RMO 2014 Coastal Andhra and Rayalaseema Region

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. 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

1.1. In an acute-angled ABC\triangle ABC, 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 ABC\triangle ABC is incenter of BXY.\triangle BXY.

2 2. Let x,y.zx,y.z be positive real numbers. Prove that y2+z2x+z2+x2y+x2+y2z2(x+y+z)\frac { { y }^{ 2 }+{ z }^{ 2 } }{ x } +\frac { { z }^{ 2 }+{ x }^{ 2 } }{ y } +\frac { { x }^{ 2 }+{ y }^{ 2 } }{ z } \ge 2(x+y+z)

3.3. Find all pairs of (x,y) of positive integers such that 2x+7y2x+7y divides 7x+2y7x+2y.

4 4. For any positive integer n>1n>1 let P(n)P(n) denote the largest prime not exceeding n. Let N(n)N(n) denote the next prime larger than P(n)P(n). (For example, P(10)=7P(10)=7 and N(10)=11N(10)=11.) If n+n+ is a prime number, prove that the value of the sum 1P(2)N(2)+1P(3)N(3)+...................+1P(n)N(n)=n12n+2\frac { 1 }{ P(2)N(2) } +\frac { 1 }{ P(3)N(3) } +...................+\frac { 1 }{ P(n)N(n) } =\frac { n-1 }{ 2n+2 }

5 5. Let ABC\triangle ABC be a triangle with AB>ACAB>AC. Let PP be a point on line beyond AA such that AP+PC=ABAP+PC=AB. Let MM be the mid-point of BCBC and let QQ be a point on the side ABAB such that CQAM CQ\bot AM. Prove that BQ=2AP.BQ=2AP.

6 6. Each square of an n×nn \times n grid is arbitrarily filled with either by 11 or by 1 -1. Let rj { r }_{ j } and ck{ c }_{ k } denote the product of all numbers in the jthj-th row and the kthk-th column respectively, 1j,kn1\le j,k\le n. Prove that j=1nrj+k=1nck0.\sum _{ j=1 }^{ n }{ { r }_{ j } } +\sum _{ k=1 }^{ n } c_ {k} \neq 0.

Note: In Question No.6, nn 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
4 years, 10 months ago

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1 vote

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No. 33

Since 2x+7y2x+7y needs to be divide in 7x+2y7x+2y, we can clearly say that (2x+7y)/(7x+2y)(2x+7y)/(7x+2y) can be 11 or any number.

So, 1st case:

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

2x+7y=7x+2y2x+7y = 7x+2y

5y=5x5y = 5x

y=xy = x

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

Second case

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

2x+7y=14x+4y2x+7y = 14x+4y

3y=12x3y = 12x

y=4xy = 4x

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

3rd case

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

2x+7y=21x+6y2x+7y = 21x + 6y

y=19xy = 19x

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

4th Case

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

2x+7y=28x+8y2x+7y = 28x+8y

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

In general,

(x,y)=(x,x),(x,19x),(x,4x)(x,y) = (x,x), (x, 19x), (x, 4x) for x is an NATURAL number...

Christian Daang - 4 years, 10 months ago

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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 ai,biR+a_i,b_i\in\mathbb{R^{+}} The following inequality always holds.

a12b1+a22b2++an2bn(a1+a2++an)2b1+b2++bn\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 - 4 years, 10 months ago

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Good and easy

Parth Lohomi - 4 years, 10 months ago

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Number 4 : Hint : Telescoping sum !

Details : Assume that the sequences of primes be {p1,p2,p3,}\{p_1,p_2, p_3, \ldots \}. Clearly, p1=2,p2=3,p_1=2,p_2=3, \ldots and so on. Now for all numbers nn such that pin<pi+1p_i\leq n < p_{i+1}, we have P(i)=piP(i)=p_i and N(i)=pi+1N(i)=p_{i+1}. How many numbers fall in this range ? Precisely pi+1pip_{i+1}-p_i of them. Since n+1n+1 is a prime, we have pk+1=n+1p_{k+1}=n+1 for some integer kk. Thus, 1P(2)N(2)+1P(3)N(3)++1P(n)N(n)=i=1kpi+1pipipi+1=i=1k(1pi1pi+1)=1p11pk+1\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 p1=2,pk+1=n+1,p_1=2, p_{k+1}=n+1,\hspace{10pt} \blacksquare.

Abhishek Sinha - 4 years, 10 months ago

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

Jon Haussmann - 4 years, 10 months ago

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AnswertoQuestionnumber2,Consider,yz(y2+z2)+xz(x2+z2)+xy(x2+y2)2(x2y2+y2z2+z2x2)Now,2(x2y2+y2z2+z2x2)=(x2y2+y2z2)+(z2x2+x2y2)+(y2z2+z2x2)2(x2yz+xy2z+xyz2)=2xyz(x+y+z)yz(y2+z2)+xz(x2+z2)+xy(x2+y2)2xyz(x+y+z)y2+z2x+x2+z2y+x2+y2z2(x+y+z)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 - 4 years, 10 months ago

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@Surya Prakash For the sixth one, a 2X2 2X2 square with exactly one 1 -1 seems to contradict the question statement. Is the question written correctly, or am I misreading something?

