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

## Problem

Let $\Gamma$ be a circle with centre $I$, and $ABCD$ a convex quadrilateral such that each of the segments $AB, BC, CD$ and $DA$ is tangent to $\Gamma$. Let $\Omega$ be the circumcircle of the triangle $AIC$. The extension of $BA$ beyond $A$ meets $\Omega$ at $X$, and the extension of $BC$ beyond $C$ meets $\Omega$ at $Z$. The extensions of $AD$ and $CD$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively. Prove that $\[AD + DT + T X + XA = CD + DY + Y Z + ZC\]$

## Video Solutions

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

## Solution

Let $O$ be the centre of $\Omega$.

For $AB=BC$ the result follows simply. By Pitot's Theorem we have $\[AB + CD = BC + AD\]$ so that, $AD = CD.$ The configuration becomes symmetric about $OI$ and the result follows immediately.

Now assume WLOG $AB < BC$. Then $T$ lies between $A$ and $X$ in the minor arc $AX$ and $Z$ lies between $Y$ and $C$ in the minor arc $YC$. Consider the cyclic quadrilateral $ACZX$. We have $\angle CZX = \angle CAB$ and $\angle IAC = \angle IZC$. So that, $\[\angle CZX - \angle IZC = \angle CAB - \angle IAC\]$ $\[\angle IZX = \angle IAB\]$ Since $I$ is the incenter of quadrilateral $ABCD$, $AI$ is the angular bisector of $\angle DBA$. This gives us, $\[\angle IZX = \angle IAB = \angle IAD = \angle IAY\]$ Hence the chords $IX$ and $IY$ are equal. So $Y$ is the reflection of $X$ about $OI$. Hence, $\[TX = YZ\]$ and now it suffices to prove $\[AD + DT + XA = CD + DY + ZC\]$ Let $P, Q, N$ and $M$ be the tangency points of $\Gamma$ with $AB, BC, CD$ and $DA$ respectively. Then by tangents we have, $AD = AM + MD = AP + ND$. So $AD + DT + XA = AP + ND + DT + XA = XP + NT$. Similarly we get, $CD + DY + ZC = ZQ + YM$. So it suffices to prove, $\[XP + NT = ZQ + YM\]$ Consider the tangent $XJ$ to $\Gamma$ with $J \ne P$. Since $X$ and $Y$ are reflections about $OI$ and $\Gamma$ is a circle centred at $I$ the tangents $XJ$ and $YM$ are reflections of each other. Hence $\[XP = XJ = YM\]$ By a similar argument on the reflection of $T$ and $Z$ we get $NT = ZQ$ and finally, $\[XP + NT = ZQ + YM\]$ as required. $QED$

~BUMSTAKA