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Difference between revisions of "2021 IMO Problems/Problem 4"
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==Problem== Let <math>\Gamma</math> be a circle with centre <math>I</math>, and <math>ABCD</math> a convex quadrilateral such that each of
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==Problem==  
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Let <math>\Gamma</math> be a circle with centre <math>I</math>, and <math>ABCD</math> a convex quadrilateral such that each of
 
the segments <math>AB, BC, CD</math> and <math>DA</math> is tangent to <math>\Gamma</math>. Let <math>\Omega</math> be the circumcircle of the triangle <math>AIC</math>.
 
the segments <math>AB, BC, CD</math> and <math>DA</math> is tangent to <math>\Gamma</math>. Let <math>\Omega</math> be the circumcircle of the triangle <math>AIC</math>.
 
The extension of <math>BA</math> beyond <math>A</math> meets <math>\Omega</math> at <math>X</math>, and the extension of <math>BC</math> beyond <math>C</math> meets <math>\Omega</math> at <math>Z</math>.
 
The extension of <math>BA</math> beyond <math>A</math> meets <math>\Omega</math> at <math>X</math>, and the extension of <math>BC</math> beyond <math>C</math> meets <math>\Omega</math> at <math>Z</math>.

Revision as of 07:52, 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. $QED$

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