Problem 4. In triangle $K L M$, the lengths of the sides are $8, 3 \sqrt{17}$, and 13. Find the area of the figure consisting of those and only those points $P$ inside triangle $K L M$ for which the condition $P K^{2}+P L^{2}+P M^{2} \leqslant 145$ is satisfied.
Answer: $49 \pi / 9$.

Solution. Let $L M=a, K M=b, K L=c, \rho^{2}=145$.

Let $G$ be the point of intersection of the medians of triangle $K L M$. Represent

$$
\overrightarrow{P K}=\overrightarrow{G K}-\overrightarrow{G P}, \overrightarrow{P L}=\overrightarrow{G L}-\overrightarrow{G P}, \overrightarrow{P M}=\overrightarrow{G M}-\overrightarrow{G P}
$$

then

$$
P K^{2}+P L^{2}+P M^{2}=G K^{2}+G L^{2}+G M^{2}-2 \cdot \overrightarrow{G P} \cdot(\overrightarrow{G K}+\overrightarrow{G L}+\overrightarrow{G M})+3 \cdot G P^{2}
$$

Since $G$ is the centroid of triangle $K L M$, then

$$
\overrightarrow{G K}+\overrightarrow{G L}+\overrightarrow{G M}=0
$$

and $G K^{2}+G L^{2}+G M^{2}=\frac{4}{9}\left(m_{a}^{2}+m_{b}^{2}+m_{c}^{2}\right)=\frac{1}{3}\left(a^{2}+b^{2}+c^{2}\right)$.

Thus,

$$
\frac{1}{3}\left(a^{2}+b^{2}+c^{2}\right)+3 \cdot G P^{2} \leqslant \rho^{2}
$$

or

$$
G P^{2} \leqslant \frac{1}{9} \cdot\left(3 \rho^{2}-a^{2}-b^{2}-c^{2}\right)
$$

Therefore, the geometric locus of points $P$ that satisfy the given condition is a circle with radius $\frac{1}{3} \sqrt{3 \rho^{2}-a^{2}-b^{2}-c^{2}}$ centered at the point of intersection of the medians of triangle $K L M$.

This circle belongs to the triangle if its radius is no greater than one third of the smallest height of $\triangle K L M$:

$$
\frac{1}{3} \sqrt{3 \rho^{2}-a^{2}-b^{2}-c^{2}} \leqslant \frac{2 S_{\Delta K L M}}{3 \max \{a, b, c\}}
$$

Thus, under the condition

$$
\frac{a^{2}+b^{2}+c^{2}}{3}\frac{7}{3}$. Therefore, the entire circle is the figure whose area is to be found, from which $S=\pi R^{2}=\frac{49 \pi}{9}$.