Among triangles $K L M$, where the radius of the circumscribed circle is 10 cm, side $K L$ is 16 cm, and height $M H$ is $\frac{39}{10}$ cm, find the angle $K M L$ of the triangle for which the median $M N$ is the smallest.
Let $M K=a, M L=b, K L=c, M N=m, \angle K M L=\alpha$. Then

$$
m^{2}=\frac{1}{4}\left(2 a^{2}+2 b^{2}-c^{2}\right)
$$

Find $\sin \alpha$, consider the cases of obtuse and acute angle $K M L$ and express $a^{2}+b^{2}$ in terms of known quantities using the cosine theorem.

## Solution

Let $M K=a, M L=b, K L=c, M N=m, \angle K M L=\alpha$.

If $R=10$ is the radius of the circumcircle of triangle $K L M$, then

$$
\sin \alpha=\sin \angle K M L=\frac{K L}{2 R}=\frac{16}{20}=\frac{4}{5}
$$

By writing the area of triangle $K L M$ in two ways, we get the equation

$$
\frac{1}{2} a b \sin \alpha=\frac{1}{2} K L \cdot M H, \text{ or } a b \cdot \frac{4}{5}=16 \cdot \frac{39}{10},
$$

from which we find that $a b=78$.

By the cosine theorem,

$$
a^{2}+b^{2}-2 a b \cos \alpha=256
$$

Therefore,

$$
a^{2}+b^{2}=256+2 a b \cos \alpha=256+2 \cdot 78 \cdot \cos \alpha
$$

From the formula for the median of a triangle, it follows that

$$
\begin{gathered}
M N^{2}=m^{2}=\frac{1}{4}\left(2 a^{2}+2 b^{2}-c^{2}\right)=\frac{1}{2}\left(a^{2}+b^{2}-\frac{c^{2}}{2}\right)= \\
=\frac{1}{2}\left(a^{2}+b^{2}-128\right)=\frac{1}{2}(256+2 \cdot 78 \cdot \cos \alpha-128)=64+78 \cdot \cos \alpha
\end{gathered}
$$

Therefore, the median $M N$ is the smallest if $\cos \alpha$ is the smallest.

Since $\sin \alpha=\frac{4}{5}$, $\cos \alpha$ is the smallest when angle $K M L$ is obtuse, i.e., when $\cos \alpha=-\frac{3}{5}$.

It remains to verify that the obtained triangle satisfies all the conditions of the problem. For this, it is sufficient to show that its height $M H$ is less than the distance from the midpoint $N$ of side $K L$ to the midpoint $P$ of the arc $K L$ containing point $M$.

Indeed,

$$
\operatorname{tg} \angle N P L=\operatorname{tg} \frac{1}{2} \angle K P L=\operatorname{tg} \frac{1}{2} \angle K M L=\operatorname{tg} \frac{\alpha}{2}=\frac{\sin \alpha}{1+\cos \alpha}=\frac{\frac{4}{5}}{1-\frac{3}{5}}=2
$$

Therefore,

$$
P N=\frac{N L}{\operatorname{tg} \frac{\alpha}{2}}=\frac{8}{2}=4>\frac{39}{10}=M H
$$

Which is what we needed to prove.

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-41.jpg?height=783&width=780&top_left_y=2144&top_left_x=642)

## Answer

$\angle K M L=\arccos \left(-\frac{3}{5}\right)=\pi-\arcsin \frac{4}{5}$.