A unit-edged parallelepiped has a base $A_{1} A_{2} A_{3} A_{4}$ and a top face $B_{1} B_{2} B_{3} B_{4}$ such that the vertices $A_{i}$ and $B_{i}$ are connected by an edge. Within what bounds does the sum of squares $A_{1} B_{2}^{2}+A_{2} B_{3}^{2}+A_{3} B_{4}^{2}+A_{4} B_{1}^{2}$ vary?
![](https://cdn.mathpix.com/cropped/2024_05_02_f7c18a33187b574dbf20g-1.jpg?height=361&width=569&top_left_y=123&top_left_x=775)

Solution. Every face of the parallelepiped is a parallelogram, and opposite faces are congruent. We also know that in a parallelogram, the sum of the squares of the sides is equal to the sum of the squares of the diagonals. The length of each edge is 1. Since

$$
B_{1} A_{2}=B_{4} A_{3} \quad \text{and} \quad A_{1} B_{4}=A_{2} B_{3}
$$

it follows that

$$
A_{3} B_{4}^{2}+A_{1} B_{2}^{2}=B_{1} A_{2}^{2}+A_{1} B_{2}^{2}=A_{1} A_{2}^{2}+A_{2} B_{2}^{2}+B_{2} B_{1}^{2}+B_{1} A_{1}^{2}=4
$$

and

$$
A_{2} B_{3}^{2}+A_{4} B_{1}^{2}=A_{1} B_{4}^{2}+A_{4} B_{1}^{2}=4
$$

Thus, \(A_{1} B_{2}^{2}+A_{2} B_{3}^{2}+A_{3} B_{4}^{2}+A_{4} B_{1}^{2}=8\), meaning that regardless of the skew angle of the parallelepiped's edges, the sum of the squares in question is always 8.