. 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.

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## Comments

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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...

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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.

$\large{\frac{a_1^{2}}{b_1}+\frac{a_2^{2}}{b_2}+\dots+\frac{a_n^{2}}{b_n}≥\frac{(a_1+a_2+\dots+a_n)^{2}}{b_1+b_2+\dots+b_n}}$

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Good and easy

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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 $p_1=2, p_{k+1}=n+1,\hspace{10pt} \blacksquare$.

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@Surya Prakash : Only mathematical expressions should be rendered in LaTeX. See point 2 of Suggestions for Sharers.

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$Answer\quad to\quad Question\quad number\quad 2,\\ \\ \qquad Consider,\\ \qquad \qquad yz({ y }^{ 2 }+{ z }^{ 2 })+xz({ x }^{ 2 }+{ z }^{ 2 })+xy({ x }^{ 2 }+{ y }^{ 2 })\ge 2({ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 2 }{ z }^{ 2 }+{ z }^{ 2 }{ x }^{ 2 })\\ \qquad Now,\\ \qquad \qquad 2({ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 2 }{ z }^{ 2 }+{ z }^{ 2 }{ x }^{ 2 })=({ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 2 }{ z }^{ 2 })+({ z }^{ 2 }{ x }^{ 2 }+{ x }^{ 2 }{ y }^{ 2 })+({ y }^{ 2 }{ z }^{ 2 }+{ z }^{ 2 }{ x }^{ 2 })\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ge 2({ x }^{ 2 }yz+x{ y }^{ 2 }z+xy{ z }^{ 2 })=2xyz(x+y+z)\\ \qquad \therefore \quad yz({ y }^{ 2 }+{ z }^{ 2 })+xz({ x }^{ 2 }+{ z }^{ 2 })+xy({ x }^{ 2 }+{ y }^{ 2 })\ge 2xyz(x+y+z)\\ \qquad \Rightarrow \frac { { y }^{ 2 }+{ z }^{ 2 } }{ x } +\frac { { x }^{ 2 }+{ z }^{ 2 } }{ y } +\frac { { x }^{ 2 }+{ y }^{ 2 } }{ z } \ge 2(x+y+z)$

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@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?

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sorry for inconvenience , actually "n" is odd in given problem

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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 $AM-HM$ .

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

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

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

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For #1, does anyone have a detailed solution?

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Can someone provide a proof for number 6? Quite an intriguing question.

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(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

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Thank you!

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I did the fourth problem using mathematical induction. Can any one suggest any other method than this?

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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)

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Thnx bro. For posting the paper.

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Can anyone suggest some tips for selecting for IMOTC? I mean to get selected in INMO

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Any idea about the cutoff? Or how many questions to qualify for the INMO?

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I think minimum of three problems (with perfect solutions) are required.

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Thanks a lot SURYA PRAKASH For Posting This Paper.

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

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Maharashtra region

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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. :)

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@Christian Daang

This proof is incorrect.

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