SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Suppose that $AI$ cuts $BC$ at $D$ and cuts $(ABC)$ again at $P$. We have known that $B,C,I,J$ lie on circle $(P,\frac{IJ}{2})$.

Suppose that $AN$ meets $BC$ at $E$, then $AE$ is the external angle bisector of $\angle BAC$, so $(DE,BC)=-1$. We have $$\overline{DE}\cdot \overline{DM}=\overline{DB}\cdot \overline{DC}=\overline{DI}\cdot \overline{DJ}$$ means that $IMJE$ is cyclic.



In the other hand, we have known that $(AD,IJ)=-1$. Combining with $D$ is orthocenter of $△NPE$, we get $$\overline{AT}\cdot \overline{AS}=-\overline{AN}\cdot \overline{AS}=\overline{AD}\cdot \overline{AP}=\overline{AI}\cdot \overline{AJ},$$ implies that $IJET$ is cyclic.

From (1) and (2), then the points $J,I,M,T,E$ lie on a circle. $\square$