XV OM - I - Problem 8

On three pairwise skew edges of a cube, choose one point on each in such a way that the sum of the squares of the sides of the triangle formed by them is minimized.
Let $ABCD$ be the base of a cube and $AA_1$, $BB_1$, $CC_1$, $DD_1$ its edges perpendicular to $ABCD$. Let $X$, $Y$, $Z$ be points chosen on three pairwise skew edges of the cube. It is known that by rotating the cube around certain axes passing through its center, the cube can be superimposed onto itself in such a way that any chosen edge can cover any other edge. Therefore, without loss of generality, we can assume that point $X$ lies on edge $A_1$. The skew edges to $AA_1$ are $BC$, $CD$, $B_1C_1$, $C_1D_1$. (Fig. 8); among them, the mutually skew ones are $BC$ and $C_1D_1$ as well as $B_1C_1$ and $CD$. Thus, we have two sets of three pairwise skew edges: $AA_1$, $BC$, $C_1D_1$ and $AA_1$, $B_1C_1$, $CD$, which transform into each other when the cube is reflected through the plane $AA_1C_1C$, which is a plane of symmetry of the cube. Therefore, we can assume that the considered set of three pairwise skew edges is $AA_1$, $BC$, $C_1D_1$ and that, for example, point $Y$ lies on $BC$ and point $Z$ on $C_1D_1$. Let $AX = x$, $BY = y$, $C_1Z = z$, $AA_1 = a$. Then


















from the obtained formula, it is clear that the sum of the squares of the sides of triangle $XYZ$ reaches a minimum when $x = y = z = \frac{a}{2}$, i.e., when points $X$, $Y$, $Z$ are chosen at the midpoints of the respective edges. The minimum value is $\frac{9}{2} a^2$.