Page # 1

28th Indian National Mathematical Olympiad-2013

Time : 4 hours Februray 03, 2013

Instructions :

� Calculators (in any form) and protractors are not allowed.

� Rulers and compasses are allowed.

� Answer all the questions. All questions carry equal marks.

� Answer to each question should start on a new page. Clearly indicate the questions number.

1. Let 1 and 2 be two circles touching each other externally at R. Let l1 be a line which is tangent to 2 at

P and passing through the centre O1 of 1 . Similarly, let l

2 be a line which is tangent to 1 at Q and passing

through the centre O2 of 2 . Suppose l

1 and l

2 are not parallel and intersect at K. If KP = KQ, prove that the

triangle PQR is equilateral.

Sol.O1 O2R

KQ P

KP = KQ, So K lies on the common tangent (Radical Axis)

Now 1 2~KPQ KO O & PQK is isoceles

1 21 2

2 1

KQP KO OPQO O is cyclic

KPQ KO O

So, 1 2KO O is also Isosceles So, KO1 = KO

2 & O

1R = O

2R, clearly in 1 2O PO ,

1 22 2

O OPO

so, 1 2 30ºPO O & similarly 2 1 30ºQO O

so, 1 2 60ºO KR O KR & PQR is equilateral.

Page # 2

2. Find all positive integers m, n and primes p 5 such thatm(4m2 + m + 12) = 3(pn � 1).

Sol. 3 24 12 3 3 nm m m p

2 3 4 1 3 nm m p ; 5p & prime

{so, 2 3m must be odd so m is even let m = 2a

24 3 8 1 3 na a p

{Now a must be 3b or 3b + 1 because 3 is a factor so,Case-1 - Let a = 3b

236 3 24 1 3 nb b p

212 1 24 1 nb b p

Now 24b + 1 must divide 12b2 + 1 & hence it must divide b - 2,so the only possibility is b = 2 & hence m = 12 & p = 7, n = 4Case-2 : if a = 3b + 1

236 24 7 24 9 3 nb b b p

236 24 7 8 3 nb b b p

so, 8 3b must divide 236 24 7b b

Hence divides 49 which is not possible for b .

so, , 12,4m n

3. Let a,b,c,d be positive integers such that a b c d. Prove that the equation x4 � ax3 � bx2 � cx � d = 0 has

no integer solution.

Sol. 4 3 2 0x ax bx cx d & a b c d

, , ,a b c d N

Let be a factor of d because other roots can�t be of the formp

q as coefficient of x4 is 1.

so, roots are either integers or unreal or irrational in pairs. Now there may be atleast one more root

(say )which is integer & it is also a factor of d.

So, ,d d

Now, 0 0f d & 1 1 0f a b c d

also ( ) 0 0, f x for x d , So there is no positive integral root.

Also. for , 1x d ; f(x) > 0 so, no integral root in [-d, -1].

Hence there is no integral root. {Though roots are in (-1, 0)}.

4. Let n be a positive integer. Call a nonempty subset S of {1,2,3,.....,n} good if the arithemtic mean of theelements of S is also an integer. Further let to denote the number of good subsets of {1,2,3,.....,n}. Provethat t

n and n are both odd or both even.

Sol. Let 1 2 3, , ,... rA x x x x be a good subset, then there must be a

set 2 31 , 1 , 1 ,... 1 rB n x n x n x n x which is also good. So, good subsets occur in a

pair.

However, there are few cases when A = B, which means if ix A 1 in x A . To count the

number of these subsets.Case-1 : If n is odd.a. If the middle element is excluded, the no. of elements in such subsets is 2k.(k before middle, & k elements after). So sum of hte elements will be k(n + 1), Apparently these sets

are good. So no. of these subsets is1

22 1n

(i.e. odd)

Page # 3

b. Similarly if mid term1

2n

is included no. of terms is 2k + 1 & sum will be 2 1 1

2

k n again

these subsets will be good.

So number os subsets will be 1

22 1n

; (odd)

so toal number of sebsets = n 122.2 2

i.e. (even)

So, if n is odd. Rest of the subsets are occuring in pair and the complete set iteself is good. so, tn is

odd.Case 2:If n in evenAgain the number of elements will be 2k & sum will be k(n+1) & these subsets are not good, sodiscarded.so, if n is even, all the good subsets occur in distinct pairs. Also, the complete set itself is not good. Sotn is even.

5. In a acute triangle ABC, O is the circumcentre, H the orthocentre and G the centroid. Let OD be perpendicularto BC and HE be perpendicular to CA, with D on BC and E on CA. Let F be the mid-point of AB. Supposethe areas of triangles ODC, HEA and GFB are equal Find all the possible values of C.

Sol.

A

B C

E

F H O

D

G

So, ar ODC ar HEA ar GFB

1 1. .2 2 6OD DC AE HE

(where ar ABC )

1cos cos 2 cos cos2 3R A c A R A C

Equ. 1 - cos cos 2 cos cos2aR A c A R A C

sin

2sin cos cos2

AC A C sin rule

tan 2sin2A C

Equ. 2 - sin1cos 2 2 3bc AaR A

2 sin . 2 sin sin1

cos sin2 3

R B R C AR A B A

3cos 2sin sin 3 cosA B C B C

3cos cos sin sinB C B C tan tan 3B C

Now, tan tan tan tan .tan .tanA B C A B C

3

2sin2 tan tan .tan .tantan

C C A B CC

Page # 4

22

2

8tan3 tan

1 tanC

CC

4 2tan 4tan 3 0C C

2tan 1 3C or tan C = 1 or 3

so, 45º 60ºC or

6. Let a,b,c,x,y,z be positive real numbers such that a + b + c = x + y + z and abc = xyz. Further, suppose thata x < y < z c and a < b < c. Prove that a = x, b = y and c = z.

Sol. 0 c x c y c z

3 2 0 c x y z c xy xz zx c xyz

3 2. 0 c a b c c xy yz zx c abc

2 2 0 c ac bc c xy yz zx ab

xy yz zx ab bc ca ...(I)

Similarly, 0a x a y c z

xy yz zx ab bc ca ...(II)

So, xy yz zx ab bc ca & &c z x a therefore y = b