# This note has been used to help create the RMO Math Contest Preparation wiki

. Regional Mathematics Olympiad-2014 Time: 3 hours December 07, 2014

Instructions: \(\bullet\) Calculators (in any form) and protractors are not allowed.

\( \bullet\) Rulers ands compasses are allowed.

\( \bullet\) Answer all the questions.

\(\bullet\) All questions carry equal marks. Maximum marks: 102

\(1.\) In an acute-angled \(\triangle ABC\), \(\angle ABC\) is the largest angle. The perpendicular bisectors of BC and BA intersect AC at X and Y respectively. Prove that circumcentre of \(\triangle ABC\) is incenter of \(\triangle BXY. \)

\( 2\). Let \(x,y.z\) be positive real numbers. Prove that \[\frac { { y }^{ 2 }+{ z }^{ 2 } }{ x } +\frac { { z }^{ 2 }+{ x }^{ 2 } }{ y } +\frac { { x }^{ 2 }+{ y }^{ 2 } }{ z } \ge 2(x+y+z) \]

\(3.\) Find all pairs of (x,y) of positive integers such that \(2x+7y\) divides \(7x+2y\).

\( 4\). For any positive integer \(n>1\) let \(P(n)\) denote the largest prime not exceeding n. Let \(N(n)\) denote the next prime larger than \(P(n)\). (For example, \(P(10)=7\) and \(N(10)=11\).) If \(n+\) is a prime number, prove that the value of the sum \[\frac { 1 }{ P(2)N(2) } +\frac { 1 }{ P(3)N(3) } +...................+\frac { 1 }{ P(n)N(n) } =\frac { n-1 }{ 2n+2 } \]

\( 5\). Let \(\triangle ABC\) be a triangle with \(AB>AC\). Let \(P\) be a point on line beyond \(A\) such that \(AP+PC=AB\). Let \(M\) be the mid-point of \(BC\) and let \(Q\) be a point on the side \(AB\) such that \( CQ\bot AM\). Prove that \(BQ=2AP. \)

\( 6\). Each square of an \(n \times n\) grid is arbitrarily filled with either by \(1\) or by \( -1\). Let \( { r }_{ j }\) and \({ c }_{ k }\) denote the product of all numbers in the \(j-th\) row and the \(k-th\) column respectively, \(1\le j,k\le n\). Prove that \(\sum _{ j=1 }^{ n }{ { r }_{ j } } +\sum _{ k=1 }^{ n } c_ {k} \neq 0. \)

Note: In Question No.6, \(n\) is an odd number.

This is RMO 2014 Coastal Andhra and Rayalaseema region. I had attempted first 4 questions. And 30 members will be selected from our region. And I want to know whether my answers are correct or not. So please try solve and keep the solutions. And, Thanks in Advance.

## Comments

Sort by:

TopNewestNo. \(3\)

Since \(2x+7y\) needs to be divide in \(7x+2y\), we can clearly say that \((2x+7y)/(7x+2y)\) can be \(1\) or any number.

So, 1st case:

\((2x+7y)/(7x+2y) = 1\)

\(2x+7y = 7x+2y\)

\(5y = 5x\)

\(y = x\)

So, Ordered pair \((x,y) = (1,1), (2,2), ...\)

Second case

\((2x+7y) = 2(7x+2y)\)

\(2x+7y = 14x+4y\)

\(3y = 12x\)

\(y = 4x\)

So, ordered pair \((x,y) = (1,4), (2,8), ...\)

3rd case

\(2x+7y = 3(7x+2y)\)

\(2x+7y = 21x + 6y\)

\(y = 19x\)

So, ordered pair \((x,y) = (1,19), (2,38), ...\)

4th Case

\(2x+7y = 4(7x+2y)\)

\(2x+7y = 28x+8y\)

\(-26x = y\) --->Rejected since there will be formed a 'Negative Integer"

In general,

\((x,y) = (x,x), (x, 19x), (x, 4x)\) for x is an NATURAL number... – Christian Daang · 2 years, 1 month ago

