[PDF] Problems and Solutions: INMO-2015

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Problems and Solutions: INMO-2015

1. LetABCbe a right-angled triangle with\B= 90. LetBDbe the

altitude fromBon toAC. LetP,QandIbe the incentres of triangles ABD,CBDandABCrespectively. Show that the circumcentre of of the trianglePIQlies on the hypotenuseAC.

Solution:We begin with the following lemma:

Lemma:LetXY Zbe a triangle with\XY Z= 90 +. Construct an isosceles triangleXEZ, externally on the sideXZ, with base angle . ThenEis the circumcentre of4XY Z.

Proof of the Lemma:DrawED?

XZ. ThenDEis the perpendicu-

lar bisector ofXZ. We also observe that\XED=\ZED= 90. Ob- serve thatEis on the perpendicu- lar bisector ofXZ. Construct the circumcircle ofXY Z. Draw per- pendicular bisector ofXYand let it meetDEinF. ThenFis the circumcentre of4XY Z. JoinXF.

Then\XFD= 90. But we know

that\XED= 90. HenceE=F.Letr1,r2andrbe the inradii of the trianglesABD,CBDandABC respectively. JoinPDandDQ. Observe that\PDQ= 90. Hence PQ

2=PD2+DQ2= 2r21+ 2r22:

Lets1= (AB+BD+DA)=2. Observe thatBD=ca=bandAD=pAB

2BD2=qc

2cab

2=c2=b. This givess1=cs=b. Butr1=

s

1c= (c=b)(sb) =cr=b. Similarly,r2=ar=b. Hence

PQ

2= 2r2c2+a2b

2 = 2r2:

Consider4PIQ. Observe that

\PIQ= 90+(B=2) = 135. HencePQ subtends90on the circumference of the circumcircle of4PIQ. But we have seen that\PDQ= 90.

Now construct a circle withPQas

diameter. Let it cutACagain inK.

It follows that\PKQ= 90and the

pointsP;D;K;Qare concyclic. We also notice\KPQ=\KDQ= 45 and\PQK=\PDA= 45. ThusPKQis an isosceles right-angled triangle withKP=KQ. Ther- foreKP2+KQ2=PQ2= 2r2and henceKP=KQ=r.

Now\PIQ= 90 + 45and\PKQ= 245= 90withKP=KQ=r.

HenceKis the circumcentre of4PIQ.

(Incidentally, This also shows thatKI=rand henceKis the point of contact of the incircle of4ABCwithAC.)

Solution 2:Here we use compu-

tation to prove that the point of contactKof the incircle withAC is the circumcentre of4PIQ. We show thatKP=KQ=r. Letr1 andr2be the inradii of triangles

ABDandCBDrespectively. Draw

PL?ACandQM?AC. Ifs1is

the semiperimeter of4ABD, then

AL=s1BD.But

s

1=AB+BD+DA2

; BD=cab ; AD=c2b Hences1=cs=b. This givesr1=s1c=cr=b,AL=s1BD=c(sa)=b.

HenceKL=AKAL= (sa)c(sa)b

=(bc)(sa)b . We observe that

2r2=(c+ab)22

=c2+a2+b22bc2ab+ 2ca2 = (b2babc+ac) = (bc)(ba):

This gives

(sa)(bc) = (sb+ba)(bc) =r(bc) + (ba)(bc) =r(bc) + 2r2=r(bc+c+ab) =ra:

ThusKL=ra=b. Finally,

KP

2=KL2+LP2=r2a2b

2+r2+c2b

2=r2:

ThusKP=r. Similarly,KQ=r. This givesKP=KI=KQ=rand

thereforeKis the circumcentre of4KIQ. (Incidentally, this also shows thatKL=ca=b=r2andKM=r1.)

2. For any natural numbern >1, write the infinite decimal expansion

of1=n(for example, we write1=2 = 0:49as its infinite decimal expan- sion, not0:5). Determine the length of the non-periodic part of the (infinite) decimal expansion of1=n. Solution:For any primep, letp(n)be the maximum power ofp dividingn; iepp(n)dividesnbut not higher power. Letrbe the length of the non-periodic part of the infinite decimal expansion of 1=n.

Write1n

= 0:a1a2arb

1b2bs:

We show thatr= max(2(n);5(n)).

Letaandbbe the numbersa1a2arandb=b1b2bsrespectively. (Herea1andb1can be both 0.) Then 1n =110 r a+X k1b(10 s)k! 110
r a+b10 s1

Thus we get10r(10s1) =n(10s1)a+b. It shows thatr

max(2(n);5(n)). Supposer >max(2(n);5(n)). Then10dividesba. Hence the last digits ofaandbare equal:ar=bs. This means 1n = 0:a1a2ar1b sb1b2bs1: This contradicts the definition ofr. Thereforer= max(2(n);5(n)).