Siddhartha Srivastava - 4 years, 10 months ago

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

Surya Prakash - 4 years, 10 months ago

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The second question can also be rewritten in the form (x2+y2+z2)(1x+1y+1z)3(x+y+z)(x^2+y^2+z^2)(\frac1x+\frac1y+\frac1z)\ge 3(x+y+z)

This can be proved by using (x2+y2+z2)(x+y+z)23(x^2+y^2+z^2)\ge \dfrac{(x+y+z)^2}{3} and AMHMAM-HM .

Rahul Saha - 4 years, 10 months ago

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

Given,1P(2)N(2)+1P(3)N(3)+1P(4)N(4)+.....+1P(n)N(n)=n12n+2wheren+1isaprime,n>1Nowabovesumistrueforn=2.Letk+1beaprimeforwhichabovesumistrue.i=2k1P(i)N(i)=k12k+2Lettheprimenexttok+1bek+r+1.P(k+1)=P(k+2)=P(k+3)=...........................=P(k+r)=k+1.Largestprimelessthanorequaltok+iisk+1,i=1,2,3....,r.lyN(k+1)=N(k+2)=N(k+3)=....................=N(k+r)=k+r+1Nowwehavetoprovethatthesumistrueforn=k+r.i=2k+r1P(i)N(i)=i=2k1P(i)N(i)+i=k+1k+r1P(i)N(i)=k12k+2+r(k+1)(k+r+1)=(k+r)12(k+r)+2(Onsimplification)ThusbyprincipleofmathematicalinductionabovesumistruenϵN,n>1,n+1isaprime.Hence,Proved.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 - 4 years, 10 months ago

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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

Shrihari B - 4 years, 9 months ago

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For #1, does anyone have a detailed solution?

Shashank Rammoorthy - 4 years, 1 month ago

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

Ryan Tamburrino - 4 years, 10 months ago

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

1) Initially, rx=cy=1 r_x =c_y = 1 . Then after the change rx=cy=1 r_x = c_y = - 1 Therefore SInitial=SFinal+4 S_{Initial} = S_{Final} + 4 .

2)Initially, rx=1,cy=1 r_x =-1, c_y = 1 . Then after the change rx=1,cy=1 r_x =1 ,c_y = - 1 Therefore SInitial=SFinal S_{Initial} = S_{Final} .

3)Initially, rx=1,cy=1 r_x = 1, c_y = -1 . Then after the change rx=1,cy=1 r_x = - 1 ,c_y = 1 Therefore SInitial=SFinal S_{Initial} = S_{Final} .

4)Initially, rx=cy=1 r_x =c_y = - 1 . Then after the change rx=cy=1 r_x = c_y = 1 Therefore SInitial=SFinal4 S_{Initial} = S_{Final} - 4 .

Therefore, we see that SinitialSfinal(mod4) S_{initial} \equiv S_{final} \pmod4 is invariant. ---- (A)

Now we prove by contradiction. Suppose there exists a configuration with S=00(mod4) S = 0 \equiv 0 \pmod4 . After changing all the 1 -1 s to 1 1 , By (A), we see that Sfinal0(mod4) S_{final} \equiv 0 \pmod4 . But S=2n2(mod4) S = 2n \equiv 2 \pmod4 since n n is odd. Thus there a contradiction and our supposition is false. Therefore there exist no configuration with S=0 S = 0 .

@Ryan Tamburrino

Siddhartha Srivastava - 4 years, 10 months ago

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Thank you!

Ryan Tamburrino - 4 years, 10 months ago

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

Surya Prakash - 4 years, 10 months ago

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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{x}y{y}z{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: ( Ai\ A_i ) and ( Bi\ B_i ) are both decreasing sequences,  a1+a2+...+an=b1+b2+...+bn and a1+a2+...+ai b1+b2+...+bi\ 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{i} < n)

Curtis Clement - 4 years, 9 months ago

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Thnx bro. For posting the paper.

SOURAV MISHRA - 4 years, 9 months ago

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

Surya Prakash - 4 years, 8 months ago

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

Swapnil Das - 3 years, 10 months ago

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

Surya Prakash - 3 years, 10 months ago

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Thanks a lot SURYA PRAKASH For Posting This Paper.

ashay wakode - 4 years, 10 months ago

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

Surya Prakash - 4 years, 10 months ago

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Maharashtra region

ashay wakode - 4 years, 10 months ago

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@Ashay Wakode Are you selected for INMO 2015

Surya Prakash - 4 years, 9 months ago

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@Surya Prakash No.

ashay wakode - 4 years, 9 months ago

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@Ashay Wakode ok

Surya Prakash - 4 years, 9 months ago

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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. :)

Christian Daang - 4 years, 9 months ago

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@Christian Daang

This proof is incorrect.

Mehul Arora - 3 years, 4 months ago

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