Let ABCD be a unit square. Draw a quadrant of circle with A as centre and B, D as end points of the arc. Similarly, Similarly, draw a quadrant of a circle with B as centre A, C as end points of the arc. Inscibe a circle touching the arcs AC and BD both externally and also touching the side CD. Find the radius of the circle

. Ans.

1 16

Sol.

By pythagores (r + R) 2 = (R – r)2 +

(r + R) 2 = (1 – r)2 +

1 4

(r + 1) 2 = (1 – r)2 +

1 4

r=

2. So l .

1 4

Here R = 1

1 . 16

Let a, b, c be positive integers such that a divides b 5, b divides c5 and c divides a5. Prove that abc divides (a + b + c) 31. Method 1 : Consider the expansion of (a + b + c) 31. All terms in it are of the form of r km ak b cm, where r km is a constant (in the form of binomial coefficient) and k, , m are are non-negative integers such that k + + m = 31. If k 1, 1, m 1 then abc always divides ak b cm. hence we have to consider terms in which one or two of k, , m are zero. Now let k = = 0, then a 31 = a. a5. a25 = a. (ck 1) (ck 1)5 = a (ck 1) (k2b k15) = abc k 16 k2 31 Clearly, Clearly, abc divides a and similarly we can say about b 31 and c31. Now let us consider two out of k, , m are not zero. Let k = 0, , m 0 then term would be rkm bcm, where + m = 31. b cm = bc b 1 cm 1, ,m 1 Now atleast one of –1 and m –1 must be greater than 25 and 5 respectively, respectively, which clearly shows the divisiblity of abc. –

–

Method 2 : a | b5 b5 = k 1 a ;

b | c 5 c5 = k 2 b ; c | a 5 a 5 = k 3 c

(a + b + c) 31

b5 a5 k b k 3 1

31

b5 b25 b 5 = k k1 k 3 1

31

31

b4 b24 31 1 =b k15k3 k1

31

b4 b24 5 25 1 = b. b . b k15 .k3 k1

= abc k1 b k1 125

4

126

k1125k830 .b24

Which is clearly divisible by abc. 3. Ans. Sol.

Let a and b be positive real numbers such that a + b = 1. Prove that aabb + abba 1. ((aa)2 + (ba)2) ((bb)2 + (ab)2) (aabb + ab ba)2

aabb + ab ba

(using cauchy skwartz s kwartz inequality)

((aa )2 (ba )2 ) ((bb )2 (ab )2 ) ......(1)

now a2a + b2a (a + b)2a and b2b + a2b (a + b)2b a2a + b2a 1 and b2b + a2b 1 (a2a + b2a)(b2b + a2b) 1 Equation (1) becomes becomes aa bb + abba 1 Hence proved

4. Ans. Sol.

Let X = {1, 2, 3,.....,10}. Find the the number of pairs {A, B} such A X. B X. A B and A B = {5, 7, 8}. 2186 X = {1, 2, 3, ......., 0} So X – (A B) has 7 elements. A will has 5, 7, 8. Rest elements can be assigned in 2 ways '1' can either go to A of B or none. So total pairs = 3 7 – (1).

(When no elements has been assigned to A or B.) 5.

Ans.

Sol.

Let ABC be a triangle. Let D, E be points on the segment BC such that BD = DE = EC. Let F be the midpoint of AC. Let BF intersect AD in P and AE in Q respectively. Determine the ratio of the area of the triangle APQ APQ to that of the quadrilateral PDEQ.

9 11

ADE =

3

x+y=

...(1)

3 Consider ADC

So

where x = a arrea of APQ

AP BD CF 1 PD CB AF

AQ BE CF 1 QE BC AF

AP 1 1 1 PD 3 1

AQ 2 1 1 QE 3

AP 3 PD

AQ 3 QE 2

AP =

3 AD 4

AQ =

3 AE 5

APQ =

3 3 AD AF sin A 4 5

APQ =

9 20 ADE

Thus (1)

x=

PQED

9 × 20 3 3 20

9 +y= y= 20 3 3 3

y=

6. Sol.

y = area of

Consider AEC

x=

x = y

&

9 1 – 20 11 3 20

3 9 20 = 11 11 3 20

Find all positive integers n such that 3 2n + 3n 3n2 + 7 is a perfect sqaure. 32n + 3n2 + 7 (a) If n is odd n = 2k + 1 n 2 is of form of 4 + 1 here 34k+2 + 3(4 + 1) + 7 9.81k + 12 + 10 (81)k 1 mod(4) 9(81)k 1 mod(4) 32n + 3n2 + 7 3 mod(4) and a perfect square can ’t be of form of 4 + 3 (b)

If n is even n = 2k 92k + 12k2 + 7 Now 92k < 92k + 12k2 + 7 (9k + 1) 1)2 where equality equality will hold only at k = 1 Rest it will be in between perfect square of 9 k and 9k +1 i.e. two consecutive consecutive no. Hence n = 2 is only solution.

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