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RMO 2016 Delhi Region

Here is the paper of RMO 2016 Delhi.Pls post answers and solutions to all questions. Also tell ur marks and estimated cut off Thanks!

Note by Kaustubh Miglani
1 year ago

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Problem 4

\(f\left( a,b,c \right) =\quad \sum { \frac { a }{ { a }^{ 3 }+{ b }^{ 2 }+{ c } } } =\quad \sum { \frac { 1 }{ { a }^{ 2 }+\frac { { b }^{ 2 } }{ a } +\frac { c }{ a } } } \\ \\ Now,\quad \frac { { a }^{ 2 }+{ \frac { { b }^{ 2 } }{ a } + }{ \frac { c }{ a } } }{ 3 } \ge \sqrt [ 3 ]{ { b }^{ 2 }{ c } } \quad \Rightarrow \quad \frac { 1 }{ { a }^{ 2 }+\frac { { b }^{ 2 } }{ a } +\frac { c }{ a } } \le \quad \frac { 1 }{ 3\sqrt [ 3 ]{ { b }^{ 2 }c } } \left\{ A.M-G.M \right\} \\ \sum { \frac { 1 }{ { a }^{ 2 }+\frac { { b }^{ 2 } }{ a } +\frac { c }{ a } } } \le \quad \sum { \frac { 1 }{ 3\sqrt [ 3 ]{ { b }^{ 2 }c } } } \le \sqrt [ 3 ]{ abc } \le \frac { a+b+c }{ 3 } =1\quad \\ \\ \Rightarrow f\left( a,b,c \right) \le 1,\quad with\quad equality\quad at\quad a=b=c=1\)

Aditya Dhawan - 1 year ago

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Will my answer be correct?

Kaustubh Miglani - 1 year ago

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I USED Titus lemma. Got same answer

Kaustubh Miglani - 1 year ago

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THEN U GOTTA PROVE IT COZ IT'S NOT IN STANDARD TEXT BOOKS

Yuvraj Singh - 9 months, 2 weeks ago

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@Yuvraj Singh It is in standard books.Atleast cauchy is

Kaustubh Miglani - 9 months, 2 weeks ago

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@Kaustubh Miglani First it's not titu's lemma ,but T2's lemma Second it's direct application of cauchy..lol

Yuvraj Singh - 4 months ago

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@Yuvraj Singh It is Titu's lemma, not T2's Lemma

Rishabh Dhiman - 1 month, 3 weeks ago

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Problem 1

Problem 1

We can see that as we move the point \(P\) on the circumference of the circle\([\)excluding \(X\) and \(Y],\)the \(\angle XPY=\angle XP_1Y\) remains constant.So this shows that \(AB=A_1B_1.\)Now we use extended sin rule to complete the problem.
Let the circum-radius of \(\triangle PAB\) be \(R\) and \(\triangle P_1A_1B_1\) be \(R_1.\)
In \(\triangle PAB, \frac{AB}{sin\angle P}=2R\) and in \(\triangle P_1A_1B_1,\frac{A_1B_1}{sin\angle P_1}=\frac{AB}{sin\angle P}=2R_1.\) Therefore \(2R=2R_1\Rightarrow\boxed {R=R_1}.\)Hence Proved.

Ayush Rai - 1 year ago

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nice solution

Abhishek Alva - 1 year ago

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Hey can u explain in detail why A1B1=AB

Kaustubh Miglani - 9 months, 3 weeks ago

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Because angle subtended by both of these chords at centre are equal. angle A1 X A =P X P1=P Y P1=B Y B1

Yuvraj Singh - 9 months, 2 weeks ago

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@Yuvraj Singh Which class are u in and which school?

Kaustubh Miglani - 9 months, 2 weeks ago

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@Kaustubh Miglani I m in class 10 And from bbps dw. But ur profile says u live in noida

Kaustubh Miglani - 9 months, 2 weeks ago

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@Kaustubh Miglani I am graduating from IIT Bombay. What is ur favourite college bro?

