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

Revision as of 12:56, 23 July 2021

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 $YC$ . 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. $QED$