4. An acute triangle $A B C$ is given. For any point $L$ on its side $A B$, denote $K$, $M$ as the feet of the perpendiculars from point $L$ to the sides $A C, B C$. Determine for which position of point $L$ the segment $K M$ is the shortest.

The regional round of category B takes place

## on Tuesday, March 27, 2007

so that it starts in the morning and the participants have 4 hours of clean time to solve the problems. For each problem, a participant can earn 6 points, and a successful solver is a student who scores 10 points or more. Allowed aids are writing and drawing supplies, school MF tables, and calculators without a graphic display. This information will be communicated to the students before the start of the competition.
4. Since angles $L K C$ and $L M C$ are right angles, points $K$ and $M$ lie on the Thales' circle over the diameter $C L$ (Fig.2). According to the inscribed angle theorem, the central angle corresponding to the chord $K M$ has a measure of $2 \gamma$, so $|K M|=|C L| \sin \gamma$ (in the right triangle $K P S$, where $P$ is the midpoint of segment $K M$ and $S$ is the midpoint of segment $C L$, it is indeed $\left.|K S|=\frac{1}{2}|C L|,|\angle K S P|=\gamma\right)$. The segment $K M$ is thus the shortest when the segment $C L$ is the shortest; this occurs precisely when $L$ is the foot of the altitude from vertex $C$ to side $A B$.

![](https://cdn.mathpix.com/cropped/2024_04_17_fb1bfc1f5e30d8251ccdg-3.jpg?height=637&width=648&top_left_y=904&top_left_x=704)

Fig. 2

For a complete solution, award 6 points. Give 2 points for noting that points $K$ and $M$ lie on the Thales' circle over the diameter $C L$, 2 points for proving the equality $|K M|=|C L| \sin \gamma$, and 2 points for determining the position of point $L$.