Seventeenth Stars of Mathematics Competition

We first show that $AVMS$ is cyclic. Since the triangle $ABC$ is isosceles, $AM\perp BC$, and since $NP$ is midline, $V$ is the midpoint of $AR$. Then $MV$ is the $M$-median of the right triangle $AMR$, so $VM=VA=VR$. Hence $V$ lies on the perpendicular bisector of $AM$. By hypothesis, $V$ also lies on the internal bisector of $\angle ASM$, so, in the circle $AMS$, $V$ is the midpoint of the arc $AM$ not containing $S$. Consequently, $AVMS$ is cyclic, as stated. Similarly, $AUMT$ is cyclic.

We now prove that $H$ lies on the circle $NUS$. Since $AVMS$ is cyclic, $\angle ASV=\angle MAR$. By hypothesis, $\angle MAR=\angle BAQ$, so $AB\parallel SV$. Similarly, $AC\parallel TU$. Then $\angle SVP=\angle APV=90^{\circ }-\frac{1}{2}\angle BAC$, and $\angle AUH=90^{\circ }-\angle UAV=90^{\circ }-\frac{1}{2}\angle BAC$, so $\angle SVP=\angle AUH$.

From the triangles $SUV$ and $AUN$, $$\frac{SU}{SV}=\frac{\sin (90^{\circ }-\frac{1}{2}\angle BAC)}{\sin \angle SUV}=\frac{AU}{AN}.$$ Since $\angle UAM=\angle UAH=\angle VAN$ and $\angle AUH=90^{\circ }-\frac{1}{2}\angle BAC=\angle ANV$, the triangles $AUH$ and $ANV$ are similar. Hence $AN/AU=NV/UH$, so $SU/SV=UH/VN$. Since $\angle SVN=180^{\circ }-\angle SVP=180^{\circ }-\angle AUM=\angle SUH$, the triangles $SUH$ and $SVN$ are similar, so $\angle USH=\angle VSN$, whence $\angle USV=\angle HSN$. Finally, $\angle USV=\angle MAV=90^{\circ }-\angle AVU=\angle HUV$, so $\angle HSN=\angle HUV=\angle HUN$. Consequently, $H$ lies on the circle $NUS$, as stated. Similarly, $H$ lies on the circle $PVT$. This ends the proof.