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!

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TopNewestProblem 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 · 9 months, 2 weeks ago

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– Kaustubh Miglani · 9 months, 2 weeks ago

Will my answer be correct?Log in to reply

– Kaustubh Miglani · 9 months, 2 weeks ago

I USED Titus lemma. Got same answerLog in to reply

– Yuvraj Singh · 6 months, 2 weeks ago

THEN U GOTTA PROVE IT COZ IT'S NOT IN STANDARD TEXT BOOKSLog in to reply

– Kaustubh Miglani · 6 months, 2 weeks ago

It is in standard books.Atleast cauchy isLog in to reply

– Yuvraj Singh · 1 month ago

First it's not titu's lemma ,but T2's lemma Second it's direct application of cauchy..lolLog in to reply

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 · 9 months, 2 weeks ago

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– Abhishek Alva · 9 months, 2 weeks ago

nice solutionLog in to reply

– Kaustubh Miglani · 6 months, 3 weeks ago

Hey can u explain in detail why A1B1=ABLog in to reply

– Yuvraj Singh · 6 months, 2 weeks ago

Because angle subtended by both of these chords at centre are equal. angle A1 X A =P X P1=P Y P1=B Y B1Log in to reply

– Kaustubh Miglani · 6 months, 2 weeks ago

Which class are u in and which school?Log in to reply

– Kaustubh Miglani · 6 months, 2 weeks ago

I m in class 10 And from bbps dw. But ur profile says u live in noidaLog in to reply

– Yuvraj Singh · 6 months, 2 weeks ago

I am graduating from IIT Bombay. What is ur favourite college bro?Log in to reply

– Kaustubh Miglani · 6 months, 2 weeks ago

IIT Is the one Bombay ,PerhapsLog in to reply

– Yuvraj Singh · 6 months, 2 weeks ago

Prefered Jee rank?mine was 17Log in to reply

– Kaustubh Miglani · 1 month, 1 week ago

Do u remember books you used for iit preparation.I am an iit aspirantLog in to reply

– Yuvraj Singh · 1 month ago

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.Log in to reply

– Yash Mehan · 1 month, 1 week ago

Hi Where Can I find some good material for RMO this year? Last time I failed miserably :'( could solve just one... Thanks :)Log in to reply

– Yuvraj Singh · 1 month ago

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 timeLog in to reply

– Ayush Rai · 1 month, 1 week ago

would u like to join my RMO/INMO group. if yes then give ur email.Log in to reply

– Yuvraj Singh · 1 month ago

is dis question addressed to me?Log in to reply

– Kaustubh Miglani · 1 month, 1 week ago

Yeah he is good.Invite himLog in to reply

– Ayush Rai · 1 month, 1 week ago

what about u?Log in to reply

– Kaustubh Miglani · 1 month, 1 week ago

what about me?Log in to reply

– Ayush Rai · 1 month, 1 week ago

wont u join our group?Log in to reply

– Abhishek Alva · 1 month ago

he is already in there @ayush raiLog in to reply

Need a solution for q 2 and 6 – Azimuddin Sheikh · 2 weeks, 6 days ago

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Can anyone post complete solution to 2nd and last problem? – Azimuddin Sheikh · 2 weeks, 6 days 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 · 1 month ago

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– Govind Ramesh · 1 month ago

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

Diagram to problem \(3\)

– Rohit Camfar · 3 months, 3 weeks agoLog in to reply

– Yuvraj Singh · 1 month ago

Instead,use A-Humpty point lemma, which proves it without anything.Log in to reply

– Rohit Camfar · 1 month ago

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

– Govind Ramesh · 1 month ago

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?Log in to reply

– Rohit Camfar · 1 month ago

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/.Log in to reply

– Govind Ramesh · 1 month ago

so by standard textbooks, do they mean ncert?Log in to reply

– Rohit Camfar · 1 month ago

What? No.......Log in to reply

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 · 3 months, 3 weeks ago

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

– Rohit Camfar · 3 months, 3 weeks agoLog in to reply

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 · 3 months, 3 weeks ago

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– Azimuddin Sheikh · 2 weeks, 6 days ago

Can anyone give complete solution of 2 and 6Log in to reply

@Azimuddin Sheikh Idk latex so cant post soln to 2 – Kaustubh Miglani · 2 weeks, 6 days 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 · 2 weeks, 6 days ago

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