RMO 2015 full paper discussion UP region

1) Let ABC be a triangle . Let P and Q denote respectively the reflection of B and C in the internal angle bisector of angle A.Show that the triangles ABC and APQ have the same incentre .

2) Let P(x) = x^2 + a * x + b be a quadratic polynomial with real coefficients. Suppose there are real numbers s not equal to t such that P(s) = t and P(t) = s .Prove that( b-st )is the root of the equation x^2 +a * x + b - st = 0.

3) Find all integers a,b,c such that a^2 = bc +1 and b^2= ac+1

4)Suppose 32 objects are placed along a circle at equal distances.In how many ways can 3 objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite .?

5)Two circles X and Y in the plane intersect at two distinct points A and B, and the centre of Y lies on X .Let points C and D be on X and Y ,respectively ,such that C,B and D are collinear .Let point E on Y be such that DE is parallel to AC. Show that AE=AB.

6)Find all real numbers a such that 4<a<5 and a(a -3{a}) is an integer .(Here {a} denotes the fractional part of a.For example {1.5} = 0.5 ; {-3.4} = 0.6)

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`italics` or `_italics_`	italics
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Note: you must add a full line of space before and after lists for them to show up correctly
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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

What is the answer to 4th and 6th....

Samarth Agarwal - 5 years, 6 months ago

I think that the answer to the last one is 3+3^1/2 and 3+2^1/2 .How did the paper go ?

Raven Herd - 5 years, 6 months ago

I was getting 3 more answers

3+(1.5)^0.5, 3+(2)^0.5, 3+(2.5)^0.5, 3+(3)^0.5 and 3+(3.5)^0.5

@Samarth Agarwal – Then yours maybe correct .

@Raven Herd – Dont know ...thats why waiting for answer keys

@Samarth Agarwal – When will it come and how did you approach the problem ?

@Raven Herd – Dont know about the answers.......

My approach was that I formed a quadratic in fractional part and found its value by Shir Dharacharya and subjected it to condition 0<f<1 and the result was that there are 5 such integers possible for each I found the value of f and got these as answer...

@Samarth Agarwal – Post the complete solution .

@Raven Herd – Please let me confirm the answer..then I will post a complete solution

@Samarth Agarwal – What about the third one?

@Raven Herd – errr....I made a silly mistake in that....I got six solutions but there are 8 solutions

@Samarth Agarwal – I too got 5 answers

Shubhendra Singh - 5 years, 6 months ago

@Shubhendra Singh – Someone was saying 6 answers ....:(

@Samarth Agarwal – he said that 6-sqrt(3) is also a solution

@Shubhendra Singh – I have checked question 6 using wolfram alpha there are 5 solutions only :)

How to do the first one

I wrote a foolish solution which said that triangles ABC and AB'C' are congurent so incentre will coincide.... :P

I wasted nearly an hour thinking on that one it looked easy but wasn't .

I thought that I will prove that angle B'IC'= 90+ B'AC' because angle BIC = 90 + BAC but solution never came.

@Raven Herd – I was terrified on seeing the figure :P

@Samarth Agarwal – Did you do the fifth one?

@Raven Herd – yes

@Samarth Agarwal – easiest !

@Raven Herd – Yeah ... it consumed only ten minutes

@Samarth Agarwal – I took 3

but that would only happen when B' coincides with B and C' coincides with C.

@Raven Herd – I know that I am wrong but only hope to get some step marks....

@Samarth Agarwal – How many questions did you do in total ? (I mean attempt )

@Raven Herd – I attemted all out of which I am sure of 3.

@Samarth Agarwal – I attempted 5 out of which I am hoping for 3.

How many marks would I get in the 3rd question if I forgot to write 2 out of 8 solutions

How are the marks divided in every question ?

I think on steps and approach

@Samarth Agarwal – As for the third one I wrote the method but didn't get the time to write the final solution .What was your approach? I formed the equation a^2+b^2 = 1-ab and used the concept of modulus 4 .

@Raven Herd – After working for long I got : a+b+c=0 a^2+b^2+c^2=2 ab+bc+ca=-1.... but there is also a condition that c=0 and a=b=1,-1.

@Samarth Agarwal – Are you sure about this one if yes then post the complete solution.

@Raven Herd – I am sure that I did it wrong.....All my friends got 8 solutions an dI got only 6 :(

@Samarth Agarwal – It does have 8 solutions my friends also said the same.

@Raven Herd – What you think they would penalize on 6 solutions?

@Samarth Agarwal – I don't think so and I did not say that infact it is better than mine . I am in no position to demoralize you.

I got very complicated answer. That is a = 4 + 1/n where n is the +ve root of mn^2 -4n-2 = 0. Ya actually it works.

easha manideep d - 5 years, 6 months ago

You are talking about which question ?

About 6th one

4th one is 32C3 - 16.26 - 32.30

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$