# INMO 2017 Board

$(1)$ Let ${ x }_{ 1 },...,{ x }_{ 2017 }$ be positive reals such that

$\huge\ \frac { 1 }{ { x }_{ 1 }+2017 } +\frac { 1 }{ { x }_{ 2 }+2017 } +...+\frac { 1 }{ { x }_{ 2017 }+2017 } =\frac { 1 }{ 2017 }$

Prove that

$\huge\ \frac { \sqrt [ 2017 ]{ { x }_{ 1 }{ x }_{ 2 }...{ x }_{ 2017 } } }{ 2016 } \ge 2017$

Two circles enclose non-intersecting areas. Common tangent lines to the two circles, one external and one internal, are drawn. Consider two straight lines each of which passes through the tangent points on one of the circles. Prove that the intersection point of the lines lies on the straight line that connects the centers of the circles.

Hello everybody. Please post solutions of these problems and post problems on your own also.

These are sample problems.

Note by Priyanshu Mishra
3 years, 7 months ago

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Come and enjoy solving problems here.

- 3 years, 7 months ago

For problem 1 apply titus lemma in the constraint given and then apply AM GM and it's done.

- 3 years, 7 months ago

Is RMO DELHI results out?

At which website?

- 3 years, 7 months ago

They send you your marks by mail. Only marks though, the cutoff hasn't been decided yet.

- 3 years, 7 months ago

At which website is the name of list of selected students declared?

- 3 years, 7 months ago

It's not declared yet but I think it will be on hbcse

- 3 years, 7 months ago

Any more problems?

- 3 years, 7 months ago

i HAve added newproblems. TRy them.

- 3 years, 6 months ago

Try the new problems

- 3 years, 6 months ago

Where are the new problems.

- 3 years, 5 months ago

- 3 years, 4 months ago

Problem 1 was just an AM-GM.

Let $y_i = \dfrac {1}{x_i + 2017}$, so $x_i = \dfrac {1 - 2017y_i}{y_i}$

We have

$\displaystyle \sum_{j=1}^{2017} y_j = \dfrac{1}{2017}$

Thus,

\begin{aligned} \dfrac{1}{2017} - y_i &= \displaystyle \sum_{\stackrel{j=1}{j \neq i}}^{2017} y_j\\ 1 - 2017 y_i &= 2017 \displaystyle \sum_{\stackrel{j=1}{j \neq i}}^{2017} y_j\\ \end{aligned}

However, we also have

$\displaystyle \sum_{\stackrel{j=1}{j \neq i}}^{2017} y_j \geq 2016 \left ( \displaystyle \prod_{\stackrel{j=1}{j \neq i}}^{2017} y_j \right ) ^{\dfrac {1}{2016}}$

Therefore,

\begin{aligned} \displaystyle \prod_{i=1}^{2017} x_i &= \displaystyle \prod_{i=1}^{2017} \dfrac {1 - 2017y_i}{y_i}\\ & = \dfrac {\displaystyle 2017^{2017} \prod_{i=1}^{2017} \left (\displaystyle \sum_{\stackrel{j=1}{j \neq i}}^{2017} y_j \right )}{\displaystyle \prod_{i=1}^{2017} y_i}\\ &\geq \dfrac {(2016 \times 2017)^{2017} \displaystyle \prod_{i=1}^{2017} \left ( \displaystyle \prod_{\stackrel{j=1}{j \neq i}}^{2017} y_j \right )^{\dfrac{1}{2016}}}{\displaystyle \prod_{i=1}^{2017} y_i}\\ &= 2016^{2017} \times 2017^{2017}\\ \end{aligned}

This just rearranges to give us the desired expression.

- 3 years, 4 months ago