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Difference between revisions of "2021 IMO Problems/Problem 4"
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We have <math>\angle CZX = \angle CAB</math> and <math>\angle IAC = \angle IZC</math>. So that, <cmath>\angle CZX - \angle IZC = \angle CAB - \angle IAC</cmath> <cmath>\angle IZX = \angle IAB</cmath>. Since <math>I</math> is the incenter of quadrilateral <math>ABCD</math>, <math>AI</math> is the angular bisector of <math>\angle DBA</math>. This gives us, <cmath>\angle IZX = \angle IAB = \angle IAD = \angle IAY</cmath>. Hence the chords  <math>IX</math> and <math>IY</math> are equal.  
 
We have <math>\angle CZX = \angle CAB</math> and <math>\angle IAC = \angle IZC</math>. So that, <cmath>\angle CZX - \angle IZC = \angle CAB - \angle IAC</cmath> <cmath>\angle IZX = \angle IAB</cmath>. Since <math>I</math> is the incenter of quadrilateral <math>ABCD</math>, <math>AI</math> is the angular bisector of <math>\angle DBA</math>. This gives us, <cmath>\angle IZX = \angle IAB = \angle IAD = \angle IAY</cmath>. Hence the chords  <math>IX</math> and <math>IY</math> are equal.  
 
So <math>Y</math> is the reflection of <math>X</math> about <math>OI</math>.
 
So <math>Y</math> is the reflection of <math>X</math> about <math>OI</math>.
 +
Similarly we get <cmath>\angle IXZ = \angle ICT</cmath> and so the chords <math>IZ</math> and <math>IT</math> are equal. Hence <math>Z</math> is the reflection of <math>T</math> about <math>OI</math>.
 +
This gives us <math>YZ</math> = <math>TX</math> immediately and now it suffices to prove, <cmath>AD + DT + XA = CD + DY + ZC</cmath>.

Revision as of 06:08, 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 $OI$. Similarly we get \[\angle IXZ = \angle ICT\] and so the chords $IZ$ and $IT$ are equal. Hence $Z$ is the reflection of $T$ about $OI$. This gives us $YZ$ = $TX$ immediately and now it suffices to prove, \[AD + DT + XA = CD + DY + ZC\].

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