Macedonian Mathematical Olympiad

First, we will show the following lemma.

Lemma. Let $H$ be an orthocenter in $\triangle ABC$, and $H'$ be the symmetric point of $H$ with respect to $AB$. Then $H'$ lies on the circle around the triangle $\triangle ABC$.

Proof.

First proof of the lemma. Let $N$ be the intersection point of $HH'$ and $AB$, and $A_1$ be the foot of the height from $A$ towards $BC$. Then $\triangle ANH \cong \triangle ANH'$, since $AN$ is a common side, $\overline{HN} = \overline{NH'}$ and $\angle ANH = \angle ANH' = 90^\circ$. From $\angle H'AB = \angle H'AN = \angle NAH = \angle BAA_1 = 90^\circ - \angle ABC = \angle NCB = \angle H'CB$ it follows that the quadrilateral $AH'BC$ is cyclic, i.e. $H'$ lies on the circle around the triangle $\triangle ABC$.

Second proof of the lemma. Let $A_1$ be the foot of the height from $A$ towards $BC$, and $B_1$ be the foot of the height from $B$ towards $AC$. From $\angle HB_1C = \angle HA_1C = 90^\circ$ it follows that the quadrilateral $B_1HA_1C$ is cyclic. Then $\angle A_1HB_1 = 180^\circ - \gamma = \alpha + \beta$, i.e. $\angle AHB = \angle A_1HB_1 = \alpha + \beta$. From $\triangle ANH \cong \triangle ANH'$ and $\triangle BNH \cong \triangle BNH'$ it follows that $\angle AH'B = \angle AHB = \alpha + \beta$. Then from
$$\angle AH'B + \angle ACB = \alpha + \beta + \gamma = 180^\circ$$
It follows that the quadrilateral $AH'BC$ is cyclic, i.e. $H'$ lies on the circle around the triangle $\triangle ABC$.

From $\angle ANH' = 90^\circ$ it follows that $AH'$ is a diameter of the circle around the triangle $\triangle ANH'$. Then $\angle H'MA = 90^\circ$, i.e.

$H'M \perp AC$$.\quad$ (1)
$$Analogously, we can show that$$
H'P $\perp BC.\quad$ (2)
From the condition of the problem we have
H'N $\perp AB.\quad$ (3)
$$$$
Using the lemma, we conclude that the quadrilateral $AH'BC$ is cyclic. From (1), (2), (3) and the theorem of Simpson for $\triangle ABC$ and having in mind that $H'$ lies on the circle around the triangle $\triangle ABC$, it follows the statement of the problem.

Using the power of point theorem, we have
$$\overline{CA} \cdot \overline{CM} = \overline{CN} \cdot \overline{CH'} = \overline{CP} \cdot \overline{CB}, \text{ i.e. } \overline{CA} \cdot \overline{CM} = \overline{CP} \cdot \overline{CB}$$
$$\frac{\overline{CM}}{\overline{CP}} = \frac{\overline{CB}}{\overline{CA}}.\quad (1)$$

From the lemma it follows that the quadrilateral $AH'BC$ is cyclic. From $\angle NAC = \alpha$ and $\angle ANC = 90^\circ$ it follows $\angle ACN = 90^\circ - \alpha$. From $\angle ANH' = 90^\circ$ it follows that $AH'$ is a diameter of the circle around the quadrilateral $AMH'N$. Then $\angle AMH' = \angle ANH' = 90^\circ$. From the last and from $\angle MCH' = \angle ACN = 90^\circ - \alpha$, it follows that $\triangle ANC \sim \triangle H'MC$. Then $\frac{AN}{AC} = \frac{H'M}{H'C}$, i.e.
$$\overline{AN} = \frac{\overline{H'M} \cdot \overline{AC}}{\overline{H'C}}.\quad (*)$$
Similarly, we have that $\triangle BNC \sim \triangle H'PC$, i.e.
$$\frac{BN}{BC} = \frac{H'P}{H'C}, \text{ so }$$
$$\overline{BN} = \frac{\overline{H'P} \cdot \overline{BC}}{\overline{H'C}}.\quad (**)$$

From () and (*) we have

$$\frac{\overline{AN}}{BN} = \frac{\frac{\overline{H' M} \cdot \overline{AC}}{\overline{H' C}}}{\frac{\overline{H' P} \cdot \overline{BC}}{\overline{H' C}}} = \frac{\overline{H' M} \cdot \overline{AC}}{\overline{H' P} \cdot \overline{BC}}.\qquad (2)$$
The quadrilateral $AH'BC$ is cyclic, so
$$\angle AH'C = \angle ABC = \beta \quad \text{and} \quad \angle BH'C = \angle BAC = \alpha.$$
Then $\angle H'BN = \angle H'BA = \angle H'CA = 90^\circ - \alpha$. Hence
$$\angle H'BP = \angle H'BN + \angle NBP = 90^\circ - \alpha + \beta = 90^\circ - (\alpha - \beta) \quad \text{ i.e. } \angle PH'B = \alpha - \beta.$$
From $\angle MH'C = \alpha$ it follows
$$\angle MH'A = \angle MH'N - \angle AH'N = \angle MH'N - \angle AH'C = \alpha - \beta \quad \text{ i.e. } \angle MAH' = 90^\circ - (\alpha - \beta).$$
It follows $\triangle H'MA \sim \triangle H'PB$, i.e. $\frac{\overline{H' M}}{MA} = \frac{\overline{H' P}}{PB}$ or
$$\frac{PB}{MA} = \frac{H'P}{H'M}.\qquad (3)$$

From (1), (2) and (3) and Menelaus' theorem for the points $M$, $N$ and $P$ and $\triangle ABC$ we have
$$\frac{\overline{CM}}{MA} \cdot \frac{\overline{AN}}{NB} \cdot \frac{\overline{BP}}{PC} = \frac{\overline{CM}}{PC} \cdot \frac{\overline{AN}}{NB} \cdot \frac{\overline{BP}}{MA} = \frac{\overline{CB}}{CA} \cdot \frac{\overline{H' M} \cdot \overline{AC}}{\overline{H' P} \cdot \overline{BC}} \cdot \frac{\overline{H' P}}{H'M} = 1,$$
i.e. $M$, $N$ and $P$ are colinear.