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 · 7 months, 2 weeks ago

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– Kaustubh Miglani · 7 months, 2 weeks ago

Will my answer be correct?Log in to reply

– Kaustubh Miglani · 7 months, 2 weeks ago

I USED Titus lemma. Got same answerLog in to reply

– Yuvraj Singh · 4 months, 2 weeks ago

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

– Kaustubh Miglani · 4 months, 2 weeks ago

It is in standard books.Atleast cauchy isLog 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 · 7 months, 2 weeks ago

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– Abhishek Alva · 7 months, 2 weeks ago

nice solutionLog in to reply

– Kaustubh Miglani · 4 months, 3 weeks ago

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

– Yuvraj Singh · 4 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 · 4 months, 2 weeks ago

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

– Kaustubh Miglani · 4 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 · 4 months, 2 weeks ago

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

– Kaustubh Miglani · 4 months, 2 weeks ago

IIT Is the one Bombay ,PerhapsLog in to reply

– Yuvraj Singh · 4 months, 2 weeks ago

Prefered Jee rank?mine was 17Log in to reply

Diagram to problem \(3\)

– Rohit Camfar · 1 month, 3 weeks agoLog 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 · 1 month, 3 weeks ago

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

– Rohit Camfar · 1 month, 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 · 1 month, 3 weeks ago

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