Q. 1 RMO 1997

Let

[BPF] = u

[BPC] = v

[CPE] = w join AP.

Let

[AFP] = x and [AEP] = y

using the triangles AFC and BFC, we get

vx – uy = uw

Again similarly using [AEB] and [CEB], we get

wx – vy = uw

Solving there equation, we get

So x + y =

Here u = 4

v = 8

B C

E

A

F WP

vu

w = 1 [AFPE] = 143

So/n Required

2

3

4

6

Q. 2 3k, K E No

O < k <

(1, 3, 9, 27, 81)

dn 20 + n2

dn 20 + (n + 1)2

Also

dn 20 + (n + 1)2 – (20 + n2)

dn 2n + 1

Also dn 4 (20 + n2) = (2n + 1) (2n - 1) + 81

dn (2n + 1) (2n - 1) so it must divide 81

By subtraction we obtain, dn 2n + 1, so 2n = -1 mod dn or

rather 4n2 = 1 mod dn or n2 = 20 mod dn, using Chinese theorem for the

residuum, using Chinese theorem for the residuum, we can say n2 = ¼ =

- 20 mod dn so dn 81 (dn = 1 mod2)

Q. 3 Clearly [x] ≠ 0 so this means these is no solutions in the interval (0,

1),

Let x = [x] + r, for some positive real number

real number

r E [0, ½ ). We have 2 cases

case 1 : r E [0, ½ ) . Then 2r E [0, 1)

and since

2x = 2 [x] + 2r

[2x] = 2[x], so,

-

since r E [0, ½ ), we find that

For [x] = 2, , and x = [x] + ,

For [x] = 3, , and ,

For [x] = 4, , and x = ,

Case 2 : r E [1/2, 1], Then

2r E [1,2)

2r – 1 E [0, 1)

and

2x = 2 [x] + 2r

= (2 [x] + 1) + (2r - 1)

[2x] = 2 [x] + 1

So

r =

since r E

we find that

then [x] < 2 since

[x] can nA be negative as well.

So [x] = 1. But, from the then end of the inequality, we have

so that [x] < 1

we member [x] ≠ 0.

Hence, there are no solutions

when r E

There fore the real solution to the original equation are

x =

Q. 4 RMO 1997

As diagonal are ┴ to each other

O

BA

AB2 + CD2 = AD2 + BC2 ………….(1)

Because AB is parallel to CD

Then CD > AB otherwise it will be square from properties A

squares or vice versa i.e. A3 > CD or it will be a square

So Δ AOB & DOC are similar

So AB2 = K2CD2

from (1) (AB + CD)2 2 ABCD

= (AD + BC)2 2 AD.BC

2 AB.CD – 2 AD BC = (AD+BC)2 – (AB+CD)2

AB.CD – ADBC = ½ (AD + BC + AB + CD)

(AD + BC – AB / CD)

In In

DB > AD + AB BC < CD + DB

AD < AB + DB (BC – CD < DB)

(AD - AB) < DB

So

AB . CD – ADBC > 0

CD

As If AD = BC

Then it because an sohmbur or square for any AD>BC ire qualify

holds

Hence AB.CD > AD. BC

for sector part

If AB.CD > AD BC

Then As per property

(AD2 + BC)2 = (AB + CD)2

AD2 + BC2 = AB2 + CD2

(AD + BC) 2 – (AB + CD)2

= 2AB . BC – 2 AB CD

= 2 (AD. BC – AB CD)

= A + xe quantity

= >

so

(AD + BC + AB + CD) {(AB + BC – (AB + CD))}

> 0

So

AB + BC > AB + CD

Q. 5

(a) let a = AP

b = BP

c = CP

d = DP

Let AB < CD

Since

AB║CD, Δ ABP ~ Δ CDCP

So

Where k E (o, 1]

Thus a = kc

b = kd

(1- k2)2 > 0

C2 (1 – k2) d2 (1 – k2) > 0

(c2 – a2) (d2 – b2) > 0

a2 b2 + c2 d2 + (a2 b2+b2a2) > a2 d2 + b2 c2 + (a2 c2 + b2 d2)

(a2 + d2) (b2 + c2) > (a2 +b2) (c2 + b2)

AD2 . BC > AB2 . CD2

(b) From part (a) we have

AD . BC > AB. CD

2AD . BC > 2 AB. CD

(a2+b2+c2+d2) + 2 AD. BC > 2 AB. CD + a2 + b2 + c2 + d2

(a2 + d2) + 2 AD . BC + (b2 + c2) > (a2 +b2) + 2 AB. CD + (c2+d2)

AD + 2 AD. BC + BC2 > AB2 + 2 AB . CD + CD2

(AD + BC)2 > (AB + CD) 2

AD + BC > AB + CD

Q. 5 RMO 1997

For triangle

equality holds if

x = y = z

But three district number real and positive

x > z

y > x Which is not possible

z > y Hence triangle in non possible.

Q. 6 From combinations

The number of ordered parts such as AUB = X is 3n =

(choose k elements from A,

elements for B – A are

composed and we can take

any subset A to complete B)

Now we eliminate case A = B (only we case)

And we divide by 2 to have ordered pair.