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INMO-2016 problems and solutions

1. LetABCbe triangle in whichAB=AC. Suppose the orthocentre of the triangle lies on the

in-circle. Find the ratioAB=BC.

Solution:Since the triangle is isosceles, the or-

thocentre lies on the perpendicularADfromAon toBC. Let it cut the in-circle atH. Now we are given thatHis the orthocentre of the triangle.

LetAB=AC=bandBC= 2a. ThenBD=a.

Observe thatb > asincebis the hypotenuse and

ais a leg of a right-angled triangle. LetBHmeet

ACinEandCHmeetABinF. By Pythagoras

theorem applied to4BDH, we get BH

2=HD2+BD2= 4r2+a2;whereris the in-radius ofABC. We want to computeBHin another way. SinceA;F;H;Eare

con-cyclic, we have

BHBE=BFBA:

ButBFBA=BDBC= 2a2, sinceA;F;D;Care con-cyclic. HenceBH2= 4a4=BE2. But BE

2= 4a2CE2= 4a2BF2= 4a22a2b

2 =4a2(b2a2)b 2:

This leads to

BH

2=a2b2b

2a2:

Thus we get

a 2b2b

2a2=a2+ 4r2:

This simplies to (a4=(b2a2)) = 4r2. Now we relatea;b;rin another way using area. We know that [ABC] =rs, wheresis the semi-perimeter ofABC. We haves= (b+b+ 2a)=2 =b+a. On the other hand area can be calculated using Heron's formula:: [ABC]2=s(s2a)(sb)(sb) = (b+a)(ba)a2=a2(b2a2): Hence r

2=[ABC]2s

2=a2(b2a2)(b+a)2:

Using this we get

a 4b

2a2= 4a2(b2a2)(b+a)2

Thereforea2= 4(ba)2, which givesa= 2(ba) or 2b= 3a. Finally, ABBC =b2a=34

Alternate Solution 1:

We use the known factsBH= 2RcosBandr= 4Rsin(A=2)sin(B=2)sin(C=2), whereRis the circumradius of4ABCandrits in-radius. Therefore

HD=BHsin\HBD= 2RcosBsin2

C = 2Rcos2B; since\C=\B. But\B= (\A)=2, sinceABCis isosceles. Thus we obtain

HD= 2Rcos22

A2 HoweverHDis also the diameter of the in circle. ThereforeHD= 2r. Thus we get

2Rcos22

A2 = 2r= 8Rsin(A=2)sin2((A)=4):

This reduces to

sin(A=2) = 2(1sin(A=2)): Therefore sin(A=2) = 2=3. We also observe that sin(A=2) =BD=AB. Finally ABBC =AB2BD=12sin(A=2)=34

Alternate Solution 2:

LetDbe the mid-point ofBC. ExtendADto meet the circumcircle inL. Then we know that HD=DL. ButHD= 2r. ThusDL= 2r. ThereforeIL=ID+DL=r+2r= 3r. We also know thatLB=LI. ThereforeLB= 3r. This gives BLLD =3r2r=32

But4BLDis similar to4ABD. So

ABBD =BLLD =32

Finally,

ABBC =AB2BD=34

Alternate Solution 3:

LetDbe the mid-point ofBCandEbe the mid-point ofDC. SinceDI=IH(=r) andDE=EC, the mid-point theorem implies thatIEkCH. ButCH?AB. ThereforeEI?AB. LetEImeet ABinF. ThenFis the point of tangency of the incircle of4ABCwithAB. Since the incircle is also tangent toBCatD, we haveBF=BD. Observe that4BFEis similar to4BDA. Hence ABBD =BEBF =BEBD =BD+DEBD = 1 +DEBD =32

This gives

ABBC =34

2. For positive real numbersa;b;c, which of the following statements

necessarily impliesa=b=c: (I)a(b3+c3) =b(c3+a3) =c(a3+b3), (II)a(a3+b3) =b(b3+c3) =c(c3+a3) ? Justify your answer. Solution:We show that (I) need not imply thata=b=cwhere as (II) always impliesa=b=c. Observe thata(b3+c3) =b(c3+a3) givesc3(ab) =ab(a2b2). This gives eithera=bor ab(a+b) =c3. Similarly,b=corbc(b+c) =a3. Ifa6=bandb6=c, we obtain ab(a+b) =c3; bc(b+c) =a3:

