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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
7 months, 2 weeks 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 · 7 months, 2 weeks ago

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@Aditya Dhawan Will my answer be correct? Kaustubh Miglani · 7 months, 2 weeks ago

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@Aditya Dhawan I USED Titus lemma. Got same answer Kaustubh Miglani · 7 months, 2 weeks ago

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@Kaustubh Miglani THEN U GOTTA PROVE IT COZ IT'S NOT IN STANDARD TEXT BOOKS Yuvraj Singh · 4 months, 2 weeks ago

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@Yuvraj Singh It is in standard books.Atleast cauchy is Kaustubh Miglani · 4 months, 2 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 · 7 months, 2 weeks ago

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@Ayush Rai nice solution Abhishek Alva · 7 months, 2 weeks ago

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@Ayush Rai Hey can u explain in detail why A1B1=AB Kaustubh Miglani · 4 months, 3 weeks ago

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@Kaustubh Miglani 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 · 4 months, 2 weeks ago

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@Yuvraj Singh Which class are u in and which school? Kaustubh Miglani · 4 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 · 4 months, 2 weeks ago

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@Kaustubh Miglani I am graduating from IIT Bombay. What is ur favourite college bro? Yuvraj Singh · 4 months, 2 weeks ago

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@Yuvraj Singh IIT Is the one Bombay ,Perhaps Kaustubh Miglani · 4 months, 2 weeks ago

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@Kaustubh Miglani Prefered Jee rank?mine was 17 Yuvraj Singh · 4 months, 2 weeks ago

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Diagram to problem \(3\)

Rohit Camfar · 1 month, 3 weeks ago

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@Rohit Camfar 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 · 1 month, 3 weeks ago

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

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

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