500. In the plane of a triangle, find a point for which the sum of the squares of the distances from the sides of the triangle is minimal.

Problems from "Journal de Mathématiques elementaires," by M. Vuibert.
500. Let $x, y, z$ be the distances from an arbitrary point (in the plane of the triangle) to its sides $a, b$, and $c$. We will use the Lagrange identity:

$$
\begin{gathered}
\left(x^{2}+y^{2}+z^{2}\right)\left(a^{2}+b^{2}+c^{2}\right)=(a x+b y+c z)^{2}+(a y-b x)^{2}+ \\
+(b z-c y)^{2}+(c x-a z)^{2}
\end{gathered}
$$

Since $a x+b y+c z=2 S$, where $S$ is the area of the given triangle, $x^{2}+y^{2}+z^{2}$ will be minimum, as it is easy to see, under the condition

$$
\begin{aligned}
& a y-b x=0 \\
& b z-c y=0 \\
& c x-a z=0
\end{aligned}
$$

i.e., when

$$
\frac{x}{a}=\frac{y}{b}=\frac{z}{c}
$$

Note. The point defined by these conditions is located at the intersection of the symmedians of the triangle and is called the Lemoine point.