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

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==Problem==  
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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
 
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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<cmath>AD + DT + T X + XA = CD + DY + Y Z + ZC</cmath>
 
<cmath>AD + DT + T X + XA = CD + DY + Y Z + ZC</cmath>
  
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==Video Solutions==
+
== Video Solutions ==
 
https://youtu.be/vUftJHRaNx8 [Video contains solutions to all day 2 problems]
 
https://youtu.be/vUftJHRaNx8 [Video contains solutions to all day 2 problems]
  
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https://youtu.be/WkdlmduOnRE
 
https://youtu.be/WkdlmduOnRE
  
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==Solution==
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== Solution 1 ==
  
 
Let <math>O</math> be the centre of <math>\Omega</math>.
 
Let <math>O</math> be the centre of <math>\Omega</math>.
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~BUMSTAKA
 
~BUMSTAKA
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==Solution2==
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== Solution 2 ==
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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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Denote <math>AD</math> tangents to the circle <math>I</math> at <math>N</math>, <math>CD</math> tangents to the same circle at <math>M</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>
 
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>.
 
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>.
 
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
 
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
 
~bluesoul
  
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==Solution 3 (Visual)==
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== Solution 3 (Visual) ==
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[[File:2021 IMO 4b.png|450px|right]]
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[[File:2021 IMO 4b.png|420px|right]]
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[[File:2021 IMO 4.png|450px|right]]
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[[File:2021 IMO 4.png|420px|right]]
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[[File:2021 IMO 4a.png|450px|right]]
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[[File:2021 IMO 4a.png|420px|right]]
 
We use the equality of the tangent segments and symmetry.
 
We use the equality of the tangent segments and symmetry.
 
   
 
   
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Therefore <math>\hspace{10mm} TX = ZY.</math>
 
Therefore <math>\hspace{10mm} TX = ZY.</math>
  
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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.
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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.
  
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Using Lemma 2 we get <math>TM = QZ, PX = NY.</math>
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Using <i><b>Claim 2</b></i> we get <math>TM = QZ, PX = NY.</math>
  
 
<cmath>AD + DT + XA = AN+ND + TM – MD +XP-PA =</cmath>
 
<cmath>AD + DT + XA = AN+ND + TM – MD +XP-PA =</cmath>
 
<cmath>= XP + TM = QZ + NY = MC + ZC + MD + DY =</cmath>
 
<cmath>= XP + TM = QZ + NY = MC + ZC + MD + DY =</cmath>
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<cmath>=CD + DY +  ZC.</cmath>
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<cmath>=CD + ZC + DY.</cmath>
 
<i><b>Claim 1</b></i>
 
<i><b>Claim 1</b></i>
  
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Let <math>O</math> be the center of <math>\Omega.</math> Then point <math>T(</math> point <math>X)</math> is symmetryc to <math>Z(Y)</math> with respect <math>IO.</math>
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Let <math>O</math> be the center of <math>\Omega.</math> Then point <math>T</math> is symmetric to <math>Z</math> with respect <math>IO,</math> point <math>X</math> is symmetric to <math>Y</math> with respect <math>IO.</math>
  
 
<i><b>Proof</b></i>
 
<i><b>Proof</b></i>
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<math>\overset{\Large\frown} {IZ}= 2\pi - \overset{\Large\frown} {IT} - \overset{\Large\frown} {TZ}= 2\gamma= \overset{\Large\frown} {IT}\implies</math> symmetry <math>T</math> and <math>Z.</math>
 
<math>\overset{\Large\frown} {IZ}= 2\pi - \overset{\Large\frown} {IT} - \overset{\Large\frown} {TZ}= 2\gamma= \overset{\Large\frown} {IT}\implies</math> symmetry <math>T</math> and <math>Z.</math>
  
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<i><b>Lemma 2</b></i>
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<i><b>Claim 2</b></i>
  
