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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 week, 4 days ago

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Problem 4

$$f\left( a,b,c \right) =\quad \sum { \frac { a }{ { a }^{ 3 }+{ b }^{ 2 }+{ c } } } =\quad \sum { \frac { 1 }{ { a }^{ 2 }+\frac { { b }^{ 2 } }{ a } +\frac { c }{ a } } } \\ \\ Now,\quad \frac { { a }^{ 2 }+{ \frac { { b }^{ 2 } }{ a } + }{ \frac { c }{ a } } }{ 3 } \ge \sqrt [ 3 ]{ { b }^{ 2 }{ c } } \quad \Rightarrow \quad \frac { 1 }{ { a }^{ 2 }+\frac { { b }^{ 2 } }{ a } +\frac { c }{ a } } \le \quad \frac { 1 }{ 3\sqrt [ 3 ]{ { b }^{ 2 }c } } \left\{ A.M-G.M \right\} \\ \sum { \frac { 1 }{ { a }^{ 2 }+\frac { { b }^{ 2 } }{ a } +\frac { c }{ a } } } \le \quad \sum { \frac { 1 }{ 3\sqrt [ 3 ]{ { b }^{ 2 }c } } } \le \sqrt [ 3 ]{ abc } \le \frac { a+b+c }{ 3 } =1\quad \\ \\ \Rightarrow f\left( a,b,c \right) \le 1,\quad with\quad equality\quad at\quad a=b=c=1$$ · 1 week, 4 days ago

Will my answer be correct? · 1 week, 3 days ago

I USED Titus lemma. Got same answer · 1 week, 3 days 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. · 1 week, 4 days ago