4B. Let $M$ be a point on the side $AB$ of triangle $ABC$. Express the distance between the orthocenters of triangles $AMC$ and $BMC$ in terms of $\overline{AB}=c$ and $\measuredangle CMA=\varphi$.
Solution. Let $H$ and $H_{1}$ be the orthocenters of triangles AMC and BMC, respectively (diagram). From the similarity of the right triangles $A D H$ and $A E M$, it follows that $\measuredangle A H D = \measuredangle A M E = \varphi$. Therefore, $\overline{H D} = \overline{A D} \cdot \operatorname{ctg} \varphi$. From the similarity of the right triangles $B F M$ and $B D H_{1}$, it follows that $\measuredangle B H_{1} D = \varphi$, hence $\overline{D H_{1}} = \overline{D B} \cdot \operatorname{ctg} \varphi$. Therefore:

![](https://cdn.mathpix.com/cropped/2024_06_05_84a5520b68919fd51f4cg-06.jpg?height=421&width=509&top_left_y=443&top_left_x=971)

$\overline{H H_{1}} = \overline{H D} + \overline{D H_{1}} = (\overline{A D} + \overline{D B}) \operatorname{ctg} \varphi = c \cdot \operatorname{ctg} \varphi$.

## III year