Let's select four vertices of a cube such that no two of them lie on the same edge. The distances from an internal point of the cube to these four vertices are $\sqrt{50}, \sqrt{70}, \sqrt{90}$, and $\sqrt{110}$. What is the length of the edge of the cube?
Let the edge of the cube be $a$, the internal point mentioned in the problem be $M$, and let's choose the labeling of the vertices such that $M A= \sqrt{50}, M B=\sqrt{70}, M C=\sqrt{90}, M D=\sqrt{110}$. Let the distances from the point $M$ to the three faces containing vertex $A$ be denoted by $x, y, z$ as shown in the figure.

# 1993-04-165-1.eps 

Based on the Pythagorean theorem,

$$
\begin{aligned}
x^{2}+y^{2}+z^{2} & =50 \\
(a-x)^{2}+(a-y)^{2}+z^{2} & =70 \\
(a-x)^{2}+y^{2}+(a-z)^{2} & =90 \\
x^{2}+(a-y)^{2}+(a-z)^{2} & =110
\end{aligned}
$$

By expanding the squares in the second, third, and fourth equations and substituting $50$ for $x^{2}+y^{2}+z^{2}$:

$$
\begin{aligned}
& 2 a(x+y)=2 a^{2}-20 \\
& 2 a(x+z)=2 a^{2}-40 \\
& 2 a(y+z)=2 a^{2}-60
\end{aligned}
$$

From this system of equations, we get $x=\frac{a}{2}, y=\frac{a}{2}-\frac{10}{a}, z=\frac{a}{2}-\frac{20}{a}$. Substituting these into the first equation:

$$
\left(\frac{a}{2}\right)^{2}+\left(\frac{a}{2}-\frac{20}{a}\right)^{2}+\left(\frac{a}{2}-\frac{10}{a}\right)^{2}=50
$$

This simplifies to

$$
\frac{3 a^{2}}{4}-80+\frac{500}{a^{2}}=0
$$

Letting $t=\frac{a^{2}}{4}$, we get the equation

$$
3 t^{2}-80 t+125=0
$$

The roots of this equation are $t=25$ and $t=\frac{5}{3}$, so $a=10$, or $a=\sqrt{\frac{20}{3}}$. In the second case, $y<0$ which means $M$ cannot be an internal point, so the only solution to the problem is: $a=10$ units.

Based on the work of Kristóf Matuszka (Szeged, Ságvári E. Gymnasium, 2nd grade)