We Know That If =

Q. 1 RMO 1991

We know that if =

then

Further areas of two triangles having the same height are in rate of Their bases.

In the figure,

………………..(1)

and

…………(2)

so …….(3)

Again

Þ ……(4)

so from (3) and (4)

Q. 2

2nd Proof:

Applying cava’s Theorem to The caviars

[A caviar mean a line segment joining a vertex of a triangle to any given point on the opposite side]

AD, BE, CF which are can current at P, we get

Hence

= …………….(1)

Now, by Menelaus Theorem to triangle ABD,

whose sides are cut by the tine FPC, we have

Þ …………..(2)

from (1) and (2) QED

Q. 2 RMO 1991

As a, b, c and d are 4 positive members then are also +ve we know, AM GM

Q.3 Let the 4 digit number is a b c d

As per question a=b & c=d

So number is a & d d [1 a 9 & 1 d 9]

We know,

102 = 100

1002 = 10000

so, it is two digit number

(xy)2 = aa bb

aabb is multiple A 11

being a perfect square it is multiple of 112 so possible numbers are

(11)2 x 1 = 121

(11)2 x 4

(11)2 x 9

(11)2 x 16

(11)2 x 25

(11)2 x 36

(11)2 x 49

(11)2 x 64 = 7744 in of form aabb

(11)2 x 8

Q. 4 Let us say urns X and Y. Both of them contain same number of balls, Then we can empty both the urns removing same number of balls form each urn.

Else, we remove the same number of balls form each of the urns so that one of the urns contain exactly one ball.

If x and y denote the number of balls is urns, say x > y, there take out y-1 balls from each.

We now double the number of balls of urn Y which contain only one ball and remove one ball form each of the urn. This forcers decreases the number of balls in the other urn x by 1. continuing this way we reach a stage when both the urns contain one ball each, whence we can empty the urns removing one ball from each of the two urns.

COMBI NATIORICS

Q. 5

Ref above figure

Let

Since P1P2//AC, so triangle BP1P2 is similar to triangle BCA.

Þ BP2 = k.AB

\ AP2 = (1- k) AB

since P2 P3 // BC and P2 A = (1 - k) AB

So triangle P2 A P3 and ABC are similar

Þ AP3 = (1 - k) AC

so P3C = kAC

Since P3P4 // AB and CP3 = k AC and triangles c P3P4 and CAB is similar

Þ P4B = (1- k) CB

Since P4 P5 // CA and

P4 B = (1 - k) CB

Therefore BP5 = (1 - k) BA So AP5 = k AB

Since P5 P6 // BC and

So AP5P6 and ABC are similar

\ P6 C = (1 - 5) AC

In Δ CAB P1 divides CB in the ratio 1-K : K

and P6 dives CA in the ratio 1-K : K

i.e.

so P6 P1 // AB

Q. 6 RMO 1991

(x + a) (x + 1991) +1

= x2 + (1991+a) x + 1991a + 1

we know D for a2 + bx + c is b2 – 4ac

Here discriminate = (1991+a)2 – 4 (1991a+1)

= 19912 + 2. 1991.a + a2

- 4 1991 a – 4

= 19912 – 2.1991.a + a2 – 4

= (1991 - a)2 -4

The expression (x - a) (x + 1991) +1 cab be factorized as (x + b) (x + a) with b and a as integers if and only if the discriminate as above is a perfect square

i. e. (1991 - a)2 – 4 = m2

Þ n2 – 4 = m2

Þ n2 – m2 = 4

Þ (n - m) (n + m) = 4

since m + n is positive that divides 4 n + m = 1,2 or 4

if n + m = 1 or 4 then n – m = 4 or ,

for that 1991 – a = + 2

Þ a = 1993 or 1989

Q. 7 RMO 1991

If n is even, n4 + 4n is divisible by 4

\ It is a composite number

If n is odd, Let n = 2p + 1 where p is positive integer

Then n4 + 4n

= n4 + (4)2p+1

= n4 + 4 (2p)4

Which is of the form n4 + 4b4, where b is a positive integer (= 2p)

n4 + 4b4

= (n4 + 4b2 + 4b4) – 4b2

= (n2 – 2b2)2 – (2b)2

= (n2 + 2b + 2b2) (n2 - 2b + 2b2)

We find that n4 + 4b4 is a composite number.

Consequently n4 + 4n is composite for all integers value n > 1

Q. 8 RMO 1991

Choose the square (any one) which is filled up with the smallest of all the positive integers filled in the 64 squares. It any of the neighbors is filled with positive integer greater then the smallest positive integer, then the average of all the integers in the squares occupying the neighbor’s positions will be greater than the said smallest integer. This is contradiction

Hence all the neighboring squares are filled with same smallest integer continuing is this way till all the squares are filled in finite number of steps.