Problem 120. Find the point for which the sum of the squares of the distances to the four vertices of a tetrahedron is minimal.
Solution. Since the center of mass of the given tetrahedron is fixed, according to problem 119, the expression $T A^{2}+T B^{2}+T C^{2}+T D^{2}$ will be minimal when $T M=0$, or $T \equiv M$.

Answer. This point is the center of mass of the tetrahedron.

We will also demonstrate the application of the method of masses in solving a series of classical geometry problems. In this context, we will note another important fact about the method of masses, which is very useful in solving problems:

Let $K$ be an arbitrary point inside the triangle $A B C$. It is always possible to choose three masses at the vertices of the triangle such that the point $K$ is the center of mass of the system of three resulting material points. (Prove this fact independently!)