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# 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.

Note by Surya Prakash
3 years ago

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No. $$3$$

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...

- 3 years ago

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.

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

- 3 years ago

Good and easy

- 3 years ago

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

- 3 years ago

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

- 3 years ago

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

- 3 years ago

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

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

- 2 years, 12 months ago

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 $$AM-HM$$ .

- 3 years ago

@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?

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

- 3 years ago

For #1, does anyone have a detailed solution?

- 2 years, 2 months ago

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

- 2 years, 11 months ago

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

- 2 years ago

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

- 2 years ago

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

- 2 years, 10 months ago

Thnx bro. For posting the paper.

- 2 years, 11 months ago

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)

- 2 years, 11 months ago

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

- 3 years ago

(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 $$S = 0$$.

Thank you!

- 3 years ago

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

- 3 years ago

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. :)

- 2 years, 11 months ago

This proof is incorrect.

- 1 year, 6 months ago

Thanks a lot SURYA PRAKASH For Posting This Paper.

- 2 years, 12 months ago

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

- 2 years, 12 months ago

Maharashtra region

- 2 years, 11 months ago

Are you selected for INMO 2015

- 2 years, 11 months ago

No.

- 2 years, 11 months ago

ok

- 2 years, 11 months ago

Comment deleted Dec 24, 2014