28th Indian National Mathematical Olympiad-2013

28th Indian National Mathematical Olympiad-2013

Time : 4 hoursFebruray 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 Gand G be two circles touching each other externally at R. Let l

be a line which is tangent to G at

12

12

P and passing through the centre O

of G . Similarly, let l

be a line which is tangent to G at Q and passing

11

2

1

through the centre O

of G . Suppose l and l

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

22

12

triangle PQR is equilateral.

Q

P

K

O1

R

O2

Sol.

KP = KQ, So K lies on the common tangent (Radical Axis)Now D KPQ ~ DKO1O2 & PQK is isoceles

KQP = KO1O2PQO1O2 is cyclic KPQ = KO2O1So, DKO1O2 is also Isosceles So, KO1 = KO2 & O1R = O2R, clearly in DO1PO2 ,

PO2 =O1O2

2

so,PO1O2 = 30 & similarly QO2O1= 30

so,O1KR = O2KR = 60 & DPQR is equilateral.

Page # 12.Find all positive integers m, n and primes p 5 such that

m(4m2 + m + 12) = 3(pn 1).

Sol.4m3+ m2 + 12m + 3 = 3pn

m2+ 3

4m + 1 = 3pn

(

)(

); p 5 & prime

2

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

4a2+ 3

8a + 1 = 3pn

(

)()

{Now a must be 3b or 3b + 1 because 3 is a factor so,

Case-1 - Let a = 3b

(36b2 + 3)(24b + 1) = 3pn

(12b2 + 1)(24b + 1) = pn

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

(36b2 + 24b + 7)(24b + 9) = 3pn

(36b2 + 24b + 7)(8b + 3) = pn

so, (8b + 3) must divide 36b2

+ 24b + 7

Hence divides 49 which is not possible for b I+.

so,

m,n = 12,4

()()

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

32

x

- ax

- bx - cx - d = 0 & a b c d

a,b,c,d N

Let a be a factor of d because other roots cant be of the formp

q as coefficient of x4is 1.

so, roots are either integers or unreal or irrational in pairs. Now there may be atleast one more root (say b )which is integer & it is also a factor of d.

So,{-d a,b 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. forx -d,-1 ; f(x) > 0 so, no integral root in [-d, -1].

[]

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

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

B ={(n + 1 x,n + 1- x ,n + 1- x ,... n + 1- xr } which is also good. So, good subsets occur in a

set

) ()2 ()3()

pair.

However, there are few cases when A = B, which means if x A

n + 1 - x A . To count the

i ()i

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 -1are good. So no. of these subsets is 2 2 1(i.e. odd)-

Page # 2

5.

Sol. n + 1

2k + 1n + 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.

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 HO

G

BDC

So, ar (DODC ) = ar (DHEA) = ar (DGFB)

12(OD.DC ) = 12 AE.HE = D 6 (where D = ar (DABC ) )

(R cos A)(12) = (c cos A)(2R cos A cosC ) = D 3

R cos A a=

c cos A 2R cos A cosC

Equ. 1 -(

)

2

(

)(

)

sin A

(

)

= 2sinC cos A cosC

sin rule

2

tan A = 2sin2C

Equ. 2 -(R cos A)(a

)=1bc sin A

2

23

R cos A

B sin A =1 (2R sinB ).(2R sinC )sin A

(

)(

)2

3

3cos A = 2sin B sinC = 3 - cos (B + C )

3cos B cosC = sinB sinCtanB tanC = 3

Now,tan A + tan B + tanC = tan A.tan B.tanC

2sin2C +3+ tanC = tan A.tan B.tanC

tanC

Page # 3

28 tan2 C

3 + tan C =

42

1+ tan2 C

tan C - 4tan C + 3 = 0

tan2 C = 1or 3tan C = 1 or3

so,C = 45 or 60

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

c3-

x + y + z c2+

xy + xz + zx c - xyz 0

(

)

(

)

c 3- a + b + c .c2+

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