$(1)$ 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.

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## Comments

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TopNewest@Harsh Shrivastava, @Sharky Kesa, @Svatejas Shivakumar, @rajdeep das, @Racchit Jain,@Vaibhav Prasad @Kalash Verma @Nihar Mahajan @Adarsh Kumar @Akshat Sharda @AkshayYadav @Swapnil Das @Rajdeep Dhingra @Anik Mandal @Lakshya Sinha @Abhay Kumar @Dev Sharma and everyone.

Come and enjoy solving problems here.

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

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@Racchit Jain

Is RMO DELHI results out?

At which website?

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

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Any more problems?

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@Sharky Kesa, @Harsh Shrivastava,

i HAve added newproblems. TRy them.

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@Sharky Kesa,

Try the new problems

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Where are the new problems.

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INMO 2017 Solution- www.zeal.academy http://www.zeal.academy/INMO%202017%20solutions(HM).pdf

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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.

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check this one

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