Log in to reply

question number 2 is solved directly by the use of Titu's Lemma or Cauchy-Schwarz in Engel form

It states that, for any any pairs of numbers \(a_i,b_i\in\mathbb{R^{+}}\) The following inequality always holds.

\[\large{\frac{a_1^{2}}{b_1}+\frac{a_2^{2}}{b_2}+\dots+\frac{a_n^{2}}{b_n}≥\frac{(a_1+a_2+\dots+a_n)^{2}}{b_1+b_2+\dots+b_n}}\] – Aritra Jana · 2 years, 1 month ago

Log in to reply

– Parth Lohomi · 2 years, 1 month ago

Good and easyLog in to reply

Number 4 : Hint : Telescoping sum !

Details : Assume that the sequences of primes be \(\{p_1,p_2, p_3, \ldots \}\). Clearly, \(p_1=2,p_2=3, \ldots\) and so on. Now for all numbers \(n\) such that \(p_i\leq n < p_{i+1}\), we have \(P(i)=p_i\) and \(N(i)=p_{i+1}\). How many numbers fall in this range ? Precisely \(p_{i+1}-p_i\) of them. Since \(n+1\) is a prime, we have \(p_{k+1}=n+1\) for some integer \(k\). Thus, \[\frac{1}{P(2)N(2)}+\frac{1}{P(3)N(3)}+\ldots + \frac{1}{P(n)N(n)}=\sum_{i=1}^{k}\frac{p_{i+1}-p_{i}}{p_ip_{i+1}}=\sum_{i=1}^{k}\big(\frac{1}{p_i}-\frac{1}{p_{i+1}}\big)=\frac{1}{p_1}-\frac{1}{p_{k+1}}\] The result follows by noting that \(p_1=2, p_{k+1}=n+1,\hspace{10pt} \blacksquare\). – Abhishek Sinha · 2 years, 1 month ago

Log in to reply

@Surya Prakash : Only mathematical expressions should be rendered in LaTeX. See point 2 of Suggestions for Sharers. – Jon Haussmann · 2 years, 1 month ago

Log in to reply

\(Answer\quad to\quad Question\quad number\quad 2,\\ \\ \qquad Consider,\\ \qquad \qquad yz({ y }^{ 2 }+{ z }^{ 2 })+xz({ x }^{ 2 }+{ z }^{ 2 })+xy({ x }^{ 2 }+{ y }^{ 2 })\ge 2({ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 2 }{ z }^{ 2 }+{ z }^{ 2 }{ x }^{ 2 })\\ \qquad Now,\\ \qquad \qquad 2({ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 2 }{ z }^{ 2 }+{ z }^{ 2 }{ x }^{ 2 })=({ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 2 }{ z }^{ 2 })+({ z }^{ 2 }{ x }^{ 2 }+{ x }^{ 2 }{ y }^{ 2 })+({ y }^{ 2 }{ z }^{ 2 }+{ z }^{ 2 }{ x }^{ 2 })\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ge 2({ x }^{ 2 }yz+x{ y }^{ 2 }z+xy{ z }^{ 2 })=2xyz(x+y+z)\\ \qquad \therefore \quad yz({ y }^{ 2 }+{ z }^{ 2 })+xz({ x }^{ 2 }+{ z }^{ 2 })+xy({ x }^{ 2 }+{ y }^{ 2 })\ge 2xyz(x+y+z)\\ \qquad \Rightarrow \frac { { y }^{ 2 }+{ z }^{ 2 } }{ x } +\frac { { x }^{ 2 }+{ z }^{ 2 } }{ y } +\frac { { x }^{ 2 }+{ y }^{ 2 } }{ z } \ge 2(x+y+z)\) – Surya Prakash · 2 years, 1 month ago

