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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. 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

 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

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

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