On $\triangle ABC$, let $M$ the midpoint of $AB$ and $N$ the midpoint of $CM$. Let $X$ a point such that $\angle XMC=\angle MBC$ and $\angle XCM=\angle MCB$ with $X,B$ in opposite sides of line $CM$. Let $\Omega$ the circumcircle of triangle $\triangle AMX$ a) Show that $CM$ is tangent to $\Omega$ b) Show that the lines $NX$ and $AC$ meet at $\Omega$
Problem
Source: Bolivia IMO TST 2022 Day 2 P1
Tags: geometry, circumcircle, Bolivia, TST, midpoints
Blastoor
28.07.2022 14:53
a) Denote by $B'$ the reflection of $B$ over $CM$ and by $K$ the intersection of lines $BB'$ and $CM$.
Claim. $AB'MX$ is cyclic.
Proof. Notice that $AM=MB$ and $B'K=KB$, thus $MK \parallel AB'$ as the midline in $\triangle BAB'$ $\implies$ $CM \parallel AB'$
$\implies$ $\angle XB'A = \angle XCM$.
By taking a look at $\triangle MCB$, we notice $\angle CMA = \angle MCB + \angle CBM \implies \angle CMX + \angle XMA= \angle MCB + \angle CBM$.
But $\angle CMX = \angle CBM$ and $\angle MCB = \angle XCM$, thus $\angle XMA=\angle XCM$.
$\implies \angle XB'A=\angle XMA$
$\implies AB'MX$ is cyclic $\blacksquare$.
So now we know that $B'$ lies on $\odot (AXM)$. But by symmetry $\angle MB'X=\angle MB'C=\angle CBM = \angle CMX$
$\implies CM$ is tangent to $\odot (AXM)$ $\square$.
b) Denote by $T$ the intersection of $\odot (AXM)$ and $AC$.
Then $\angle XTC=\angle XB'A=\angle XCM \implies MC$ is tangent to $\odot (CXT)$.
Look at the radical axis $TX$ of $\odot (AXM)$ and $\odot (CXT)$. It is well-known by Power of a Point that the radical axis cuts the common tangent in half $\implies TX$ goes through $N$ (the midpoint of $CM$) $\square$.