On the line containing side $A B$ of an acute-angled triangle $A B C$, construct a point $M$ such that the distance between its projections onto the lines $A C$ and $B C$ is minimal. What is this distance?
Point $M$, its projections on lines $A C$ and $B C$, and vertex $C$ lie on the same circle.

## Solution

Let $P$ and $Q$ be the projections of the point $M$ on the lines $A C$ and $B C$. Then the points $P, M, Q$, and $C$ lie on a circle with diameter $M C$. Therefore, $P Q = M C \sin \angle A C B$. Consequently, $P Q$ is minimized when $M C$ is minimized, i.e., when $M$ is the foot of the altitude of triangle $A B C$ drawn from vertex $A$.

If $h$ is the length of this altitude, then the minimum distance between the projections of the point $M$ is $h \sin \angle A C B$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-37.jpg?height=537&width=712&top_left_y=1132&top_left_x=679)

Problem