Cyclic quadrilateral $ABCD$ with circumcircle $\omega$ is given. Let $E$ be a fixed point on segment $AC$. $M$ is an arbitrary point on $\omega$, lines $AM$ and $BD$ meet at a point $P$. $EP$ meets $AB$ and $AD$ at points $R$ and $Q$, respectively, $S$ is the intersection of $BQ,DR$ and lines $MS$ and $AC$ meet at a point $T$. Prove that as $M$ varies the circumcircle of triangle $\bigtriangleup CMT$ passes through a fixed point other than $C$. Proposed by Chunlai Jin - China
Problem
Source: IGO 2024 Advanced Level - Problem 5
Tags: geometry
22.11.2024 02:45
A very tricky question . Even characterizing the fixed point is extremely difficult. I am bad at protective geometry so I just did a lot of length bashing . Would enjoy to see a better solution . We begin with the following lemma: Lemma: If $ABCD$ is a harmonic quadrilateral, then $T$ is fixed. (Let the tangents to $w$ at $B$ and $D$ meet at $Y$, then $T$ is the unique point such that $\left(YE; AT\right) = -1$). Proof: Consider a projective transformation that maps $ABCD$ to a unit square while preserving its circumcircle. It suffices to show that $E$ is the midpoint of $AT$. Let the line through $M$ parallel to $AC$ meet $w$ again at $M'$, and let $BM'$ meet $AD$ at $F$. Notice that \[ \left(AM, AM'; AB, AD\right) = -1 \]and, by the cevian configuration, \[ \left(AP, AS; AB, AD\right) = -1. \]Thus, $A$, $M'$, and $S$ are collinear. Next, since \[ \measuredangle BM'P = \measuredangle PMB = \measuredangle AMB = \measuredangle ADB = \measuredangle FDB, \]we conclude that $DPM'F$ is concyclic. Since \[ \measuredangle DPF = \measuredangle DM'F = 90^\circ, \]it follows that $FP \parallel AC$. Now, we compute $AT$ using Menelaus's theorem: \[ AT = MM' \cdot \frac{SA}{SM'} = MM' \cdot \frac{BF}{BM'} \cdot \frac{QA}{QF}. \]Substituting values, we get: \[ AT = MM' \cdot \left( \frac{1}{BM' \cdot \cos\left(\angle ABM'\right)} \right) \cdot \left( \frac{QE}{QP} \right). \]Simplifying using the ratio lemma: \[ AT = \left( \frac{2BM\cdot \sin\left(\angle DBM\right)}{BM' \cdot \cos\left(\angle ACM'\right)} \right) \cdot \left( \frac{AE \cdot \sin\left(45^\circ\right)}{AP \cdot \sin\left(\angle DAP\right)} \right). \]Expanding further: \[ AT = 2AE\cdot \left( \frac{1}{AP \cdot \left(\sqrt{2} \cdot \cos\left(\angle ACM'\right)\right)} \right) = 2AE \cdot \left( \frac{1}{AP \cdot AM} \right) = 2AE. \] Thus, $E$ is the midpoint of $AT$, as desired. \(\Box\) Now let $ABCD$ be a non-harmonic quadrilateral. Let the tangents to $w$ at $B$ and $C$ meet at $Y$, and let $AY$ meet $w$ again at $C'$. Let $PE$ meet $AC'$ at $E'$. Let $T'$ be the unique point along $AC'$ such that $\left(YE'; AT'\right) = -1$. Then, by the lemma, $M$, $T$, and $T'$ are collinear because $S$ is the same in $ABC'D$ and $ABCD$. We now state an important claim: Claim: The circumcircle of $\triangle CMT$, the circumcircle of $\triangle AMT'$, and the circle passing through $C$ tangent to $AC'$ at $A$ are concurrent. Proof: Let the circumcircles of $\triangle CMT$ and $\triangle AMT'$ meet again at $X$. Then, we compute: \[ \measuredangle AXC = \measuredangle AXM + \measuredangle MXC = \measuredangle AT'M + \measuredangle MTC = \measuredangle AT'T + \measuredangle T'TA = \measuredangle T'AT = \measuredangle C'AC. \] Thus, the three circles are concurrent. \(\Box\) Now it suffices to show that the intersection of the circumcircle of $\triangle AMT'$ with the circle passing through $C$ tangent to $AC'$ at $A$ is fixed as $M$ varies. To simplify this problem, we