RMO #1990

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)(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)1)Two boxes contain between them 6565 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 22 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)2)For all positive real numbers a,b,ca, b, c prove that ab+c+ba+c+ca+b32\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}.

3)3)A square sheet of paper ABCDABCD is so folded that BB falls on the mid-point of MM of CDCD. Prove that the crease will divide BCBC in the ratio 5:35:3.

4)4)Find the remainder when 219902^{1990} is divided by 19901990.

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

6)6)NN is a 5050-digit number (in a decimal scale). All digits except the 26th26^{th} digit (from the left) are 1.If NN is divisible by 1313, find the 26th26^{th} digit.

7)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: 30,33,41,58,66,6930, 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)8)If the circumcenter and centroid of a triangle coincide, prove that the triangle must be equilateral.

Note by Akshat Sharda
4 years ago

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It is easy to use the collory of C-S for Q2. however here is the solution with AM-HM: ab+c+ba+c+ca+b32\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}, ab+c+1+ba+c+1+ca+b+13\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}+1-3 (a+b+c)(1b+c+1a+c+1a+b)3(a+b+c)(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}) -3 12[(a+b)+(b+c)+(a+c)][1b+c+1a+c+1a+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 12[(a+b)+(b+c)+(a+c)][1b+c+1a+c+1a+b]312[32]3=923=32\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}

Sualeh Asif - 4 years ago

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RMO then contained 8 questions!

Aakash Khandelwal - 4 years ago

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

Nihar Mahajan - 4 years ago

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

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

Adding the above inequalitiees we get 2(AP+BP+CP)>(AB+BC+AC)(AP+BP+CP)>(AB+BC+AC)2(AP+BP+CP)>s2(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

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

AB>AP\Rightarrow AB>AP and AB>BPAB>BP

Similar reasoning gives us

AC>AP,AC>PCAC>AP , AC>PC and BC>BP,BC>PCBC>BP, BC>PC

Adding we get 2(AP+BP+CP)<2(AB+BC+AC)(AP+BP+CP)<2s2(AP+BP+CP)<2(AB+BC+AC) \Rightarrow (AP+BP+CP)<2s

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

Anish Roy - 2 years, 1 month ago

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2) Immediate using Titu's Lemma

4) Remainder is 10241024

Dev Sharma - 4 years ago

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Q7 Ages are 47 ,22 ,19 and 11

rajdeep das - 3 years ago

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Q6 it is 3

rajdeep das - 3 years ago

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