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# 2021 IMO Problems/Problem 4

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$Problem:$ 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 $\[AD + DT + T X + XA = CD + DY + Y Z + ZC\]$

$Solution:$

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 $YC$.