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
5 years, 9 months 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 - 5 years, 9 months ago

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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


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 - 3 years, 10 months ago

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RMO then contained 8 questions!

Aakash Khandelwal - 5 years, 9 months ago

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This is probably one of the easiest RMO.

A Former Brilliant Member - 5 years, 9 months ago

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

Nihar Mahajan - 5 years, 9 months ago

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2) Immediate using Titu's Lemma

4) Remainder is 10241024

Dev Sharma - 5 years, 9 months ago

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Q7 Ages are 47 ,22 ,19 and 11

rajdeep das - 4 years, 10 months ago

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Q6 it is 3

rajdeep das - 4 years, 10 months ago

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First was easy, did contradiction. You can use pigeon hole too. 2nd easy Titu Lema... 3rd easy Pythagoras 4th Fermat's little theorem and a clever observation that $2¹⁹⁹⁰-2¹⁰$ is divisible by 10 5th medium problem. Had to proof another result beforehand. 6th Really easy! Divisibility rule of 13 or you could see a nice pattern in residues $modulo 13$ for 10^n. They repeat after every 6qth power of 10. 7th Medium. One just needs to pay attention and do not assign any equation to any sum. Take assumption that she is oldest, then she is 2nd oldest bla bla... 8th problem Tooooo easy! Took hardly 5 minutes. Use basic definition of centroid, circumcenter. Properties of median, Pythagoras (at least I used) Properties of isoceles triangle. Congruency can help too I think. That's it. Give it a try.

Tech Toppers - 10 months ago

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