This document provides instructions for the 28th Indian National Mathematical Olympiad exam to be held on February 03, 2013. It states that calculators and protractors are not allowed, but rulers and compasses are. It includes 6 multi-part math problems to be solved on separate pages with clear numbering. The problems cover topics like properties of circles touching externally, positive integer solutions to equations, properties of polynomial equations, subsets with integer mean averages, relationships between areas of triangles formed by triangle centers, and inequalities relating positive real numbers.

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 l2 be a line which is tangent to 1 at Q and passing
through the centre O2 of 2 . Suppose l1 and l2 are not parallel and intersect at K. If KP = KQ, prove that the
triangle PQR is equilateral.
Q P
K
O1 R O2
Sol.
KP = KQ, So K lies on the common tangent (Radical Axis)
Now  KPQ ~ KO1O2 & PQK is isoceles
KQP  KO1O2
 PQO1O2 is cyclic 

KPQ  KO2O1

 
So, KO1O2 is also Isosceles So, KO1 = KO2 & O1R = O2R, clearly in O1PO2 ,
O1O2
PO2 
2

so, PO1O2  30º & similarly QO2O1  30º
so, O1KR  O2 KR  60º & PQR is equilateral.
Page # 1

2. 2. Find all positive integers m, n and primes p  5 such that
m(4m2 + m + 12) = 3(pn – 1).
Sol. 4m 3  m 2  12m  3  3 p n
2
 m  3   4m  1  3 p n ; p  5 & prime
{so, m 2  3 must be odd so m is even let m = 2a
2
  4a  3   8a  1  3 p n
{Now a must be 3b or 3b + 1 because 3 is a factor so,
Case-1 - Let a = 3b
2
 36b  3   24b  1  3p n
2
 12b  1  24b  1  p n
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 = 4
Case-2 : if a = 3b + 1
2
 36b  24b  7   24b  9   3 p n
2
  36b  24b  7   8b  3   p n
so,  8b  3  must divide 36b2  24b  7
Hence divides 49 which is not possible for b    .
so,  m, n   12,4 
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. x 4  ax 3  bx 2  cx  d  0 & a  b  c  d
a, b, c, d  N
p
Let  be a factor of d because other roots can’t be of the form 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, f  0   d  0 & f  1  1  a  b  c  d  0
also f (x)  0 for x 0,d , So there is no positive integral root.
Also. for x   d, 1 ; 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 the
elements of S is also an integer. Further let to denote the number of good subsets of {1,2,3,.....,n}. Prove
that tn and n are both odd or both even.
Sol. Let A  x1, x2 , x3 ,...xr  be a good subset, then there must be a
set B   n  1 x,  n  1  x2 ,  n  1  x3 ,...  n  1  xr  which is also good. So, good subsets occur in a
pair.
However, there are few cases when A = B, which means if xi  A   n  1  xi  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
n 1
are good. So no. of these subsets is 2 2
 1 (i.e. odd)
Page # 2

3.  n  1  2k  1 n  1
b. Similarly if mid term  is included no. of terms is 2k + 1 & sum will be again
 2   2
these subsets will be good.
n 1
So number os subsets will be 2 2
 1 ; (odd)
n 1
so toal number of sebsets = 2.2 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 even
Again the number of elements will be 2k & sum will be k(n+1) & these subsets are not good, so
discarded.
so, if n is even, all the good subsets occur in distinct pairs. Also, the complete set itself is not good. So
tn is even.
5. In a acute triangle ABC, O is the circumcentre, H the orthocentre and G the centroid. Let OD be perpendicular
to BC and HE be perpendicular to CA, with D on BC and E on CA. Let F be the mid-point of AB. Suppose
the areas of triangles ODC, HEA and GFB are equal Find all the possible values of C.
A
E
F H O
Sol. G
B D C
So, ar  ODC   ar  HEA   ar  GFB 
1 OD.DC   1 AE.HE  
 2 2 6
(where   ar  ABC  )
  R cos A   12   c cos A  2R cos A cos C    3
Equ. 1 - R cos A  a 2  c cos A  2R cos A cos C 
sin A
 2 sin C cos A cos C  sin rule 
2

 tan A  2 sin 2C
Equ. 2 - R cos A   a 2   1 bc sin A
2 3
1  2R sin B  .  2R sin C  sin A
 R cos A  B sin A  
2 3

 3 cos A  2 sin B sin C  3   cos  B  C  
 
 3 cos B cos C  sin B sin C  tan B tan C  3
Now, tan A  tan B  tan C  tan A.tan B.tan C
3
2 sin 2C   tan C  tan A.tan B.tan C
tan C

Page # 3

4. 8 tan2 C
3  tan2 C  tan4 C  4 tan2 C  3  0
1  tan2 C
 
 tan2 C  1or 3  tan C = 1 or 3
so, C  45º or 60º
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 that
a  x < y < z  c and a < b < c. Prove that a = x, b = y and c = z.
Sol.  c  x  c  y  c  z   0
 c 3   x  y  z  c 2   xy  xz  zx  c  xyz  0
 c 3   a  b  c  .c 2   xy  yz  zx  c  abc  0
 c 2   ac  bc  c 2    xy  yz  zx   ab  0
 xy  yz  zx  ab  bc  ca ...(I)
Similarly,  a  x  a  y  c  z   0
  xy  yz  zx   ab  bc  ca ...(II)
So, xy  yz  zx  ab  bc  ca & c  z & x  a therefore y = b
Page # 4