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Let circles <math>\omega</math> centered at <math>I</math> and <math>\Omega</math> centered at <math>O</math> be given. Let points <math>A</math> and <math>A'</math> lies on  <math>\Omega</math> and symmetrical with respect <math>OI.</math> Let <math>AC</math> and <math>A'B</math> be tangents to  <math>\omega</math>. Then <math>AC = A'B.</math>
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Let circles <math>\omega</math> centered at <math>I</math> and <math>\Omega</math> centered at <math>O</math> be given. Let points <math>A</math> and <math>A'</math> lies on  <math>\Omega</math> and <math>A</math> be symmetric to <math>A'</math> with respect <math>OI.</math> Let <math>AC</math> and <math>A'B</math> be tangents to  <math>\omega</math>. Then <math>AC = A'B.</math>
 
   
 
   
 
<i><b>Proof</b></i>
 
<i><b>Proof</b></i>
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'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 +
== See also ==
 +
{{IMO box|year=2021|num-b=3|num-a=5}}
 +
 +
[[Category:Olympiad Geometry Problems]]

Latest revision as of 09:30, 18 June 2023

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 1

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

Solution 2

Denote $AD$ tangents to the circle $I$ at $N$, $CD$ tangents to the same circle at $M$; $XB$ 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

Solution 3 (Visual)

2021 IMO 4b.png
2021 IMO 4.png
2021 IMO 4a.png

We use the equality of the tangent segments and symmetry.

Using Claim 1 we get $\overset{\Large\frown} {TX}$ symmetric to $\overset{\Large\frown} {ZY}$ with respect $IO.$

Therefore $\hspace{10mm} TX = ZY.$

Let $P, Q, N$ and $M$ be the tangency points of $\Gamma$ with $AB, BC, CD,$ and $DA,$ respectively.

Using Claim 2 we get $TM = QZ, PX = NY.$

\[AD + DT + XA = AN+ND + TM – MD +XP-PA =\] \[= XP + TM = QZ + NY = MC + ZC + MD + DY =\] \[=CD + ZC + DY.\] Claim 1

Let $O$ be the center of $\Omega.$ Then point $T$ is symmetric to $Z$ with respect $IO,$ point $X$ is symmetric to $Y$ with respect $IO.$

Proof

Let $\angle BAD =2\alpha, \angle ABC =2\beta, \angle BCD =2\gamma, \angle ADC =2\delta.$

We find measure of some arcs: \[\overset{\Large\frown} {IT}= 2\angle ICT = 2\gamma,\] \[\overset{\Large\frown} {IY}= 2\angle IAY = 2\alpha,\] \[\overset{\Large\frown} {XY}= 2\angle XAY = 2\pi - 4\alpha,\] $\overset{\Large\frown} {IX}= 2\pi - \overset{\Large\frown} {IY} - \overset{\Large\frown} {XY} = 2\alpha =\overset{\Large\frown} {IY}\implies$ symmetry $X$ and $Y.$ \[\overset{\Large\frown} {TZ}= 2\angle DCZ = 2\pi – 4\gamma,\] $\overset{\Large\frown} {IZ}= 2\pi - \overset{\Large\frown} {IT} - \overset{\Large\frown} {TZ}= 2\gamma= \overset{\Large\frown} {IT}\implies$ symmetry $T$ and $Z.$

Claim 2

Let circles $\omega$ centered at $I$ and $\Omega$ centered at $O$ be given. Let points $A$ and $A'$ lies on $\Omega$ and $A$ be symmetric to $A'$ with respect $OI.$ Let $AC$ and $A'B$ be tangents to $\omega$. Then $AC = A'B.$

Proof \[AI = A'I, IB = IC, \angle ACI = \angle A'BI = 90^\circ \implies\] \[\triangle AIC = \triangle A'IB \implies A'B = AC.\]

vladimir.shelomovskii@gmail.com, vvsss

See also

2021 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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