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

Line 1: Line 1:
<math>Problem:</math> Let <math>\Gamma</math> be a circle with centre <math>I</math>, and <math>ABCD</math> a convex quadrilateral such that each of
+
==Problem== 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>.
Line 6: Line 6:
  
  
<math>Solution:</math>
+
==Solution==
  
 
Let <math>O</math> be the centre of <math>\Omega</math>
 
Let <math>O</math> be the centre of <math>\Omega</math>
Line 13: Line 13:
 
Now assume WLOG <math>AB < BC</math>. Then <math>T</math> lies between <math>A</math> and <math>X</math> in the minor arc <math>AX</math> and <math>Z</math> lies between <math>Y</math> and <math>C</math> in the minor arc <math>YC</math>.
 
Now assume WLOG <math>AB < BC</math>. Then <math>T</math> lies between <math>A</math> and <math>X</math> in the minor arc <math>AX</math> and <math>Z</math> lies between <math>Y</math> and <math>C</math> in the minor arc <math>YC</math>.
 
Consider the cyclic quadrilateral <math>ACZX</math>.
 
Consider the cyclic quadrilateral <math>ACZX</math>.
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>.
 
Hence, <cmath>TX = YZ</cmath> and now it suffices to prove <cmath>AD + DT + XA = CD + DY + ZC</cmath>
 
Hence, <cmath>TX = YZ</cmath> and now it suffices to prove <cmath>AD + DT + XA = CD + DY + ZC</cmath>
Line 20: Line 20:
 
Consider the tangent <math>XJ</math> to <math>\Gamma</math> with <math>J \ne P</math>. Since <math>X</math> and <math>Y</math> are reflections about <math>OI</math> and <math>\Gamma</math> is a circle centred at <math>I</math> the tangents <math>XJ</math> and <math>YM</math> are reflections of each other. Hence <cmath>XP = XJ = YM</cmath>. By a similar argument on the reflection of <math>T</math> and <math>Z</math> we get <math>NT = ZQ</math> and finally,
 
Consider the tangent <math>XJ</math> to <math>\Gamma</math> with <math>J \ne P</math>. Since <math>X</math> and <math>Y</math> are reflections about <math>OI</math> and <math>\Gamma</math> is a circle centred at <math>I</math> the tangents <math>XJ</math> and <math>YM</math> are reflections of each other. Hence <cmath>XP = XJ = YM</cmath>. By a similar argument on the reflection of <math>T</math> and <math>Z</math> we get <math>NT = ZQ</math> and finally,
 
<cmath> XP + NT = ZQ + YM</cmath> as required.
 
<cmath> XP + NT = ZQ + YM</cmath> as required.
<cmath>QED</cmath>
+
<math>QED</math>

Revision as of 07:51, 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. $QED$

Invalid username
Login to AoPS