3. Find all real functionsffromR!Rsatisfying the relation

f(x2+yf(x)) =xf(x+y): Solution:Putx= 0and we getfyf(0)= 0. Iff(0)6= 0, thenyf(0) takes all real values whenyvaries over real line. We getf(x)0. Supposef(0) = 0. Takingy=x, we getfx2xf(x)= 0for all real x. Suppose there existsx06= 0inRsuch thatf(x0) = 0. Puttingx=x0 in the given relation we get f x20=x0f(x0+y); for ally2R. Now the left side is a constant and hence it follows thatfis a constant function. But the only constant function which satisfies the equation is identically zero function, which is already obtained. Hence we may consider the case wheref(x)6= 0for all x6= 0. Sincefx2xf(x)= 0, we conclude thatx2xf(x) = 0for allx6= 0. This implies thatf(x) =xfor allx6= 0. Sincef(0) = 0, we conclude thatf(x) =xfor allx2R. Thus we have two functions:f(x)0andf(x) =xfor allx2R.

4. There are four basket-ball playersA;B;C;D. Initially, the ball is

withA. The ball is always passed from one person to a different person. In how many ways can the ball come back toAaftersevenpasses? (For exampleA!C!B!D!A!B!C!Aand

A!D!A!D!C!A!B!Aare two ways in which the ball

can come back toAafter seven passes.) Solution:Letxnbe the number of ways in whichAcan get back the ball afternpasses. Letynbe the number of ways in which the ball goes back to a fixed person other thanAafternpasses. Then x n= 3yn1; and y n=xn1+ 2yn1:

We also havex1= 0,x2= 3,y1= 1andy2= 2.

Eliminatingynandyn1, we getxn+1= 3xn1+ 2xn. Thus

x

3= 3x1+ 2x2= 23 = 6;

x

4= 3x2+ 2x3= (33) + (26) = 9 + 12 = 21;

x

5= 3x3+ 2x4= (36) + (221) = 18 + 42 = 60;

x

6= 3x4+ 2x5= (321) + (260) = 63 + 120 = 183;

x

7= 3x5+ 2x6= (360) + (2183) = 180 + 366 = 546:

Alternate solution:Since the ball goes back to one of the other 3 persons, we have x n+ 3yn= 3n; since there are3nways of passing the ball innpasses. Usingxn=

3yn1, we obtain

x n1+xn= 3n1; withx1= 0. Thus x

7= 36x6= 3635+x5= 3635+ 34x4= 3635+ 3433+x3

= 3

635+ 3433+ 32x2= 3635+ 3433+ 323

= (235) + (233) + (23) = 486 + 54 + 6 = 546:

5. LetABCDbe a convex quadrilateral. Let the diagonalsACandBD

intersect inP. LetPE,PF,PGandPHbe the altitudes fromPon to the sidesAB,BC,CDandDArespectively. Show thatABCDhas an incircle if and only if 1PE +1PG =1PF +1PH

CD=candDA=d. Let\APB=\CPD=. Then\BPC=\DPA=

. Let us also writePE=h1,PF=h2,PG=h3andPH=h4.

Observe that

h

1a=pqsin; h2b=qrsin; h3c=rssin; h4d=spsin:

Hence 1h 1+1h 3=1h 2+1h 4: is equivalent to apq +crs =bqr +dsp

This is the same as

ars+cpq=bsp+dqr: Thus we have to prove thata+c=b+dif and only ifars+cpq=bsp+dqr.

Now we can writea+c=b+das

a

2+c2+ 2ac=b2+d2+ 2bd:

But we know that

a

2=p2+q22pqcos; c2=r2+s22rscos

b

2=q2+r2+ 2qrcos; d2=p2+s2+ 2pscos;

Hencea+c=b+dis equivalent to

pqcos+rscos+ac=pscos+qrcos+bd: Similarly, by squaringars+cpq=bsp+dqrwe can show that it is equivalent to pqcos+rscos+ac=pscos+qrcos+bd: We conclude thata+c=b+dis equivalent tocpq+ars=bps+dqr.

HenceABCDhas an in circle if and only if

1h 1+1h 3=1h 2+1h 4:

6. From a set of11square integers, show that one can choose6num-

bersa2;b2;c2;d2;e2;f2such that a

2+b2+c2d2+e2+f2(mod 12):

Solution:The first observation is that we can find 5 pairs of squares such that the two numbers in a pair have the same parity. We can see this as follows:Odd numbersEven numbersOdd pairsEven pairsTotal pairs

011055

110055

29145
38145
47235
56235
65325
74325
83415
92415

101505

110505

Let us take such 5 pairs: say(x21;y21);(x22;y22);:::;(x25;y25). Thenx2jy2jis divisible by4for1j5. Letrjbe the remainder whenx2jy2jis divisible by 3,1j3. We have 5 remaindersr1;r2;r3;r4;r5.

But these can be 0, 1 or 2. Hence either one of the remainders occur 3 times or each of the remainders occur once. If, for example r

1=r2=r3, then 3 dividesr1+r2+r3; ifr1= 0;r2= 1andr3= 2, then

again 3 dividesr1+r2+r3. Thus we can always find three remainders whose sum is divisible by 3. This means we can find 3 pairs, say, (x21;y21);(x22;y22);(x23;y23)such that 3 divides(x21y21)+(x22y22)+(x23y23). Since each difference is divisible by 4, we conclude that we can find

6 numbersa2;b2;c2;d2;e2;f2such that

a

2+b2+c2d2+e2+f2(mod 12):

quotesdbs_dbs46.pdfusesText_46

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