# 2021 IMO Problems/Problem 4

## Contents

## 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

## Video Solutions

https://youtu.be/vUftJHRaNx8 [Video contains solutions to all day 2 problems]

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

## 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

## Solution2

Denote tangents to the circle at , tangents to the same circle at ; tangents at and tangents at . We can get that .Since Same reason, we can get that We can find that . Connect separately, we can create two pairs of congruent triangles. In $\trangle{XIF},\triangle{YIM}$ (Error compiling LaTeX. ! Undefined control sequence.), since After getting that , we can find that . Getting that , same reason, we can get that . Now the only thing left is that we have to prove . Since we can subtract and get that ,means and we are done ~bluesoul