SAUDI ARABIAN IMO Booklet 2023

On $MN$, take the point $D$ such that $ND=AM$. Then, $AN=DM$. According to the sine theorem then $$\frac{BM}{\sin \angle BNM}=\frac{BN}{\sin \angle BMN}=\frac{AC}{\sin \angle AMC}=\frac{MC}{\sin \angle MAC},$$

so $\angle BNM=\angle MAC<90^{\circ }$, which implies that $△AMC\cong △NDB$ so $$\angle BDM=180^{\circ }-\angle BDN=180^{\circ }-\angle AMC=\angle AMB=60^{\circ }.$$ Also, $BD=MC=MB$ so triangle $BDM$ is isosceles, with $\angle BDM=60^{\circ }$ so equilateral. Therefore $AN=DM=BM=\frac{1}{2}BC$.



Let $X,Y$ be the centers of the circumcircles of triangles $AMC,BMN$ and $H,K$ respectively their orthocenters. We have a familiar result: triangle $ABC$ has circumradius $R$ and orthocenter $H$ then $AH=2R|\cos A|$. Applying to this problem, notice that two triangles $AMC$ and $BMN$ have the same radius of circumcircle $R$, so $$MH=2R\cos 60^{\circ }=R\text{ and }MK=2R|\cos 120^{\circ }|=R.$$ Therefore, four points $X,Y,H,K$ are on the same circle with center $M$. Next, denote $\ell$ as the internal bisector of $\angle BMN$, then $MH,MX$ are symmetric through $\ell$. And $MH=MX$ so we have $HX\perp \ell$. Similarly $KY\perp \ell$ so $HX\parallel KY$. From this it follows that $HXYK$ is an isosceles trapezoid. $\square$