$A B C$ is a triangle, and we take a point $D$ inside it, whose distances from the vertices of the triangle are $p, q, r$. What is the distance of this point from the centroid of the triangle, if the sides of the triangle are $a, b, c$.
Let $S$ be the centroid of the triangle, the median corresponding to side $a$ is $A A_{1}=K_{1}, D S A_{1} \angle=\varepsilon$; from triangle $B D C$ (1st K. M. L. 204. problem)

$$
{\overline{D A_{1}}}^{2}=\frac{2 q^{2}+2 r^{2}-a^{2}}{4}
$$

From triangle $D S A$:

$$
p^{2}=\overline{D S}^{2}+\frac{4 K_{1}^{2}}{9}+\frac{4 D S \cdot K_{1}}{3} \cos \varepsilon
$$

From triangle $D S A_{1}$:

$$
2{\overline{D A_{1}}}^{2}=2 \overline{D S}^{2}+\frac{2 K_{1}^{2}}{9}+\frac{4 D S \cdot K_{1}}{3} \cos \varepsilon
$$

Adding (2) and (3):

$$
p^{2}+2{\overline{D A_{1}}}^{2}=3 \overline{D S}^{2}+\frac{2}{3} K_{1}^{2}
$$

Substituting the value of ${\overline{D A_{1}}}^{2}$ from (1) into (4) and replacing $K_{1}^{2}$ with the equal value $\frac{2 b^{2}+2 c^{2}-a^{2}}{4}$, we obtain:

$$
9 \overline{D S}^{2}=3\left(p^{2}+q^{2}+r^{2}\right)-\left(a^{2}+b^{2}+c^{2}\right)
$$

(Odön Kornis, Pécs.)

Note. From this equation, we can see that the sum of the squares of the distances $p, q, r$ is minimized when $D S=0$, i.e., when:

$$
p^{2}+q^{2}+r^{2}=\frac{a^{2}+b^{2}+c^{2}}{3}
$$

In this case, $D$ is the centroid of the triangle. Thus, we have found a point - the centroid - for which the sum of the squares of the distances from the vertices is minimal.

Furthermore, we can see that if we draw a circle around the centroid, the sum of the squares of the distances from the vertices to any point on the circumference of the circle is constant.

The problem was also solved by. Dénes A., Eislitzer Gy., Friedmann B., Goldstein Zs., Grünhut B., Hofbauer E., Kántor N., Klein M., Mihálkovics E., Prakatur T., Riesz F., Schiffer H., Spitzer Ö., Szabó I., Szabó K., Weisz L.