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)
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.
(a)
(c)
(d)
(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
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
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
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
3. Letus trysuccessivelyt om akeu pe achofthegivenamountsus ing6 coins :
inga right -angle dtrianglewith hyp otenus e40 m and one side of
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
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,
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:
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
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
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