Therefore

b(a2c2) +b2(ac) =c3a3: This gives (ac)(a2+b2+c2+ab+bc+ca) = 0. Sincea;b;care positive, the only possibility is a=c. We have therefore 4 possibilities:a=b=c;a6=b,b6=candc=a;b6=c,c6=aanda=b; c6=a,a6=bandb=c. Supposea=bandb;a6=c. Thenb(c3+a3) =c(a3+b3) givesac3+a4= 2ca3. This implies that a(ac)(a2acc2) = 0. Thereforea2acc2= 0. Puttinga=c=x, we get the quadratic equationx2x1 = 0. Hencex= (1 +p5)=2. Thus we get a=b= 1 +p5 2 c; carbitrary positive real number:

Similarly, we get other two cases:

b=c= 1 +p5 2 a; aarbitrary positive real number; c=a= 1 +p5 2 b; barbitrary positive real number:

Anda=b=cis the fourth possibility.

Consider (II):a(a3+b3) =b(b3+c3) =c(c3+a3). Supposea;b;care mutually distinct. We may assumea= maxfa;b;cg. Hencea > banda > c. Usinga > b, we get from the rst relation that a

3+b3< b3+c3. Thereforea3< c3forcinga < c. This contradictsa > c. We conclude thata;b;c

cannot be mutually distinct. This means some two must be equal. Ifa=b, the equality of the rst two expressions givea3+b3=b3+c3so thata=c. Similarly, we can show thatb=cimpliesb=a andc=agivesc=b.

Alternate for (II) by a contestant:We can write

a 3c +b3c =c3a +a2; b 3a +c3a =a3b +b2; c 3b +a3b =b3c +c2:

Adding, we get

a 3c +b3a +c3b =a2+b2+c2:

Using C-S inequality, we have

(a2+b2+c2)2= pa 3pc pac+pb 3pa pba+pc 3pb pcb 2 a3c +b3a +c3b ac+ba+cb = (a2+b2+c2)(ab+bc+ca):

Thus we obtain

a

2+b2+c2ab+bc+ca:

However this implies (ab)2+ (bc)2+ (ca)20 and hencea=b=c.

3. LetNdenote the set of all natural numbers. Dene a functionT:N!NbyT(2k) =kand

T(2k+1) = 2k+2. We writeT2(n) =T(T(n)) and in generalTk(n) =Tk1(T(n)) for anyk >1. (i) Show that for eachn2N, there existsksuch thatTk(n) = 1. (ii) Fork2N, letckdenote the number of elements in the setfn:Tk(n) = 1g. Prove that c k+2=ck+1+ck, fork1.

Solution:

(i) Forn= 1, we haveT(1) = 2 andT2(1) =T(2) = 1. Hence we may assume thatn >1. Supposen >1 is even. ThenT(n) =n=2. We observe that (n=2)n1 forn >1. Supposen >1 is odd so thatn3. ThenT(n) =n+1 andT2(n) = (n+1)=2. Again we see that (n+ 1)=2(n1) forn3. Thus we see that in at most 2(n1) stepsTsendsnto 1. Hencek2(n1). (Here 2(n1) is only a bound. In reality, less number of steps will do.) (ii) We show thatcn=fn+1, wherefnis then-th Fibonacci number. Letn2Nand letk2Nbe such thatTk(n) = 1. Herencan be odd or even. Ifnis even, it can be either of the form 4d+ 2 or of the form 4d. Ifnis odd, then 1 =Tk(n) =Tk1(n+1). (Observe thatk >1; otherwise we getn+1 = 1 which is impossible sincen2N.) Heren+ 1 is even. Ifn= 4d+ 2, then again 1 =Tk(4d+ 2) =Tk1(2d+ 1). Here 2d+ 1 =n=2 is odd. Thus each solution ofTk1(m) = 1 produces exactly one solution ofTk(n) = 1 andnis either odd or of the form 4d+ 2. Ifn= 4d, we see that 1 =Tk(4d) =Tk1(2d) =Tk2(d). This shows that each solution of T k2(m) = 1 produces exactly one solution ofTk(n) = 1 of the form 4d. Thus the number of solutions ofTk(n) = 1 is equal to the number of solutions ofTk1(m) = 1 and the number of solutions ofTk2(l) = 1 fork >2. This shows thatck=ck1+ck2fork >2. We also observe that 2 is the only number which goes to 1 in one step and 4 is the only number which goes to 1 in two steps. Hencec1= 1 andc2= 2. This proves thatcn=fn+1for alln2N.

