SAUDI ARABIAN IMO Booklet 2023

Let $\omega ,{\omega }_c,{\omega }_b,\Omega$ be the circumcircles of triangles $ABC,AB{C}^{\prime },A{B}^{\prime }C,A{B}^{\prime }{C}^{\prime }$ respectively. Denote $O_c,O_b$ as the circumcenters of ${\omega }_c,{\omega }_b$ and $R$ as the radius of $\omega$. Let $f(X)$ be the reflection of the figure $X$ over the line $AB$.

It is easy to check that $f(\Delta ABC)=\Delta A^{\prime }BC$, so $f(\omega )={\omega }_c$ and $f(O)=O_c$. Since $C^{\prime }H$ is the common chord of ${\omega }_c,\Omega$, hence $K{O}_c$ is the perpendicular bisector of $H{C}^{\prime }$. Now the homothety of center $O$ and ratio $2$ will send $AB$ to a parallel line through $O_c$, which is $O_{c}K$ and also send $AC$ to $O_{b}K$ as well. So this homothety sends $A$ to $K$, which implies that $KA=AO=R$.

Let $B{O}_c$ intersects ${\omega }_c$ at $S^{\prime }$ then $\angle BA{S}^{\prime }=90^{\circ }$. We will prove that $S^{\prime }\equiv S$. Since $A{O}_{c}BO$ is the rhombus, hence $KO\parallel S^{\prime }B$. But $KO=S^{\prime }B=2R$, hence $K{S}^{\prime }BO$ is a parallelogram, then $O{S}^{\prime }$ passes through $D$, implies that $S^{\prime }\equiv S$. Finally, we have $SK=OK=R=KA$ which finishes the solution. $\square$