# Difference between revisions of "2021 IMO Problems/Problem 4"

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==Video Solution== | ==Video Solution== | ||

+ | Alternate solution: | ||

https://www.youtube.com/watch?v=U95v_xD5fJk | https://www.youtube.com/watch?v=U95v_xD5fJk | ||

+ | Almost same solution: | ||

+ | https://youtu.be/WkdlmduOnRE |

## Revision as of 15:09, 28 July 2021

## Problem

Let be a circle with centre , and a convex quadrilateral such that each of the segments and is tangent to . Let be the circumcircle of the triangle . The extension of beyond meets at , and the extension of beyond meets at . The extensions of and beyond meet at and , respectively. Prove that

## Solution

Let be the centre of .

For the result follows simply. By Pitot's Theorem we have so that, The configuration becomes symmetric about and the result follows immediately.

Now assume WLOG . Then lies between and in the minor arc and lies between and in the minor arc . Consider the cyclic quadrilateral . We have and . So that, Since is the incenter of quadrilateral , is the angular bisector of . This gives us, Hence the chords and are equal. So is the reflection of about . Hence, and now it suffices to prove Let and be the tangency points of with and respectively. Then by tangents we have, . So . Similarly we get, . So it suffices to prove, Consider the tangent to with . Since and are reflections about and is a circle centred at the tangents and are reflections of each other. Hence By a similar argument on the reflection of and we get and finally, as required.

~BUMSTAKA

## Video Solution

Alternate solution: https://www.youtube.com/watch?v=U95v_xD5fJk Almost same solution: https://youtu.be/WkdlmduOnRE