Log in to reply

Here's my solution for question no.4, I did it using mathematical induction,

\(Given,\\ \qquad \frac { 1 }{ P(2)N(2) } +\frac { 1 }{ P(3)N(3) } +\frac { 1 }{ P(4)N(4) } +.....+\frac { 1 }{ P(n)N(n) } =\frac { n-1 }{ 2n+2 } \\ \qquad where\quad n+1\quad is\quad a\quad prime,\quad n>1\\ Now\quad above\quad sum\quad is\quad true\quad for\quad n=2.\\ Let\quad k+1\quad be\quad a\quad prime\quad for\quad which\quad above\quad sum\quad is\quad true.\\ \qquad \Rightarrow \sum _{ i=2 }^{ k }{ \frac { 1 }{ P(i)N(i) } } =\frac { k-1 }{ 2k+2 } \\ Let\quad the\quad prime\quad next\quad to\quad k+1\quad be\quad k+r+1.\\ \qquad \Rightarrow P(k+1)=P(k+2)=P(k+3)=...........................=P(k+r)=k+1.\\ \qquad \because Largest\quad prime\quad less\quad than\quad or\quad equal\quad to\quad k+i\quad is\quad k+1,\quad i=1,2,3....,r.\\ \qquad |||ly\quad N(k+1)=N(k+2)=N(k+3)=....................=N(k+r)=k+r+1\\ Now\quad we\quad have\quad to\quad prove\quad that\quad the\quad sum\quad is\quad true\quad for\quad n=k+r.\\ \qquad \sum _{ i=2 }^{ k+r }{ \frac { 1 }{ P(i)N(i) } } =\sum _{ i=2 }^{ k }{ \frac { 1 }{ P(i)N(i) } } +\sum _{ i=k+1 }^{ k+r }{ \frac { 1 }{ P(i)N(i) } } \\ \qquad \qquad \qquad \qquad \quad \quad =\frac { k-1 }{ 2k+2 } \quad +\quad \frac { r }{ (k+1)(k+r+1) } \\ \qquad \qquad \qquad \qquad \quad \quad =\frac { (k+r)-1 }{ 2(k+r)+2 } \quad (On\quad simplification)\\ Thus\quad by\quad principle\quad of\quad mathematical\quad induction\quad above\quad sum\quad is\\ true\quad \forall \quad n\epsilon N,\quad n>1,\quad n+1\quad is\quad a\quad prime.\\ \qquad \qquad \qquad \qquad \qquad Hence,\quad Proved. \) – Surya Prakash · 2 years, 1 month ago

Log in to reply

The second question can also be rewritten in the form \((x^2+y^2+z^2)(\frac1x+\frac1y+\frac1z)\ge 3(x+y+z)\)

This can be proved by using \((x^2+y^2+z^2)\ge \dfrac{(x+y+z)^2}{3}\) and \(AM-HM\) . – Rahul Saha · 2 years, 1 month ago

Log in to reply

@Surya Prakash For the sixth one, a \( 2X2 \) square with exactly one \( -1 \) seems to contradict the question statement. Is the question written correctly, or am I misreading something? – Siddhartha Srivastava · 2 years, 1 month ago

Log in to reply

– Surya Prakash · 2 years, 1 month ago

sorry for inconvenience , actually "n" is odd in given problemLog in to reply

For #1, does anyone have a detailed solution? – Shashank Rammoorthy · 1 year, 4 months ago

Log in to reply

Question 1 is really easy. Just involves understanding two triangles are congruent and so their corresponding angles are equal. Hardly a 2 liner solution – Shrihari B · 2 years ago

Log in to reply

Any idea about the cutoff? Or how many questions to qualify for the INMO? – Swapnil Das · 1 year, 1 month ago

Log in to reply

– Surya Prakash · 1 year, 1 month ago

I think minimum of three problems (with perfect solutions) are required.Log in to reply

Can anyone suggest some tips for selecting for IMOTC? I mean to get selected in INMO – Surya Prakash · 1 year, 12 months ago

