After seeing that my friends on Brilliant are eagerly preparing for RMO, I wish to help them and therefore I am posting the RMO$(1990)$ question paper, I also want to know about the correct way of solving these problems (my second purpose for posting these questions). Please post solution also.

$1)$Two boxes contain between them $65$ balls of several different sizes. Each ball is white, black, red or yellow. If you take any five balls of the same colour, at least $2$ of them will always be of the same size (radius). Prove that there are at least three balls which lie in the same box have the same colour and have the same size (radius).

$2)$For all positive real numbers $a, b, c$ prove that $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$.

$3)$A square sheet of paper $ABCD$ is so folded that $B$ falls on the mid-point of $M$ of $CD$. Prove that the crease will divide $BC$ in the ratio $5:3$.

$4)$Find the remainder when $2^{1990}$ is divided by $1990$.

$5)$$P$ is any point inside a $\triangle ABC$. The perimeter of the triangle $AB +BC +CA=2s$. Prove that $s < AP + BP + CP < 2s.$

$6)$$N$ is a $50$-digit number (in a decimal scale). All digits except the $26^{th}$ digit (from the left) are 1.If $N$ is divisible by $13$, find the $26^{th}$ digit.

$7)$A census-man on duty visited a house in which the lady inmates declined to reveal their individual ages, but said “We do not mind giving you the sum of the ages of any two ladies you may choose”. There upon the census man said, “In that case, please give me the sum of the ages of every possible pair of you”. They gave the sums as follows: $30, 33, 41, 58, 66, 69$. The census-man took these figures and happily went away. How did he calculate the individual ages of the ladies from these figures?

$8)$If the circumcenter and centroid of a triangle coincide, prove that the triangle must be equilateral.

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## Comments

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TopNewestIt is easy to use the collory of C-S for Q2. however here is the solution with AM-HM: $\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$, $\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}+1-3$ $(a+b+c)(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}) -3$ $\dfrac{1}{2}[(a+b)+(b+c)+(a+c)][\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}]-3$ Apply AM-HM to find that $\frac{1}{2}[(a+b)+(b+c)+(a+c)][\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}]-3 \geq \dfrac{1}{2}[3^2]-3=\dfrac{9}{2}-3=\dfrac{3}{2}$

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RMO then contained 8 questions!

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@Akshat Sharda Thanks for posting! And of course keep posting! :)

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This is probably one of the easiest RMO.

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Solution of Q.5

We know that in a triangle (sum of two sides) $>$ (third side)

In $\triangle APB, AP + PB>AB$

similarly, In $\triangle APC, AP + PC>AC$ and In $\triangle BPC, BP + PC>BC$

Adding the above inequalitiees we get $2(AP+BP+CP)>(AB+BC+AC) \Rightarrow (AP+BP+CP)>\frac{(AB+BC+AC)}{2} \Rightarrow (AP+BP+CP)>s$

Now, it is easy to see that

$\angle APB>\angle BAP , \angle APB>\angle ABP$

$\Rightarrow AB>AP$ and $AB>BP$

Similar reasoning gives us

$AC>AP , AC>PC$ and $BC>BP, BC>PC$

Adding we get $2(AP+BP+CP)<2(AB+BC+AC) \Rightarrow (AP+BP+CP)<2s$

$\therefore s<(AP+BP+CP)<2s$, as desired

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2) Immediate using Titu's Lemma

4) Remainder is $1024$

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Q7 Ages are 47 ,22 ,19 and 11

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Q6 it is 3

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