Yuvraj Singh - 9 months, 2 weeks ago

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@Yuvraj Singh IIT Is the one Bombay ,Perhaps

Kaustubh Miglani - 9 months, 2 weeks ago

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@Kaustubh Miglani Prefered Jee rank?mine was 17

Yuvraj Singh - 9 months, 2 weeks ago

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@Yuvraj Singh Do u remember books you used for iit preparation.I am an iit aspirant

Kaustubh Miglani - 4 months, 1 week ago

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@Kaustubh Miglani hcv,Bmsharma,resnick,irodov,krotov,kolenkow -Phy atkins,rc mukerje,bahadur,jd lee-Chm sl loney,hallknight,mostly tatamcgraw buks and tons of practise question papers here at allen.u can download them online also.Ncert is must .Hope it helps.

Yuvraj Singh - 4 months ago

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@Yuvraj Singh Hi Where Can I find some good material for RMO this year? Last time I failed miserably :'( could solve just one... Thanks :)

Yash Mehan - 4 months, 1 week ago

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@Yash Mehan first solve all previous year question papers, then use brilliant if ur aspiration is limited to rmo not inmo or selection camp P.S. i did 4 at my time

Yuvraj Singh - 4 months ago

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@Yash Mehan would u like to join my RMO/INMO group. if yes then give ur email.

Ayush Rai - 4 months, 1 week ago

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@Ayush Rai My email is adityak1135@gmail.com I know that this is a really late reply but I didn't know about it earlier ;)

Aditya Khurmi - 2 months ago

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@Ayush Rai My email yashmehan@gmail.com Sorry for the late reply I do not often open up Brilliant Thanks!!!

Yash Mehan - 2 months ago

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@Ayush Rai is dis question addressed to me?

Yuvraj Singh - 4 months ago

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@Ayush Rai Yeah he is good.Invite him

Kaustubh Miglani - 4 months, 1 week ago

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@Kaustubh Miglani what about u?

Ayush Rai - 4 months, 1 week ago

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@Ayush Rai what about me?

Kaustubh Miglani - 4 months, 1 week ago

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@Kaustubh Miglani wont u join our group?

Ayush Rai - 4 months, 1 week ago

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@Ayush Rai he is already in there @ayush rai

Abhishek Alva - 4 months ago

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Problem 5 1)its the easiest- use the identity \({ (x+y+z) }^{ 2 }+{ (x+y-z) }^{ 2 }+{ (x-y+z) }^{ 2 }+{ (-x+y+z) }^{ 2 }={ (2x) }^{ 2 }+{ (2y) }^{ 2 }+{ (2z) }^{ 2 }\) which gives infinite solutions in integers x,y,z.u can check it satisfis all other cond.s 2)solve this one other way round, assume 2016 = \({ (2x) }^{ 2 }+{ (2y) }^{ 2 }+{ (2z) }^{ 2 }\) then it implies 504 = \({ (x) }^{ 2 }+{ (y) }^{ 2 }+{ (z) }^{ 2 }\) by inspection x=22 y=4 z=6, then use the identity Problem 2 -we obtain nth term in terms of n,a,b then use strong induction.

Yuvraj Singh - 4 months ago

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the question says pairwise distinct positive integers with no common factor, doesn't that mean they're pairwise coprime?

