APMO

In the sequel, we denote $\angle BAC=\alpha$, $\angle CBA=\beta$, $\angle ACB=\gamma$. Let $O^{\prime }$ be the circumcenter of the triangle $MNH$. The lengths of line segments starting from the point $H$ will be treated as signed quantities. Let us denote by $M^{\prime },N^{\prime }$ the point of intersection of $CH,BH$, respectively, with the circumcircle of the triangle $ABC$ (distinct from $C,B$, respectively.) From the fact that 4 points $A,M,H,C$ lie on the same circle, we see that $\angle MH{M}^{\prime }=\alpha$ holds. Furthermore, $\angle B{M}^{\prime }C$, $\angle B{N}^{\prime }C$ and $\alpha$ are all subtended by the same arc $\text{overparen}BC$ of the circumcircle of the triangle $ABC$ at points on the circle, and therefore, we have $\angle B{M}^{\prime }C=\alpha$, and $\angle B{N}^{\prime }C=\alpha$ as well. We also have $\angle ABH=\angle AC{N}^{\prime }$ as they are subtended by the same arc $\text{overparen}A{N}^{\prime }$ of the circumcircle of the triangle $ABC$ at points on the circle. Since $H{M}^{\prime }\perp BM$, $H{N}^{\prime }\perp AC$, we conclude that $$\angle M^{\prime }HB=90^{\circ }-\angle ABH=90^{\circ }-\angle AC{N}^{\prime }=\alpha$$ is valid as well. Putting these facts together, we obtain the fact that the quadrilateral $HB{M}^{\prime }M$ is a rhombus. In a similar manner, we can conclude that the quadrilateral $HC{N}^{\prime }N$ is also a rhombus. Since both of these rhombuses are made up of 4 right triangles with an angle of magnitude $\alpha$, we also see that these rhombuses are similar. Let us denote by $P,Q$ the feet of the perpendicular lines on $HM$ and $HN$, respectively, drawn from the point $O^{\prime }$. Since $O^{\prime }$ is the circumcenter of the triangle $MNH$, $P,Q$ are respectively, the midpoints of the line segments $HM,HN$. Furthermore, if we denote by $R,S$ the feet of the perpendicular lines on $HM$ and $HN$, respectively, drawn from the point $O$, then since $O$ is the circumcenter of both the triangle $M^{\prime }BC$ and the triangle $N^{\prime }BC$, we see that $R$ is the intersection point of $HM$ and the perpendicular bisector of $B{M}^{\prime }$, and $S$ is the intersection point of $HN$ and the perpendicular bisector of $C{N}^{\prime }$. We note that the similarity map $\varphi$ between the rhombuses $HB{M}^{\prime }M$ and $HC{N}^{\prime }N$ carries the perpendicular bisector of $B{M}^{\prime }$ onto the perpendicular bisector of $C{N}^{\prime }$, and straight line $HM$ onto the straight line $HN$, and hence $\varphi$ maps $R$ onto $S$, and $P$ onto $Q$. Therefore, we get $HP:HR=HQ:HS$. If we now denote by $X,Y$ the intersection points of the line $H{O}^{\prime }$ with the line through $R$ and perpendicular to $HP$, and with the line through $S$ and perpendicular to $HQ$, respectively, then we get $$H{O}^{\prime }:HX=HP:HR=HQ:HS=H{O}^{\prime }:HY$$ so that we must have $HX=HY$, and therefore, $X=Y$. But it is obvious that the point of intersection of the line through $R$ and perpendicular to $HP$ with the line through $S$ and perpendicular to $HQ$ must be $O$, and therefore, we conclude that $X=Y=O$ and that the points $H,O^{\prime },O$ are collinear.

Alternate Solution:

Deduction of the fact that both of the quadrilaterals $HB{M}^{\prime }M$ and $HC{N}^{\prime }N$ are rhombuses is carried out in the same way as in the preceding proof. We then see that the point $M$ is located in a symmetric position with the point $B$ with respect to the line $CH$, we conclude that we have $\angle CMB=\beta$. Similarly, we have $\angle CNB=\gamma$. If we now put $x=\angle AH{O}^{\prime }$, then we get $$\angle O^{\prime }=\beta -\alpha -x,\angle MNH=90^{\circ }-\beta -\alpha +x,$$ from which it follows that $$\angle ANM=180^{\circ }-\angle MNH-(90^{\circ }-\alpha )=\beta -x.$$ Similarly, we get $$\angle NMA=\gamma +x.$$ Using the laws of sines, we then get $$\begin{array}{rl}\frac{\sin (\gamma +x)}{\sin (\beta -x)}& =\frac{AN}{AM}=\frac{AC}{AM}\cdot \frac{AB}{AC}\cdot \frac{AN}{AB}\\ & =\frac{\sin \beta }{\sin (\beta -\alpha )}\cdot \frac{\sin \gamma }{\sin \beta }\cdot \frac{\sin (\gamma -\alpha )}{\sin \gamma }=\frac{\sin (\gamma -\alpha )}{\sin (\beta -\alpha )}\end{array}$$ On the other hand, if we let $y=\angle AHO$, we then get $$\angle OHB=180^{\circ }-\gamma -y,\angle CHO=180^{\circ }-\beta +y,$$ and since $$\angle HBO=\gamma -\alpha ,\angle OCH=\beta -\alpha ,$$ using the laws of sines and observing that $OB=OC$, we get $$\begin{array}{rl}\frac{\sin (\gamma -\alpha )}{\sin (\beta -\alpha )}=\frac{\sin \angle HBO}{\sin \angle OCH}& =\frac{\sin (180^{\circ }-\gamma -y)\cdot \frac{OH}{OB}}{\sin (180^{\circ }-\beta +y)\cdot \frac{OH}{OC}}\\ & =\frac{\sin (180^{\circ }-\gamma -y)}{\sin (180^{\circ }-\beta +y)}=\frac{\sin (\gamma +y)}{\sin (\beta -y)}\end{array}$$ We then get $\sin (\gamma +x)\sin (\beta -y)=\sin (\beta -x)\sin (\gamma +y)$. Expanding both sides of the last identity by using the addition formula for the sine function and after factoring and using again the addition formula we obtain that $\sin (x-y)\sin (\beta +\gamma )=0$. This implies that $x-y$ must be an integral multiple of $180^{\circ }$, and hence we conclude that $H,O,O^{\prime }$ are collinear.