Macedonian Mathematical Olympiad

First, we will show the following lemma.

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

Proof.

First proof of the lemma. Let $N$ be the intersection point of $H{H}^{\prime }$ and $AB$, and $A_1$ be the foot of the height from $A$ towards $BC$. Then $△ANH\cong △AN{H}^{\prime }$, since $AN$ is a common side, $\overline{HN}=\overline{N{H}^{\prime }}$ and $\angle ANH=\angle AN{H}^{\prime }=90^{\circ }$. From $\angle H^{\prime }AB=\angle H^{\prime }AN=\angle NAH=\angle BA{A}_1=90^{\circ }-\angle ABC=\angle NCB=\angle H^{\prime }CB$ it follows that the quadrilateral $A{H}^{\prime }BC$ is cyclic, i.e. $H^{\prime }$ lies on the circle around the 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 $\angle H{B}_{1}C=\angle H{A}_{1}C=90^{\circ }$ it follows that the quadrilateral $B_{1}H{A}_{1}C$ is cyclic. Then $\angle A_{1}H{B}_1=180^{\circ }-\gamma =\alpha +\beta$, i.e. $\angle AHB=\angle A_{1}H{B}_1=\alpha +\beta$. From $△ANH\cong △AN{H}^{\prime }$ and $△BNH\cong △BN{H}^{\prime }$ it follows that $\angle A{H}^{\prime }B=\angle AHB=\alpha +\beta$. Then from $$\angle A{H}^{\prime }B+\angle ACB=\alpha +\beta +\gamma =180^{\circ }$$ It follows that the quadrilateral $A{H}^{\prime }BC$ is cyclic, i.e. $H^{\prime }$ lies on the circle around the triangle $△ABC$.

From $\angle AN{H}^{\prime }=90^{\circ }$ it follows that $A{H}^{\prime }$ is a diameter of the circle around the triangle $△AN{H}^{\prime }$. Then $\angle H^{\prime }MA=90^{\circ }$, i.e.



$H^{\prime }M\perp AC$.\quad (1) $$Analogously,wecanshowthat$$ H'P \perp BC.\quad (2) $$Fromtheconditionoftheproblemwehave$$ 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. --- **Alternative solution.** 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 (*) $$From(\ast )and(\ast \ast )wehave$$ \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.