SAUDI ARABIAN IMO Booklet 2023

a) Redefine the point $K$ as the orthocenter of the triangle $BCD$, then $B$ is also the orthocenter of the triangle $KCD$, we will show that $K\in PQ$. Construct the parallelogram $BCDT$.



Since $K$ is the orthocenter of triangle $BCD$, $BC\perp KD$, and $BC\parallel TD$ so $TD\perp KD$. Similarly, $TC\perp KC$ so $KCTD$ is inscribed in a circle of diameter $KT$. Since $CD$ is a common tangent to $(O_1),(O_2)$, $\angle BCD=\angle CAB$, $\angle BDC=\angle DAB$. Therefore $$180^{\circ }-\angle CBD=\angle BCD+\angle BDC=\angle CAB+\angle DAB=\angle CAD.$$ Since $BCTD$ is a parallelogram, $\angle CBD=\angle CTD$, hence $180^{\circ }-\angle CTD=\angle CAD$ entails $ACTD$ internal. Therefore, the points $A,C,T,D,K$ belong to the circle of diameter $KT$. Since $BP,BQ$ are the diameters of $(O_1),(O_2)$, $\angle BAP=\angle BAQ=90^{\circ }$, so $A,P,Q$ are collinear. It is left to prove that $KA\perp AB$. From $KACD$ is inscribed, we have $\angle KAC=\angle KDC=90^{\circ }-\angle BCD=90^{\circ }-\angle BAC$ implies that $$\angle KAB=\angle KAC+\angle BAC=90^{\circ }.$$ Hence $K,P,Q$ are collinear and thus the original problem is solved.

b) Draw the altitude $BH$ of the triangle $BCD$ and $M$ is the midpoint of $CD$. We have $$\angle KBP=180^{\circ }-(\angle CBP+\angle CBH)=180^{\circ }-(90^{\circ }-\angle BPC+90^{\circ }-\angle BCH)=2\angle BCH.$$ Since $BX$ is the bisector of $\angle KBP$, then $\angle KBX=\angle BCM$. We can see that $AKMH$ is cyclic since $\angle KAM=\angle KHM=90^{\circ }$ so $\angle BKX=\angle BMC$. This implies that $$\Delta BKX\sim \Delta CMB⟹\frac{BK}{CM}=\frac{KX}{MB}.$$ Similarly, $\Delta BKY\sim \Delta DMB⟹\frac{BK}{DM}=\frac{KY}{MB}$. And $CM=DM$ so we get $$\frac{KX}{MB}=\frac{KY}{MB}⟹KX=KY.$$ $\square$