Q. 1 RMO 1993
A
E
D
E
Let E and F be the midpoints of AD and BC respectively.
FG ┴ AB, G is the midpoint of DB.

EG
= ED + DG
= ½ (AD+DB)
= ½ AB
=4
FG
= 3 = (½ CD)
Right angled triangle
EF2
EGF
= EG2 + GF2
= 42 + 32
= 52

EF = 5
Q. 2 RMO 1993
For any n EN, if unit digit of 3n = 4n
Let
3 if n = 1 mod 4
9 if n = 2 mod 4
7 if n = 3 mod 4
1 if n = 4 mod 4
Not That 3 Un = unit mod 20
Since 20 / 3 – U1
and for any n EN
20/ 3n – Un

20/ 3n+1 – 3 Un
It follow by induction,
20/3n – Un for all n EN
Hence Ten’s digit of 3n is even
Q. 3 RMO 1993
Let us consider the circumscribing circle of the given regular polygon A1
A2 A3 …. A20. The twenty vertices divide the circle into twenty equal arcs
(because equal chords of a circle cut off equal arcs). The given problem
amounts to finding the number of ways in which we can select Three out of
the twenty arcs in such a way that :
(a)
No two of the vertices are consecutive (because two consecutive
vertices give rise to a side of the polygon).
(b)
No two of the arcs determined by the three vertices are of equal
length (because two equal arcs mean that the vaulting triangle is
isosceles).
Let us denote the are length between two correctives vertices by x.
(1)
The condition arises are is of length 2x let us being by choosing
A1 to be first of the Tree vertices. Let the second vertex chosen
be A3. The third vertex can be any one of the remaining vertices
with exception of A2, A4, A5, A20 or A12.
I and is A2, A4 or A20 it is consecutive to A1 or A3, it is as, then A1
A3 = A3 A5, if it is A19, then A19 = A1 A3, it is A12 then A3 A12
A1. Thus there are 12 choices (A6 to A11 and A13 to A18) for the
third vertex.
Since we could have begun by choosing any one of the twenty
vertices, there are 12 x 20 = 240 ways in which the smallest are is of
length 2x.
(2)
Let us now consider the cases in which the smallest are is of
length 3x.
As before, let us begin by choosing A1 and A4 as two of the
vertices. The remaining vertex can be any of the ten vertices as to
A4. Thus there are 20 choices for the third vertex, there are
20 x 10 = 200 ways in which the smallest are is of length 3x.
(3)
For smallest are of length 4x, as above there are 20 x 6 = 120
ways.
(4)
For smallest are 5x, it is evident the no of ways = 20 x 4 = 80
(5)
For length of smallest are can not exceed 5x. For it the length of
smallest are is at least 6x, the at least 7x and at least 5x, which is
not possible since the sum of all the Three arcs will then be at
least 22x which is > 20x.
So there are 240 + 200 + 120 + 80 = 640 ways and consequently 640
scalene triangles can be formed within given condition.
Q.4
E
D
C
b
A
F
B
a
O is the centre of the circle. The circle touches CD and passes Through
A and B E is the pint of tangency and F the point at which OE meets
AB.
OE ┴ AB
OF = b – r1
FB =
a
2
Δ OFB,
OB2 = OF2 + FB2

r

r1 =
2
1
= OF2 + FB2
b a2

2 8b
Similarly
a b2
r 2 
2 8a
a  b a 3  b3
r1  r2 

2
8ab
(a  b) (a  b)( a 2  ab  b 2 )


2
5ab
 1 (a  b) 2  ab) 
 ( a  b)  

8ab
2

 1 1 ( a  b) 2 
 ( a  b)   

8ab 
2 8
 4  1 ( a  b) 2 

 (a  b)

8
8
ab


5
 ( a  b)
8
So r1+r2 >
5
( a  b)
8
Q.5 RMO 1993
1993  (1  18) 93 93C0 (1) 93 C1  18 93C2 (18) 2  ......  1832
= 1 + 93.18 + multiple of 182
= 55 + multiple of 162
(13)99 = (1+12)99
= 1 + 99.12 +
99.98
99.98.97
.(12) 2 
(12) 3 + multiple of (12)4
2
6
= 1+18.12+18.12+a multiple of 162
Therefore 1993 – 1399
= (55-55) + multiple of 162
= multiple of 162
2
361
 19 
Also   
 13  169
 19 
So  
 13 
92
 2 46
Consequently
1923 > 19.1992
> 19.1392.246
> 1393.246
> 1393.224
>1399
Since 224 = (24)6 = 166 > 136
Thus 1993 – 1399 is a positive integer divisible by 162
Q. 6 RMO 1993
We know
A.M.> GM
1 a
 1.a
2
1 b
 1.b
2
1 c
 1.c
2
1 b
 1.b
2
(1+a) (1+b) (1+c) (1+d) > 16
a.b.c.d
> 16
as abcd = 1
Q. 7 RMO 1993
Let the ages are x1, x2 …… x10 respectively.
S = x1+x2+ ……+ x10
Since X1, x2 ….. x10 are whole numbers S in also whole number.
The sum written by 10 persons are S-x1, S-x2 ….. S-x10
By adding
(S-x1) + (S-x2) + ……
=
10S – (x1+x2+……
=
9S
+ (S – x10)
+x10)
So the sum of all the ten sums written by individual foresworn is multiple
of 9.
Since S-x, S-x2…..
S - x10 form a nine element set, therefore two of
them must be identical.
The total of nine sums given
82+83+84+…….
+92 = 783
which is a multiple of 9
There fore, the 10th sum, which must be one of the nine given sums must be
a multiple of 9. sine the only number one of given number which is
multiple of 9 is 90. so 10th number in 90.
This gives 9S = 783+90 = 873
 S = 97
Now subtracting the sum 82, 83…… to 92 gives individual ages 15, 14, 13,
12, 10, 8, 7, 7,6 and 5 respective
Q. 8 RMO 1993
Let us call the friends a, b, c, d, e and f.
Since I dinned with all the six on exactly once day, therefore at
exactly one dinner I had all the six friends.
Since, I dinned with every five of them on two days, I must have had
sis dinners in which I dinned with
bcdef;
acdef;
abdef;
a b c e f;
abcdf;
and
a b c d e;
respectively
The above seven (1+6) dinners are such that every four of the friends were
there in exactly 3 of these, and every there of them were There is exactly 4
of there.
Two of the friends say a and b, were present at the first dinner, and at four
other dinners, that it at five dinners. Like wise every other pair of the
friends was present at five of the above dinners.
Each of the friends were present at the first dinner and five others that is at
six dinners.
Since each one of the friends was present at seven dinners in all, I must
have had me dinner exclusively with each me of the six friends. It is clear
that each one of the friends was absent at five of these six dinners, and
therefore present at seven dinners in all, and absent at six out the thirteen
dinners considered so far.
Since each friend was absent at seven dinners, therefore I must have had
one dinner alone.