A cube has two skew edges on which a unit-length segment moves. In which position of the segments will the volume of the tetrahedron determined by the endpoints of the segments be maximal?
Let the edge of the cube be of length $a$, and let two skew edges of the cube be $A B$ and $C D$ (the points $B$ and $C$ are directly above each other). Furthermore, let the two unit-length segments be $P Q$ (on the edge $A B$) and $X Y$ (on the edge $C D$). Let's examine what happens when we move the segment $P Q$ along $A B$ and fix $X Y$. The height of the triangle $P Q X$ is the distance from the line $P Q$ to the point $X$, which is the same as the distance from the line $A B$ to the point $X$, so it is constant. The base corresponding to this height is clearly constant (one unit), so the area of the triangle $P Q X$ is constant. The triangle $P Q X$ is always in the same plane (the plane $A B X$), and since the height of the tetrahedron $P Q X Y$ is the distance from this plane to the point $Y$, the height of the tetrahedron is constant. We have already shown that the base area of the tetrahedron does not change, so the volume of the tetrahedron is constant.

We can proceed similarly if we fix $P Q$ and move $X Y$. Therefore, starting from any position of the two segments, by first moving $P Q$ and then $X Y$, we can arrive at a tetrahedron of the same volume, where $P$ coincides with $B$ and $X$ coincides with $C$. The base area of the tetrahedron $P Q X$ is $a / 2$ and the height is 1 (the length of the segment $X Y$). This implies that the volume of the tetrahedron is $a / 6$ for any position of the segments.

Gábor Székelyhidi (Kuwait, New English School, 11th grade)

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