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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

$$f\left( a,b,c \right) =\quad \sum { \frac { a }{ { a }^{ 3 }+{ b }^{ 2 }+{ c } } } =\quad \sum { \frac { 1 }{ { a }^{ 2 }+\frac { { b }^{ 2 } }{ a } +\frac { c }{ a } } } \\ \\ Now,\quad \frac { { a }^{ 2 }+{ \frac { { b }^{ 2 } }{ a } + }{ \frac { c }{ a } } }{ 3 } \ge \sqrt [ 3 ]{ { b }^{ 2 }{ c } } \quad \Rightarrow \quad \frac { 1 }{ { a }^{ 2 }+\frac { { b }^{ 2 } }{ a } +\frac { c }{ a } } \le \quad \frac { 1 }{ 3\sqrt [ 3 ]{ { b }^{ 2 }c } } \left\{ A.M-G.M \right\} \\ \sum { \frac { 1 }{ { a }^{ 2 }+\frac { { b }^{ 2 } }{ a } +\frac { c }{ a } } } \le \quad \sum { \frac { 1 }{ 3\sqrt [ 3 ]{ { b }^{ 2 }c } } } \le \sqrt [ 3 ]{ abc } \le \frac { a+b+c }{ 3 } =1\quad \\ \\ \Rightarrow f\left( a,b,c \right) \le 1,\quad with\quad equality\quad at\quad a=b=c=1$$ · 7 months, 2 weeks ago

Will my answer be correct? · 7 months, 2 weeks ago

I USED Titus lemma. Got same answer · 7 months, 2 weeks ago

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

It is in standard books.Atleast cauchy is · 4 months, 2 weeks ago

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. · 7 months, 2 weeks ago

nice solution · 7 months, 2 weeks ago

Hey can u explain in detail why A1B1=AB · 4 months, 3 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 B1 · 4 months, 2 weeks ago

Which class are u in and which school? · 4 months, 2 weeks ago

I m in class 10 And from bbps dw. But ur profile says u live in noida · 4 months, 2 weeks ago

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

IIT Is the one Bombay ,Perhaps · 4 months, 2 weeks ago

Prefered Jee rank?mine was 17 · 4 months, 2 weeks ago

Diagram to problem $$3$$

· 1 month, 3 weeks ago

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

=> $$\angle PAC= \color{red}{\angle PQC = \angle BAQ}$$

=> $$\angle PAC = \angle BAQ$$ · 1 month, 3 weeks ago

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

· 1 month, 3 weeks ago

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 $$A_{2}P_{2}B_{2}.$$

$$K.I.P.K.I.G$$ · 1 month, 3 weeks ago