Let $k$ be a circle with centre $O$. Let $AB$ be a chord of this circle with midpoint $M\neq O$. The tangents of $k$ at the points $A$ and $B$ intersect at $T$. A line goes through $T$ and intersects $k$ in $C$ and $D$ with $CT < DT$ and $BC = BM$. Prove that the circumcentre of the triangle $ADM$ is the reflection of $O$ across the line $AD$.
Problem
Source: Swiss IMO TST 2020 P3
Tags: geometry, symmedian
rafaello
28.09.2021 00:35
Note that $DC$ is the symmedian of triangle $ADB$. Hence, we know that $\triangle AMD\sim \triangle CBD$, however as $AM=MB=CB$, we get that $\triangle AMD\cong \triangle CBD$. Thus if $E$ is the reflection of $O$ over $AD$, we get that
$$\measuredangle AED=\measuredangle DOA=2\measuredangle DBA=2\measuredangle BMD=2\measuredangle AMD,$$this shows that $M$ lies on the circle with centre $E$ and radius $EA=ED$, we are done.
VulcanForge
28.09.2021 05:23
It suffices to show the reflection of $M$ across $AD$ lies on $k$. Let $C'$ be the point on $\widehat{ACB}$ such that $AC=AM$; then since $DC$ is a symmedian in $\triangle ADB$, we have that $DC'$ is a median (i.e. it passes through $M$). Now if $M'$ is the reflection of $M$ across $AD$, we have$$\measuredangle DM'A = -\measuredangle DMA =-\measuredangle C'MA = \measuredangle MC'A= \measuredangle DC'A$$hence $M' \in k$ as desired.
ike.chen
28.09.2021 05:55
Since $DC, DM$ are isogonal wrt $\angle ADB$ and $\angle DCB = \angle DAB = \angle DAM$, we have $DBC \sim DMA$. But $BC = BM = MA$, so the two triangles are actually congruent. Now, it's easy to see $DB = DM$, so $$\angle AMD = 180^{\circ} - \angle BMD = 180^{\circ} - \angle MBD = 180^{\circ} - \angle ABD.$$Hence, the reflection of $M$ over $AD$ lies on $k$, which clearly suffices. $\blacksquare$