Govind Ramesh - 4 months ago

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The question mentions that they all don't have a factor in common, and that doesn't mean they are pairwise relatively prime. If x,y, and z are relatively prime, then we are done because then the gcd of these 7 terms would be 1 :)

Aditya Khurmi - 2 months ago

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@Aditya Khurmi oh, thank you i didn't understand the question properly then

Govind Ramesh - 1 month, 3 weeks ago

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Problem 4

By the Cauchy-Schwarz inequality, \(\left(\dfrac{1}{a}+1+c\right)(a^{3}+b^{2}+c) \geq (a+b+c)^{2}=9\)

\(\implies (ac+a+1)(a^{3}+b^{2}+c) \geq 9a\)

Thus, we get

\(f(a,b,c)=\quad \sum {\dfrac{(ac+a+1)a}{(ac+a+1)({a}^{3}+{b}^{2}+{c})}} \leq \quad \sum \dfrac{a(ac+a+1)}{9a}\)

\(\implies f(a,b,c) \leq \dfrac{1}{9}[(ac+a+1)+(ba+b+1)+(cb+c+1)]\)

\(=\dfrac{1}{9}[(a+b+c)+(1+1+1)+(ab+bc+ca)] \leq \dfrac{1}{9}[3+3+3]=\boxed{1}\)

Here we used the identity \((a+b+c)^{2} \geq 3(ab+bc+ca)\) to get \(ab+bc+ca \leq 3\)

Aditya Khurmi - 2 months ago

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Need a solution for q 2 and 6

Azimuddin Sheikh - 3 months, 2 weeks ago

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Can anyone post complete solution to 2nd and last problem?

Azimuddin Sheikh - 3 months, 2 weeks ago

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Diagram to problem \(3\)

Rohit Camfar - 6 months, 3 weeks ago

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Instead,use A-Humpty point lemma, which proves it without anything.

Yuvraj Singh - 4 months ago

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Yup, and then repeat the same procedure to prove the lemma in examination.

Rohit Camfar - 4 months ago

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@Rohit Camfar So in exam, to do any inequality, we can use only upto am-gm? not even cauchy? or can we not use even weighted am-gm?

Govind Ramesh - 4 months ago

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@Govind Ramesh You can use AM-GM and weighted AM-GM because at times some problems are not easy to prove without weighted AM-GM{RMO 2012 paper1 p3}. Use Cauchy-Schwarz too because it is of the basic inequalities. If you use a lemma like the Titu's lemma, then better prove it for RMO, I guess no need to prove that in INMO, not IMO definitely!

Aditya Khurmi - 1 month, 3 weeks ago

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@Govind Ramesh Hm... All those words go above me. Humpty-Dumpty theorem (Both different) are well-known. Although not that standard to be used directly in exams. I have seen guys proving each and every lemma, all for the sake of marks only/.

Rohit Camfar - 4 months ago

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@Rohit Camfar so by standard textbooks, do they mean ncert?

Govind Ramesh - 4 months ago

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@Govind Ramesh What? No.......

Rohit Camfar - 4 months ago

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Solution to \(3\)

Const:-- Join \(BP\) , \(BQ\) , \(CP\) & \(CQ.\)

Clearly, Using Alternate Segment Theorem , we get:\( \color{red}{\angle BAQ = \angle BQP}\) and \( \color{Blue}{\angle BAP = \angle BPQ}\).--- \([1]\)

Also Using the isosceles triangle property, we get :: \( \color{red}{\angle PQC = \angle BQP}\) & \( \color{blue}{\angle QPC = \angle BPQ}\)-----\([2]\)

Thus from \(eq^{n}\) \([1]\) and \([2]\) we get: \( \color{red}{\angle BAQ = \angle PQC}\) & \( \color{blue}{\angle BAP = \angle QPC}\).

Adding these two results we get: \(\angle BAQ + \angle BAP\) = \(\angle PQC + \angle QPC\)

Thus \(\angle PCQ\) = \(180^{\circ}-[\angle PQC + \angle QPC]\) = \(180^{\circ}-[\angle BAQ + \angle BAP]\)

Therefore, \(\angle PCQ + PAQ\) = \(180^{\circ}-[\angle BAQ + \angle BAP] + [\angle BAP + \angle BAQ]\) = \(180^{\circ}\)

=>\(\quad APCQ\) is cyclic

=> \(\angle PAC= \color{red}{\angle PQC = \angle BAQ}\)

=> \(\angle PAC = \angle BAQ\)

Rohit Camfar - 6 months, 2 weeks ago

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Diagram to problem 1.........

