Problem 4. In triangle $P Q R$, the lengths of the sides are 4, 7, and 9. Find the area of the figure consisting of those and only those points $M$ inside triangle $P Q R$ for which the condition $M P^{2}+$ $M Q^{2}+M R^{2} \leqslant 50$ is satisfied.
Answer: $4 \pi / 9$.

Solution. Let $Q R=a, P R=b, P Q=c, \rho^{2}=50$.

Let $G$ be the point of intersection of the medians of triangle $P Q R$. Represent

$$
\overrightarrow{M P}=\overrightarrow{G P}-\overrightarrow{G M}, \overrightarrow{M Q}=\overrightarrow{G Q}-\overrightarrow{G M}, \overrightarrow{M R}=\overrightarrow{G R}-\overrightarrow{G M}
$$

then

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

Since $G$ is the centroid of triangle $P Q R$, then

$$
\overrightarrow{G P}+\overrightarrow{G Q}+\overrightarrow{G R}=0
$$

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

Thus

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

or

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

Therefore, the geometric locus of points $M$ satisfying 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 $P Q R$.

This circle belongs to the triangle if its radius is no greater than one third of the smallest height of $\triangle P Q R$:

$$
\frac{1}{3} \sqrt{3 \rho^{2}-a^{2}-b^{2}-c^{2}} \leqslant \frac{2 S_{\triangle P Q R}}{3 \max \{a, b, c\}}
$$

Thus, under the condition

$$
\frac{a^{2}+b^{2}+c^{2}}{3}\frac{2}{3}$. Therefore, the entire circle is the figure whose area is to be found, from which $S=\pi R^{2}=\frac{4 \pi}{9}$.