Folklore

In a cube with an edge length of 1, two sections in the form of regular hexagons were made. Find the length of the segment where these sections intersect.
Let $P Q K L M N$ be a section of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, which is a regular hexagon (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_476f7e28c93a861c2173g-28.jpg?height=626&width=712&top_left_y=990&top_left_x=707)

Let the lines $P N$ and $K Q$ intersect at point $T$ on the line $A D$. Clearly, triangle $K T N$ is isosceles. Therefore, $T P = T Q$, and the right triangles $A P T$ and $A Q T$ are congruent by the leg and hypotenuse. Hence, triangle $P A Q$ is an isosceles right triangle. The same is true for triangle $K B Q$. These triangles are congruent by the hypotenuse, so $A Q = B Q$. Similarly, it can be shown that the other vertices of the section are also midpoints of the edges of the cube.

The center of the section is the midpoint of its diagonal $K N$, which coincides with the midpoint of the diagonal $B D_{1}$ of the cube, that is, with the center of the cube.

Note that the two sections mentioned in the problem condition have a common segment connecting the midpoints of opposite edges of the cube. Indeed, the cube has 12 edges. If the midpoints of six of them are the vertices of the section, then no more than four of the remaining six points lie in the same plane. Therefore, the two such sections have two common vertices, which lie on opposite edges of the cube. Since both sections are symmetric with respect to the center of the cube, these vertices lie on opposite edges of the cube. Consequently, the sections intersect along a segment connecting the midpoints of these edges. The length of such a segment is equal to the diagonal of a face of the cube, which is $\sqrt{2}$.

## Answer