4. Suppose 2016 points of the circumference of a circle are coloured red and the remaining points are

coloured blue. Given any natural numbern3, prove that there is a regularn-sided polygon all of whose vertices are blue. Solution:LetA1;A2;:::;A2016be 2016 points on the circle which are colouredredand the remain- ing blue. Letn3 and letB1;B2;:::;Bnbe a regularn-sided polygon inscribed in this circle with the vertices chosen in anti-clock-wise direction. We placeB1atA1. (It is possible, in this position, some otherB's also coincide with some otherA's.) Rotate the polygon in anti-clock-wise direction gradually till someB's coincide with (an equal number of)A's second time. We again rotate the polygon in the same direction till someB's coincide with an equal number ofA's third time, and so on until we return to the original position, i.e.,B1atA1. We see that the number of rotations will not be more than 2016n, that is, at most these many times someB's would have coincided with an equal number ofA's. Since the interval (0;360) has innitely many points, we can nd a value2(0;360) through which the polygon can be rotated from its initial position such that noBcoincides with anyA. This gives an-sided regular polygon having only blue vertices. Alternate Solution:Consider a regular 2017n-gon on the circle; say,A1A2A3A2017n. For eachj, 1j2017, consider the pointsfAk:kj(mod 2017)g. These are the vertices of a regularn-gon, saySj. We get 2017 regularn-gons;S1;S2;:::;S2017. Since there are only 2016 red points, by pigeon-hole principle there must be somen-gon among these 2017 which does not contain any red point. But then it is a bluen-gon.

5. LetABCbe a right-angled triangle with\B= 90. LetDbe a point onACsuch that the in-radii

of the trianglesABDandCBDare equal. If this common value isr0and ifris the in-radius of triangleABC, prove that1r 0=1r +1BD

Solution:LetEandFbe the incentres of tri-

anglesABDandCBDrespectively. Let the in- circles of trianglesABDandCBDtouchACin

PandQrespectively. If\BDA=, we see that

r

0=PDtan(=2) =QDcot(=2):

Hence

PQ=PD+QD=r0

cot2 + tan2 =2r0sin:But we observe that

DP=BD+DAAB2

; DQ=BD+DCBC2

ThusPQ= (bca+ 2BD)=2. We also have

ac2 = [ABC] = [ABD] + [CBD] =r0(AB+BD+DA)2 +r0(CB+BD+DC)2 =r0(c+a+b+ 2BD)2 =r0(s+BD): But r

0=PQsin2

=PQh2BD; wherehis the altitude fromBon toAC. But we know thath=ac=b. Thus we get ac= 2r0(s+BD) = 2PQh2BD(s+BD) =(bca+ 2BD)ca(s+BD)2BDb:

Thus we get

2BDb= 2BD(sb))(s+BD):

This givesBD2=s(sb). SinceABCis a right-angled triangler=sb. Thus we getBD2=rs. On the other hand, we also have [ABC] =r0(s+BD). Thus we get rs= [ABC] =r0(s+BD): Hence 1r 0=1r +BDrs =1r +1BD

Alternate Solution 1:Observe that

r 0r =APAX =CQCX =AP+CQAC whereXis the point at which the incircle ofABCtouches the sideAC. Ifs1ands2are respectively the semi-perimeters of trianglesABDandCBD, we knowAP=s1BDandCQ=s2BD.

Thereforer0r

=(s1BD) + (s2BD)AC =s1+s22BDb But s

1+s2=AD+BD+c2

+CD+BD+a2 =(a+b+c) + 2BD2 =s+BD2

This gives

r0r =s+BD2BDb =sBDb

We also have

r

0=[ABD]s

1=[CBD]s

2=[ABD] + [CBD]s

1+s2=[ABC]s+BD=rss+BD:

This implies that

r0r =ss+BD: Comparing the two expressions forr0=r, we see that sBDb =ss+BD: Therefores2BD2=bs, orBD2=s(sb). Thus we getBD=ps(sb).

We know now that

r 0r =ss+BD=sBDb =BD(sb) +BD=ps(sb)(sb) +ps(sb)=pspsb+ps

Therefore

rr

0= 1 +rsbs

This gives

1r 0=1r rsbs 1r But rsbs 1r sbps(sb)! 1r =sbBD 1r

If\B= 90, we know thatr=sb. Therfore we get

1r 0=1r +sbBD 1r =1r +1BD

Alternate Solution 2 by a contestant:Ob-

serve that\EDF= 90. Hence4EDPis similar to4DFQ. ThereforeDPDQ=EPFQ. Tak- ingDP=y1andDQ=x1, we getx1y1= (r0)2.

We also observe thatBD=x1+x2=y1+y2.

