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

 
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<cmath>AD + DT + T X + XA = CD + DY + Y Z + ZC</cmath>
 
<cmath>AD + DT + T X + XA = CD + DY + Y Z + ZC</cmath>
  
 +
==Video Solutions==
 +
https://youtu.be/vUftJHRaNx8 [Video contains solutions to all day 2 problems]
 +
 +
https://www.youtube.com/watch?v=U95v_xD5fJk
 +
 +
https://youtu.be/WkdlmduOnRE
  
 
==Solution==
 
==Solution==
  
Let <math>O</math> be the centre of <math>\Omega</math>
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Let <math>O</math> be the centre of <math>\Omega</math>.
 +
 
 
For <math>AB=BC</math> the result follows simply. By Pitot's Theorem we have <cmath>AB + CD = BC + AD</cmath> so that, <math>AD = CD.</math> The configuration becomes symmetric about <math>OI</math> and the result follows immediately.
 
For <math>AB=BC</math> the result follows simply. By Pitot's Theorem we have <cmath>AB + CD = BC + AD</cmath> so that, <math>AD = CD.</math> The configuration becomes symmetric about <math>OI</math> and the result follows immediately.
  
 
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.  
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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.  
 
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>
 
Let <math>P, Q, N</math> and <math>M</math> be the tangency points of <math>\Gamma</math> with <math>AB, BC, CD</math> and <math>DA</math> respectively. Then by tangents we have, <math>AD = AM + MD = AP + ND</math>. So <math>AD + DT + XA = AP + ND + DT + XA = XP + NT</math>.
 
Let <math>P, Q, N</math> and <math>M</math> be the tangency points of <math>\Gamma</math> with <math>AB, BC, CD</math> and <math>DA</math> respectively. Then by tangents we have, <math>AD = AM + MD = AP + ND</math>. So <math>AD + DT + XA = AP + ND + DT + XA = XP + NT</math>.
Similarly we get, <math>CD + DY + ZC = ZQ + YM</math>. So it suffices to prove, <cmath>XP + NT = ZQ + YM.</cmath>
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Similarly we get, <math>CD + DY + ZC = ZQ + YM</math>. So it suffices to prove, <cmath>XP + NT = ZQ + YM</cmath>
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,
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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.
 
<math>QED</math>
 
<math>QED</math>
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~BUMSTAKA
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==Solution2==
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Denote <math>AD</math> tangents to the circle <math>I</math> at <math>M</math>, <math>CD</math> tangents to the same circle at <math>N</math>; <math>XB</math> tangents at <math>F</math> and <math>ZB</math> tangents at <math>J</math>. We can get that <math>AD=AM+MD;CD=DN+CN</math>.Since <math>AM=AF,XA=XF-AM;ZC=ZJ-CN</math> Same reason, we can get that <math>DT=TN-DM;DY=YM-DM</math>
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We can find that <math>AD+XA+DT=XF+TN;CD+DY+ZC=ZJ+YM</math>. Connect <math>IM,IT,IT,IZ,IX,IN,IJ,IF</math> separately, we can create two pairs of congruent triangles. In <math>\triangle{XIF},\triangle{YIM}</math>, since <math>\widehat{AI}=\widehat{AI},\angle{FXI}=\angle{MYI}</math> After getting that <math>\angle{IFX}=\angle{IMY};\angle{FXI}=\angle{MYI};IF=IM</math>, we can find that <math>\triangle{IFX}\cong \triangle{IMY}</math>. Getting that <math>YM=XF</math>, same reason, we can get that <math>ZJ=TN</math>.
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Now the only thing left is that we have to prove <math>TX=YZ</math>. Since <math>\widehat{IX}=\widehat{IY};\widehat{IT}=\widehat{IZ}</math> we can subtract and get that <math>\widehat{XT}=\widehat{YZ}</math>,means <math>XT=YZ</math> and we are done
 +
~bluesoul

Latest revision as of 14:05, 19 December 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 \[AD + DT + T X + XA = CD + DY + Y Z + ZC\]

Video Solutions

https://youtu.be/vUftJHRaNx8 [Video contains solutions to all day 2 problems]

https://www.youtube.com/watch?v=U95v_xD5fJk

https://youtu.be/WkdlmduOnRE

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. $QED$

~BUMSTAKA

Solution2

Denote $AD$ tangents to the circle $I$ at $M$, $CD$ tangents to the same circle at $N$; $XB$ tangents at $F$ and $ZB$ tangents at $J$. We can get that $AD=AM+MD;CD=DN+CN$.Since $AM=AF,XA=XF-AM;ZC=ZJ-CN$ Same reason, we can get that $DT=TN-DM;DY=YM-DM$ We can find that $AD+XA+DT=XF+TN;CD+DY+ZC=ZJ+YM$. Connect $IM,IT,IT,IZ,IX,IN,IJ,IF$ separately, we can create two pairs of congruent triangles. In $\triangle{XIF},\triangle{YIM}$, since $\widehat{AI}=\widehat{AI},\angle{FXI}=\angle{MYI}$ After getting that $\angle{IFX}=\angle{IMY};\angle{FXI}=\angle{MYI};IF=IM$, we can find that $\triangle{IFX}\cong \triangle{IMY}$. Getting that $YM=XF$, same reason, we can get that $ZJ=TN$. Now the only thing left is that we have to prove $TX=YZ$. Since $\widehat{IX}=\widehat{IY};\widehat{IT}=\widehat{IZ}$ we can subtract and get that $\widehat{XT}=\widehat{YZ}$,means $XT=YZ$ and we are done ~bluesoul

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