2021 IMO Problems/Problem 4

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

Solution 3 (Visual)

Lemma 1 Let $O$ be the center of $\Omega.$ Then point $T($ point $X)$ is symmetryc to $Z(Y)$ 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 $Y.$ $\overset{\Large\frown} {TZ}= 2\angle DCZ = 2\pi – 4\gamma,$ $\overset{\Large\frown} {IZ}= 2\pi - \overset{\Large\frown} {AT} - \overset{\Large\frown} {TZ}= = 2\gamma= \overset{\Large\frown} {IT}\implies$ symmetry $T$ and $Z.$

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