Rohit Camfar - 6 months, 3 weeks ago

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Pls do provide the solution of the last problem

Pranav Pundeer - 2 months ago

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Can anyone give complete solution of 2 and 6

Azimuddin Sheikh - 3 months, 2 weeks ago

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@Azimuddin Sheikh At least two cards have same suit let P1, P2 has same suit then the assistant will turn down one of P1, P2 (say P1) and arrange P2, P3, P4, P5 in such a way that P2 takes first position so the magician know that the card turned down and the card having first position have same suit. This will help him to identify suit of P1 now there are 6 ways to arrange P3 P4 P5 as LMH, LHM, MLH, MHL, HLM, HML (low number, middle number, high number) If difference of numbers obtained on P1 and P2 is 6 then turn down the higher one and give first position to lower one. Let us assume P1 has number k and P2 has no k + 1, k + 2, k + 3, k + 4, k + 5 or k + 6. If P2 has k + 1 then show LMH. If P2 has k + 2 then show LHM and soon. If difference is > 6 then turn down lower one and give first position to higher one and add 1,2,3,4,5,6 for arrangements LMH, LHM, MLH, MHL, HLM, HML in the higher number for example adding 4 to 11 indicates 2.

Kaustubh Miglani - 3 months, 2 weeks ago

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@Kaustubh Miglani The last combinatorics question is more or less the same as the first question in the chapter combinatorics from the book 'The art and craft of problem-solving'

Aditya Khurmi - 2 months, 1 week ago

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@Azimuddin Sheikh Idk latex so cant post soln to 2

Kaustubh Miglani - 3 months, 2 weeks ago

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Solution to Problem 1. Let \(P_{1}\) and \(P_{2}\) be the two points on \(\odot\omega_{1}\) as \(P\) varies across the circumference of \(\odot \omega_{1}.\) Similarly, let \(P_{1}X\) and \(P_{1}Y\) meet \(\odot\omega_{2}\) at \(A_{1}\) , \(B_{1}\) and \(P_{2}X\) and \(P_{2}Y\) meet \(\odot\omega2\) at \(A_{2}\) , \(B_{2}\) respectively.

Clearly , \(\angle XP_{1}Y = \angle XP_{2}Y\) and \(\angle XB_{1}Y = \angle XB_{2}Y\)

Adding these two \(eq^{n}\) we get:: \(\angle A_{1}XB_{1} = \angle A_{2}XB_{2}\) => \(A_{1}B_{1} = A_{2}B_{2}.\)

Also, \(\Delta XP_{1}Y \sim \Delta A_{1}P_{1}B_{1}\) => \(\dfrac{Radius XP_{1}Y}{Radius A_{1}P_{1}B_{1}}\) = \(\dfrac{XY}{A_{1}B_{1}}\) ........[ \(1\) ]

Similarly, \(\Delta XP_{2}Y \sim \Delta A_{2}P_{2}B_{2}\)

=> \(\dfrac{Radius XP_{2}Y}{Radius A_{2}P_{2}B_{2}}\) = \(\dfrac{XY}{A_{2}B_{2}}\)

=> \(\dfrac{Radius XP_{1}Y}{Radius A_{2}P_{2}B_{2}}\) = \(\dfrac{XY}{A_{1}B_{1}}\) ........[ \(2\) ]

[Because \(\Delta XP_{1}Y\) and \(\triangle XP_{2}Y\) have the same circumcircle. & \(A_{1}B_{1} = A_{2}B_{2}\)]

From \(eq^{n}\) \([1]\) and \([2]\) we get that :: Radius \(A_{1}P_{1}B_{1}\) = Radius \(A_{2}P_{2}B_{2}.\)

\(K.I.P.K.I.G\)

Rohit Camfar - 6 months, 3 weeks ago

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