Log in to reply

Thnx bro. For posting the paper. – Sourav Mishra · 2 years ago

Log in to reply

For Q2 we can use Muirhead's Theorem/inequality, which is very easy to use on Symmetric inequalities. Firstly, multiply both sides by \({x}\)\({y}\)\({z}\) and expand both sides. using bracket notation the problem reduces to showing that: [3,1,0] 'maximises' [2,1,1]. Well 3 conditions have to hold for A to maximise B: (\(\ A_i \)) and (\(\ B_i \)) are both decreasing sequences, \[\ a_1 + a_2 +...+ a_n = b_1 +b_2 + ...+ b_ n \ and \ a_1+a_2 +...+a_i \geq\ b_1 +b_2 +...+b_i \] (for 0 < \({i}\) < n) – Curtis Clement · 2 years ago

Log in to reply

I did the fourth problem using mathematical induction. Can any one suggest any other method than this? – Surya Prakash · 2 years, 1 month ago

Log in to reply

(6).Let \( \sum r_j + \sum c_k = S \).Now, take any random configuration with at least one \( -1 \). Suppose the \( -1 \) has coordinates \( (x,y) \). Changing the \( -1 \) to \( 1 \), we see that all \( r_j \) and \( c_k \) remain unchanged except for \( r_x \) and \( c_y \). These both change their signs. Now there are 4 cases,

1) Initially, \( r_x =c_y = 1 \). Then after the change \( r_x = c_y = - 1\) Therefore \( S_{Initial} = S_{Final} + 4 \).

2)Initially, \( r_x =-1, c_y = 1 \). Then after the change \( r_x =1 ,c_y = - 1\) Therefore \( S_{Initial} = S_{Final} \).

3)Initially, \( r_x = 1, c_y = -1 \). Then after the change \( r_x = - 1 ,c_y = 1\) Therefore \( S_{Initial} = S_{Final} \).

4)Initially, \( r_x =c_y = - 1 \). Then after the change \( r_x = c_y = 1\) Therefore \( S_{Initial} = S_{Final} - 4 \).

Therefore, we see that \( S_{initial} \equiv S_{final} \pmod4 \) is invariant. ---- (A)

Now we prove by contradiction. Suppose there exists a configuration with \( S = 0 \equiv 0 \pmod4 \). After changing all the \( -1 \)s to \( 1 \), By (A), we see that \( S_{final} \equiv 0 \pmod4 \). But \( S = 2n \equiv 2 \pmod4 \) since \( n \) is odd. Thus there a contradiction and our supposition is false. Therefore there exist no configuration with \( S = 0 \).

@Ryan Tamburrino – Siddhartha Srivastava · 2 years, 1 month ago

Log in to reply

– Ryan Tamburrino · 2 years, 1 month ago

Thank you!Log in to reply

Can someone provide a proof for number 6? Quite an intriguing question. – Ryan Tamburrino · 2 years, 1 month ago

Log in to reply

Number 2:

Let's say that x = y = z so that,

(2x^2)/x + (2x^2)/x + (2x^2)/x >/= 2(3x)

2x + 2x + 2x >/= 6x

6x >/= 6x

We can clearly see that 6x >/= 6x So, that inequality is correct. :) – Christian Daang · 2 years ago

Log in to reply

@Christian Daang

This proof is incorrect. – Mehul Arora · 7 months, 2 weeks ago

Log in to reply

Thanks a lot SURYA PRAKASH For Posting This Paper. – Ashay Wakode · 2 years, 1 month ago

Log in to reply

– Surya Prakash · 2 years, 1 month ago

Your welcome........................May I know to which state you belong to??Log in to reply

– Ashay Wakode · 2 years, 1 month ago

Maharashtra regionLog in to reply

– Surya Prakash · 2 years, 1 month ago

Are you selected for INMO 2015Log in to reply

– Ashay Wakode · 2 years, 1 month ago

No.Log in to reply

– Surya Prakash · 2 years, 1 month ago

okLog in to reply

Log in to reply

– Siddhartha Srivastava · 2 years, 1 month ago

You've literally copied this from someone else's solution.The only thing you've changed is the LaTeXing. The least you could do is mention whose solution this was.Log in to reply