesofboth6and7twic e;that

is,t hemultiple sof 42wouldbecounte dtwic e. Thu swene edtosub tractfrom thissumthenumb erof

p ositiveinte ge rsbetween200an d2000whicharemult iplesof42. That numb e ris

2000 200

0 =4704=43

42 42

The refore, the number of p ositive inte gers b e tween 200 and 2000 which are multipl e s of 6 or 7 is

300+257043=514. Butwear easkedforthenu mb e rofp ositiveinte gerswhicha remultip le sof6or7,

but NOTBOTH. Thusweneedto again subt ractthenumb e r ofmult iples of42in thisr ange ,name ly

43. Th enal answe ris514043=471. Answer is(B )

Part B

6

1. First draw the radius OD. Le t E = . Since D E = r = OD, 4DOE is isos celes. The refore,

6 6

DOE = . Since BDO is an e xt erior angle to 4DOE, it is equal in me as ure to the s um of th e

6

opp ositeint eriorangles ofthetriangle,i.e . BDO=2 . Now4BODisisos cele s sin cetwoofitss ide s

6 6 6

areradiiofthecircle. Thus D BO= BDO=2 . S inc e BOAisanexterior angleto4BOE, itis

6 6 6

equal in m easure to thes umof E =and EBO =2 . Thus BOA=3 , whichm eansthat th e

1

valuek int heproblemis .

3

2. Sincethere are 5 re gionsofe qu alare awh ich sumto180 s quare units ,each region has area36 s qu ar e

units. The dime nsions of theinne r square are clearly 6 unit s on a side , and of the out er square ar e

p p

180=6 5unitsonaside. Letxandyb et hedim ensionsofoneofthefou rcongruentre gions,wher e

p p

x<y. Thenx+y=6 5andy0x=6. Onaddingtheseanddividingby2wege ty=3( 5+1),and

p

the nite as ilyfollowsthat x=3( 501).

: : :

3. (a) Notethat 5!=5432 =120, which en ds in the digit0. Thus n!, wher e n>5,must als o e ndin

thedigit0,since5!even ly dividesn!, for n>5. Thusn!+1 e ndsin thedigit1 wheneve rn5,

which m eansthat 25canne vere ve nlyd ividen!+1whe nn5. Weareleft toe xam ine thecases

n=4,3,2,and1. S inc e 4!+1=24+1=25,wese et hat25divide s eve nly4!+1. Thusn=4 is

thelarges tvalueofnsuchthat25even ly dividesn!+1.

(b) Noters tthat (x+2y)+( y+2x )=3x+3 y=3(x+y). Thusthr eeeve nlydividesthesumofth e

twonumb er s. Thisc anberewritte nasy+2x=3(x+y)0(x+2 y ) . Suppos ethat3e ve nlydivide s

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4. (a) SameasProblem#5(b)ontheJu niorPap er(PartB).

(b) Let r b e the radius of each of the four large r c oins which

sur roundt hecoinofradius1. Thenbycons ide ringtwosuch

ne ighb ouringcoinsandthecoinofradius1(asinthediagram

b e low)wehavearight-angledtrianglewhenweconne ctthe

thr eec entr es. TheThe or emofPythagorasthenim pliest hat

2 2 2

5. (a) Thes liceisinthes hapeofanis os cele strianglewithtwosidesequaltothealtitu deoftheequilater al

triangularface s ofside aand thethir d sideof lengtha. B yusing theTheoremof Pythagorason

p

halfofoneequilater altr iangularfac ewese et hatitsalt itudeisgive nbya 3= 2. Thusthep er im eter

p p

ofth etrian gularslic e isa+2(a 3 =2)=a( 1+ 3) .

(b) Wemustnowndtheareaofthetrian gulars licewhoses ide swefoundinpart(a)ab ove. Cons ide r

thealtitude hwhich splitsthe isosc eless lice into 2 congruenthalve s. E achhalf isa right -angle d

Figure

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