4B. On three non-adjacent edges of a cube with edge length a, vertices of a triangle are chosen, different from the vertices of the cube, such that the sum of the squares of the sides of that triangle is the smallest. What is that sum?
Solution. On the edges $A A_{1}, B C$ and $C_{1} D_{1}$, we choose the points $X, Y$ and $Z$ respectively. We denote $\overline{A X}=x, \overline{B Y}=y, \overline{C_{1} Z}=z$, and $\overline{A B}=a$. Then:

$$
\begin{aligned}
\overline{X Y}^{2} & =\overline{X B}^{2}+\overline{B Y}^{2} \\
& =\overline{X A}^{2}+\overline{A B}^{2}+\overline{B Y}^{2} \\
& =x^{2}+a^{2}+y^{2} \\
\overline{Y Z}^{2} & =\overline{Y C}_{1}^{2}+{\overline{C_{1} Z}}^{2} \\
& =\overline{Y C}^{2}+{\overline{C C_{1}}}^{2}+{\overline{C_{1} Z}}^{2} \\
& =(a-y)^{2}+a^{2}+z^{2} \\
\overline{Z X}^{2} & =\overline{Z A}_{1}^{2}+{\overline{A_{1} X}}^{2} \\
& =\overline{Z D}_{1}^{2}+{\overline{D_{1} A_{1}}}^{2}+{\overline{A_{1} X}}^{2} \\
& =(a-z)^{2}+a^{2}+(a-x)^{2}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_e95f2dbd17df8adc2c7cg-07.jpg?height=426&width=466&top_left_y=1229&top_left_x=976)

By summing the last three equations, we get:

$$
\begin{aligned}
\overline{X Y}^{2}+\overline{Y Z}^{2}+\overline{Z X}^{2} & =2\left(x^{2}+y^{2}+z^{2}\right)-2 a(x+y+z)+6 a^{2} \\
& =2\left[\left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{a}{2}\right)^{2}+\left(z-\frac{a}{2}\right)^{2}+\frac{9}{4} a^{2}\right]
\end{aligned}
$$

This sum will be the smallest if $x=y=z=\frac{a}{2}$, that is, if the points $X, Y$ and $Z$ are the midpoints of the respective edges. Therefore, the smallest sum is $\frac{9}{4} a^{2}$.

## 4th Year