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

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)

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

@@ 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 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 5: / Line 7: @@
 <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]
-<math>Solution:</math>
+https://www.youtube.com/watch?v=U95v_xD5fJk
-Let <math>O</math> be the centre of <math>\Omega</math>
+https://youtu.be/WkdlmduOnRE
-For <math>AB=BC</math> the result follows simply. By Pitot's Theorem we have <cmath>AB + CD = BC + AD</cmath> so that, <cmath>AD = CD.</cmath> The configuration becomes symmetric about <math>OI</math> and the result follows immediately.
+== Solution 1 ==
+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.
 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>.
-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>.
-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>.
+Hence, <cmath>TX = YZ</cmath> and now it suffices to prove <cmath>AD + DT + XA = CD + DY + ZC</cmath>
-This gives us <math>YZ</math> = <math>TX</math> immediately 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>.
+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,
+<cmath> XP + NT = ZQ + YM</cmath> as required.
+<math>QED</math>
+~BUMSTAKA
+== Solution 2 ==
+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>.
+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
+== Solution 3 (Visual) ==
+[[File:2021 IMO 4b.png|420px|right]]
+[[File:2021 IMO 4.png|420px|right]]
+[[File:2021 IMO 4a.png|420px|right]]
+We use the equality of the tangent segments and symmetry.
+Using <i><b>Claim 1</b></i> we get <math>\overset{\Large\frown} {TX}</math> symmetric to <math>\overset{\Large\frown} {ZY}</math> with respect <math>IO.</math>
+Therefore <math>\hspace{10mm} TX = ZY.</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.
+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>= XP + TM = QZ + NY = MC + ZC + MD + DY =</cmath>
+<cmath>=CD + ZC + DY.</cmath>
+<i><b>Claim 1</b></i>
+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>
+Let <math>\angle BAD =2\alpha, \angle ABC =2\beta, \angle BCD =2\gamma, \angle ADC =2\delta.</math>
+We find measure of some arcs:
+<cmath>\overset{\Large\frown} {IT}= 2\angle ICT = 2\gamma,</cmath>
+<cmath>\overset{\Large\frown} {IY}= 2\angle IAY = 2\alpha,</cmath>
+<cmath>\overset{\Large\frown} {XY}= 2\angle XAY = 2\pi - 4\alpha,</cmath>
+<math>\overset{\Large\frown} {IX}= 2\pi - \overset{\Large\frown} {IY} - \overset{\Large\frown} {XY} = 2\alpha =\overset{\Large\frown} {IY}\implies</math> symmetry <math>X</math> and <math>Y.</math>
+<cmath>\overset{\Large\frown} {TZ}= 2\angle DCZ = 2\pi – 4\gamma,</cmath>
+<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>
+<i><b>Claim 2</b></i>
+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>
+<cmath>AI = A'I, IB = IC, \angle ACI = \angle A'BI = 90^\circ \implies</cmath>
+<cmath>\triangle AIC = \triangle A'IB \implies A'B = AC.</cmath>
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+== See also ==
+{{IMO box|year=2021|num-b=3|num-a=5}}
+[[Category:Olympiad Geometry Problems]]

2021 IMO (Problems) • Resources
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