From a point $S$, which lies outside the circle $\Omega$, tangent lines $SA$ and $SB$ to that circle are drawn. On the chord $AB$ an arbitrary point $K$ is chosen. $SK$ intersects $\Omega$ at points $P$ and $Q$, and chords $RT$ and $UW$ pass through $K$ such that $W, Q$ and $T$ lie in the same half-plane with respect to $AB$. Lines $WR$ and $TU$ intersect chord $AB$ at $C$ and $D$, and $M$ is the midpoint of $PQ$. Prove that $\angle AMC = \angle BMD$
Problem
Source: Belarus - Iran Competition 2022
Tags: geometry
sami1618
14.09.2024 19:38
Let the tangents to $\Omega$ at $P$ and $Q$ meet at $X$. As $ABPQ$ is harmonic, $X$ lies on $AB$. Label the circumcircle of $KMX$ as $\Gamma$. It is a well known fact of harmonic quadrilaterals that $\angle KMB=\angle KMA$. Notice as $\angle KMX=90^{\circ}$, $\Gamma$ is the circles of Apollonius of $(AB;KX)$. Applying DIT on quadrilateral $RTUW$, line $AB$, and conic $\Omega$ gives that there exists an inversion preserving $K$ and swapping pairs $(A,B)$ and $(C,D)$. But this inversion must be the inversion about $\Gamma$ is it preserves $K$ and swaps $A$ and $B$. Thus $C$ and $D$ are inverses in $\Gamma$ so $\angle KMC=\angle KMD$, and the result follows.
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Haris1
16.01.2025 22:41
@above beautiful sol but also quite overkill. Heres my solution : First we invert wrt to $K$ with radius $$\sqrt{KA*KB}$$Its easy to see that $C$ and $D$ go to $$(KTU)\cap AB$$and $$(KRW) \cap AB$$. And the new condition is just to prove $$AD=BC$$After doing very basic angle chase these length ratios are equivalent to proving $$AR/sin(ABR)=BU/sin(BAU)$$which is clearly true. Nice problem