perform inversion at $A$, resulting in the following problem: Problem: Let $A$ be a point in the plane, and let $B$, $C$, and $D$ be three points lying in that order along a line $l$. Let $C'$ be the midpoint of $BD$. Let $M$ be a variable point on line $l$, and let $E$ be a fixed point on $AC$ beyond $C$. Let $AM$ meet the circumcircle of $\triangle ABD$ again at $P$, and let the circumcircle of $\triangle AEP$ meet $AC'$ again at $E'$. Let the two circles passing through $A$ that are tangent to $l$ at $B$ and $D$ intersect again at $Y$. Let $T'$ be the midpoint of $YE'$, and let $MT'$ intersect the line passing through $C$ parallel to $AC'$ at $X$. Prove that as $M$ varies, the point $X$ is fixed. To solve the problem, define the linear function: \[ f(X) = \mathbb{P}\left(X, (AEP)\right) - \mathbb{P}\left(X, ABD\right), \]where $\mathbb{P}(X, \Gamma)$ denotes the power of $X$ with respect to the circle $\Gamma$. Then: \[ f(M) = 0 \quad \text{and} \quad f(C) = BC \cdot CD - AC \cdot CE. \] Thus: \[ f(C') = BC' \cdot DC' - AC' \cdot E'C' = \frac{C'M}{CM} \left(BC \cdot CD - AC \cdot CE\right). \] Now observe: \[ CX = \frac{CM}{C'M} \cdot C'T' = \frac{CM}{C'M} \cdot \left(\frac{YC' - EC'}{2}\right). \] Simplifying further: \[ CX = \frac{CM}{C'M} \cdot \left(\frac{AC' \cdot AY' - AC' \cdot EC'}{2AC' }\right). \] Substituting values: \[ CX = \frac{CM}{C'M} \cdot \left(\frac{BC' \cdot DC' - AC' \cdot EC'}{2AC'}\right) = \frac{BC \cdot CD - AC \cdot CE}{2AC'}. \] Thus, $X$ is fixed as $M$ varies. \(\Box\)
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23.11.2024 08:07
Let $EP$ intersect the circle $\omega$ at points $U$ and $V$, and let $MT$ intersect $\omega$ again at $W$. Claim. $AUWV$ is a harmonic quadrilateral
Now, consider an inversion centered at $A$ with an arbitrary radius. This leads to a new problem: Given five points $A,B,C,D,E$ such that $A,C$ and $E$ are collinear and $B,C$ and $D$ are collinear. Let $M$ be an arbitrary point on the line $BD$. The line $AM$ intersects the circle $(ABD)$ again at $P$. The circle $(AEP)$ intersects $BD$ at points $U$ and $V$, and let $W$ be the midpoint of segment $UV$. The circle $(AMW)$ intersects the line $AC$ again at $T$. Prove that the circle $(CMT)$ passes through two fixed point. Solution. We have the following sequence of equalities\begin{align*}2\overline{CA}.\overline{CT}=2\overline{CM}.\overline{CW} &= \overline{CM}\left(\overline{CU}+\overline{CV}\right) = \left(\overline{CM} - \overline{CU}\right)\left(\overline{CV} - \overline{CM}\right) + CM^2 + \overline{CU}.\overline{CV} \\ & = CM^2 + \overline{CU}.\overline{CV} - \overline{MU}.\overline{MV} = CM^2 +\overline{CA}.\overline{CE} - \overline{MA}.\overline{MP} \\& = CM^2 + \overline{CA}.\overline{CE} - \left(\overline{MC} + \overline{CB}\right)\left(\overline{MC}+\overline{CD}\right) \\&= \overline{CM}\left(\overline{CB}+\overline{CD}\right) + \overline{CA}.\overline{CE} - \overline{CB}.\overline{CD}\end{align*}Hence, let $F$ be a point on $CA$ such that $$2\overline{CA}.\overline{CF} = \overline{CA}.\overline{CE} - \overline{CB}.\overline{CD}$$then $F$ is fixed. Additionally, we have $$2\overline{CA}.\overline{FT} = \overline{CM}\left(\overline{CB}+\overline{CD}\right)$$Therefore, if we consider the spiral similarity centered at a point $H$ that maps segment $FT$ to $CM$, then both the directed angle $\left(HF,HC\right)$ and the ratio $HF:HC$ remain unchanged. This implies that $H$ is fixed. Thus, the circle $(CMT)$ passes through a fixed point $H\neq C$.