Saudi Arabia booklet 2024

Let $I_a,I_b,I_c$ be the ex-center of angle $A,B,C$ in triangle $ABC$ respectively. Then, it is clear that $A,B,C$ are the feet of the altitude in triangle $I_a{I}_b{I}_c$, and $X,Y,Z$ are the midpoints of the three sides of triangle $I_a{I}_b{I}_c$. Thus $AX\parallel YZ$. On the other hand, $AX\perp AI$ and $AI\perp EF$ imply that $EF\parallel YZ$. Similarly, we get two triangles $DEF$ and $XYZ$ have three corresponding sides parallel.

According to Thales theorem then $$\frac{PD}{PY}=\frac{DE}{XY}\text{ and }\frac{QD}{QZ}=\frac{DF}{XZ},$$ but two triangles $DEF$, $XYZ$ are similar so $$\frac{DE}{XY}=\frac{DF}{XZ}⟹\frac{PD}{PY}=\frac{QD}{QZ}.$$ This shows that $YZ\parallel PQ$ according to Thales theorem, hence $PQ\parallel EF$. Using the trapezoidal lemma, $XT$ will bisect the segments $EF,PQ$. On the other hand $XA\parallel EF$ leads to $X(AT,EF)=-1$. Projected onto $(O)$ then one can get the harmonic quadrilateral $AMKN$. If the tangent line of $(O)$ at $A,K$ intersects at $L$, it is clear that $L\in MN$. Points $A,O,K,L$ belong to the circle of diameter $LO$, so if $(LO)$ intersects $MN$ at $H$, we get $\angle OHL=90^{\circ }$ and $H$ is the midpoint of $MN$. $\square$