}

Car: Sharvani i.f.

Side $AB$ of triangle $ABC$ is equal to $c$. A point $M$ is taken on side $AB$ such that $\angle CMA = \varphi$.

Find the distance between the orthocenters of triangles $AMC$ and $BMC$.

#
Let $D_{a}$ and $D_{b}$ be the projections of the orthocenters $H_{a}$ and $H_{b}$ of triangles $A M C$ and $B M C$ onto the line $C M$ (see figure).

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![](https://cdn.mathpix.com/cropped/2024_05_06_cdc67dfeb844af03d94fg-46.jpg?height=657&width=791&top_left_y=0&top_left_x=662)

The points $H_{a}$ and $H_{b}$ lie on the perpendicular dropped from $C$ to the line $A B$. The segment $D_{a} D_{b}$ is the projection of the segment $A B$ onto the line $C M$, and since the angle between the lines $A B$ and $C M$ is $\varphi$ or $180^{\circ}-\varphi$, we have $D_{a} D_{b}=A B|\cos \varphi|$ $=c|\cos \varphi|$.

On the other hand, the segment $D_{a} D_{b}$ is the projection of the segment $H_{a} H_{b}$ onto the line $C M$, and since the angle between the lines $H_{a} H_{b}$ and $C M$ is $90^{\circ}-\varphi$ or $90^{\circ}+\varphi$, we have $H_{a} H_{b}=\frac{D_{a} D_{b}}{\sin \varphi}=\frac{c|\cos \varphi|}{\sin \varphi}=c|\operatorname{ctg} \varphi|$.

## Answer

$c|\operatorname{ctg} \varphi|$.