Asian Pacific Mathematics Olympiad

In the sequel, we denote $\angle BAC = \alpha$, $\angle CBA = \beta$, $\angle ACB = \gamma$. Let $O'$ 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 $M', N'$ 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 $\angle MHM' = \alpha$ holds. Furthermore, $\angle BM'C$, $\angle BN'C$ and $\alpha$ are all subtended by the same arc $\text{BC}$ of the circumcircle of the triangle $ABC$ at points on the circle, and therefore, we have $\angle BM'C = \alpha$, and $\angle BN'C = \alpha$ as well. We also have $\angle ABH = \angle ACN'$ as they are subtended by the same arc $\text{AN'}$ of the circumcircle of the triangle $ABC$ at points on the circle. Since $HM' \perp BM$, $HN' \perp AC$, we conclude that
$$\angle M'HB = 90^\circ - \angle ABH = 90^\circ - \angle ACN' = \alpha$$
is valid as well. Putting these facts together, we obtain the fact that the quadrilateral $HBM'M$ is a rhombus. In a similar manner, we can conclude that the quadrilateral $HCN'N$ 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 $O'$. Since $O'$ 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 $M'BC$ and the triangle $N'BC$, we see that $R$ is the intersection point of $HM$ and the perpendicular bisector of $BM'$, and $S$ is the intersection point of $HN$ and the perpendicular bisector of $CN'$.
We note that the similarity map $\phi$ between the rhombuses $HBM'M$ and $HCN'N$ carries the perpendicular bisector of $BM'$ onto the perpendicular bisector of $CN'$, and straight line $HM$ onto the straight line $HN$, and hence $\phi$ 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 $HO'$ with the line through $R$ and perpendicular to $HP$, and with the line through $S$ and perpendicular to $HQ$, respectively, then we get
$$HO' : HX = HP : HR = HQ : HS = HO' : HY$$
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 $H, O', O$ are collinear.

Alternate Solution:

Deduction of the fact that both of the quadrilaterals $HBM'M$ and $HCN'N$ 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 $x = \angle AHO'$, then we get
$$\angle O' = \beta - \alpha - x, \quad \angle MNH = 90^\circ - \beta - \alpha + x,$$
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
$$\begin{aligned} \frac{\sin (\gamma + x)}{\sin (\beta - x)} & = \frac{AN}{AM} = \frac{AC}{AM} \cdot \frac{AB}{AC} \cdot \frac{AN}{AB} \\ & = \frac{\sin \beta}{\sin (\beta - \alpha)} \cdot \frac{\sin \gamma}{\sin \beta} \cdot \frac{\sin (\gamma - \alpha)}{\sin \gamma} = \frac{\sin (\gamma - \alpha)}{\sin (\beta - \alpha)} \end{aligned}$$
On the other hand, if we let $y = \angle AHO$, we then get
$$\angle OHB = 180^\circ - \gamma - y, \quad \angle CHO = 180^\circ - \beta + y,$$
and since
$$\angle HBO = \gamma - \alpha, \angle OCH = \beta - \alpha,$$
using the laws of sines and observing that $OB = OC$, we get
$$\begin{aligned} \frac{\sin (\gamma - \alpha)}{\sin (\beta - \alpha)} = \frac{\sin \angle HBO}{\sin \angle OCH} & = \frac{\sin (180^\circ - \gamma - y) \cdot \frac{OH}{OB}}{\sin (180^\circ - \beta + y) \cdot \frac{OH}{OC}} \\ & = \frac{\sin (180^\circ - \gamma - y)}{\sin (180^\circ - \beta + y)} = \frac{\sin (\gamma + y)}{\sin (\beta - y)} \end{aligned}$$
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 $H, O, O'$ are collinear.