Since\EBF= 45, we get

1 = tan45

= tan(1+2) =tan1+ tan21tan1tan2:But tan1=r0=y2and tan2=r0=x2. Hence we obtain 1 = (r0=y2) + (r0=x2)1(r0)2=x2y2:

Solving forr0, we get

r

0=x2y2x1y1x

2+y2:

We also know

r=AB+BCAC2 =x2+y2(x1+y1)2 =(x2x1) + (y2y1)2

Finally,

1r +1BD =2(x2x1) + (y2y1)+1x 1+x2

2x1+ 2x2+ (x2x1) + (y2y1)(x1+x2)((x2x1) + (y2y1)):

But we can write

2x1+ 2x2+ (x2x1) + (y2y1) = (x1+x2+x2x1) + (y1+y2+y2y1) = 2(x2+y2);

and (x1+x2)((x2x1) + (y2y1)) = 2(x1+x2)(x2y1) = 2(x2(x2+x1y1)x1y1) = 2(x2y2x1y1):

Therefore

1r +1BD =2(x2+y2)2(x2y2x1y1)=1r 0: Remark:One can also chooseB= (0;0),A= (0;a) andC= (1;0) and the coordinate geometry proof gets reduced considerbly.

6. Consider a non-constant arithmetic progressiona1;a2;:::;an;:::. Suppose there exist relatively

prime positive integersp >1 andq >1 such thata21,a2p+1anda2q+1are also the terms of the same arithmetic progression. Prove that the terms of the arithmetic progression are all integers.

Solution:Let us takea1=a. We have

a

2=a+kd;(a+pd)2=a+ld;(a+qd)2=a+md:

Thus we have

a+ld= (a+pd)2=a2+ 2pad+p2d2=a+kd+ 2pad+p2d2: Since we have non-constant AP, we see thatd6= 0. Hence we obtain 2pa+p2d=lk. Similarly, we get 2qa+q2d=mk. Observe thatp2qpq26= 0. Otherwisep=qand gcd(p;q) =p >1 which is a contradiction to the given hypothesis that gcd(p;q) = 1. Hence we can solve the two equations fora;d: a=p2(mk)q2(lk)2(p2qpq2); d=q(lk)p(mk)p

2qpq2:

It follows thata;dare rational numbers. We also have p

2a2=p2a+kp2d:

Butp2d=lk2pa. Thus we get

p

2a2=p2a+k(lk2pa) = (p2k)pa+k(lk):

This shows thatpasatises the equation

x

2(p2k)xk(lk) = 0:

Sinceais rational, we see thatpais rational. Writepa=w=z, wherewis an integer andzis a natural numbers such that gcd(w;z) = 1. Substituting in the equation, we obtain w

2(p2k)wzk(lk)z2= 0:

This showszdividesw. Since gcd(w;z) = 1, it follows thatz= 1 andpa=wan integer. (In fact any rational solution of a monic polynomial with integer coecients is necessarily an integer.) Similarly, we can prove thatqais an integer. Since gcd(p;q) = 1, there are integersuandvsuch thatpu+qv= 1. Thereforea= (pa)u+ (qa)v. It follows thatais an integer. Butp2d=lk2pa. Hencep2dis an integer. Similarly,q2dis also an integer. Since gcd(p2;q2) = 1,

it follows thatdis an integer. Combining these two, we see that all the terms of the AP are integers.

Alternatively, we can prove thataanddare integers in another way. We have seen thataandd are rationals; and we have three relations: a

2=a+kd; p2d+ 2pa=n1; q2d+ 2qa=n2;

wheren1=lkandn2=mk. Leta=u=vandd=x=ywhereu;xare integers andv;yare natural numbers, and gcd(u;v) = 1, gcd(x;y) = 1. Putting this in these relations, we obtain u

2y=uvy+kxv2;(1)

2puy+p2vx=vyn1;(2)

2quy+q2vx=vyn2:(3)

Now (1) shows thatvju2y. Since gcd(u;v) = 1, it follows thatvjy. Similarly (2) shows that yjp2vx. Using gcd(y;x) = 1, we get thatyjp2v. Similarly, (3) shows thatyjq2v. Thereforeydivides gcd(p2v;q2v) =v. The two resultsvjyandyjvimplyv=y, since bothv;yare positive.

Substitute this in (1) to get

u

2=uv+kxv:

This shows thatvju2. Since gcd(u;v) = 1, it follows thatv= 1. This givesv=y= 1. Finally a=uandd=xwhich are integers. |||-000000|||-quotesdbs_dbs47.pdfusesText_47

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