Problem 4. In triangle $A B C$, the lengths of the sides are 4, 5, and $\sqrt{17}$. Find the area of the figure consisting of those and only those points $X$ inside triangle $A B C$ for which the condition $X A^{2}+X B^{2}+X C^{2} \leqslant 21$ is satisfied.
Answer: $5 \pi / 9$.

Solution. Let $B C=a, A C=b, A B=c, \rho^{2}=21$.

Let $G$ be the point of intersection of the medians of triangle $A B C$. Represent

$$
\overrightarrow{X A}=\overrightarrow{G A}-\overrightarrow{G X}, \overrightarrow{X B}=\overrightarrow{G B}-\overrightarrow{G X}, \overrightarrow{X C}=\overrightarrow{G C}-\overrightarrow{G X}
$$

then

$$
X A^{2}+X B^{2}+X C^{2}=G A^{2}+G B^{2}+G C^{2}-2 \cdot \overrightarrow{G X} \cdot(\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C})+3 \cdot G X^{2}
$$

Since $G$ is the centroid of triangle $A B C$, then

$$
\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C}=0
$$

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

Thus,

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

or

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

Therefore, the geometric locus of points $X$ 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 $A B C$.

This circle belongs to the triangle if its radius is no greater than one third of the smallest height of $\triangle A B C$:

$$
\frac{1}{3} \sqrt{3 \rho^{2}-a^{2}-b^{2}-c^{2}} \leqslant \frac{2 S_{\triangle A B C}}{3 \max \{a, b, c\}}
$$

Thus, under the condition

$$
\frac{a^{2}+b^{2}+c^{2}}{3}\frac{\sqrt{5}}{3}$. Therefore, the entire circle is the figure whose area is to be found, from which $S=\pi R^{2}=\frac{5 \pi}{9}$.