IMO Team Selection Test 1

We consider the configuration as in the figure. Other configurations are treated analogously. Denote by $D$ the second intersection of $AH$ with the circumcircle of $△ABC$. Denote by $S$ the second intersection of the circumcircles of $ABC$ and $AHO$. (Because $△ABC$ is acute, both $O$ and $H$ lie in the interior of $ABC$ and also in the interior of the circumcircle, hence $D$ and $S$ both exist.) We have $$\angle OSH=\angle OAH=\angle OAD=\angle ODA=\angle ODH,$$ where we use that $|OA|=|OD|$. Moreover, we have $$\angle OHD=180^{\circ }-\angle OHA=180^{\circ }-\angle OSA=180^{\circ }-\angle OAS=\angle OHS,$$ where we use that $|OA|=|OS|$. Now we conclude that $△OHS\cong △OHD$ (SAA). This yields that $D$ and $S$ are each others reflection images in $OH$.

Therefore, if we reflect the circumcentre $K$ of $△OHS$ in $OH$, we get the circumcentre $L$ of $△OHD$. Now we must prove that $L$ lies on $BC$. Point $D$ is the reflection of $H$ in $BC$. This is a known fact, which we can prove as follows: $\angle DBC=\angle DAC=\angle HAC=90^{\circ }-\angle ACB=\angle HBC$ and analogously $\angle DCB=\angle HCB$, hence $△DBC\cong △HBC$ (ASA). Hence, $D$ is indeed the reflection of $H$ in $BC$, from which we get that $BC$ is the perpendicular bisector of $HD$. Because $L$ lies on the perpendicular bisector of $HD$, we get that $L$ lies on $BC$, which is what we wanted to prove. $\square$