Consider the planes passing through the space diagonals of a unit cube. These planes intersect the cube in a polygon. Within what bounds does the area of such intersection polygons vary?
Let the endpoints of the body diagonal in question be $A$ and $G$. These two points lie on any of the intersecting planes, just as on the perimeter of each face of the cube. Thus, each side of the intersection polygon $M$ has one of the endpoints $A$ or $G$, which is only possible if $M$ is a quadrilateral with $A$ and $G$ as opposite vertices. (Since the combination of the cube and the intersecting plane is centrally symmetric, $M$ is actually a parallelogram.) Let $P$ and $Q$ be the other two opposite vertices of $M$.

# $1987-10-312-1 . \mathrm{eps}$ 

Figure 1

Consider the projection of the cube's edges and the projection of $M$ onto a plane perpendicular to $A G$ (Figure 1). As is known, the projection of the edge network is a regular hexagon with side length $\sqrt{6} / 3$ and its three main diagonals. The projection of $M$ is then a segment $P^{\prime} Q^{\prime}$ passing through the center of the above hexagon, which is the projection of the diagonal $P Q$ of the quadrilateral.

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Figure 2

Due to the orthogonal projection, $P^{\prime} A$ and $Q^{\prime} A$ are equal to the altitudes corresponding to the common side $A G$ in the triangles $A P G$ and $G Q A$ (Figure 2), so the area of the quadrilateral $M$, $t=\frac{1}{2} P^{\prime} Q^{\prime} \cdot A G$.

In this product, $A G$ is constant, so $t$ is the smallest, or the largest, when the projection segment $P^{\prime} Q^{\prime}$ is the shortest, or the longest. As the intersecting plane rotates, $P^{\prime} Q^{\prime}$ rotates around the center of the hexagon and clearly takes on every possible position. It will be the shortest when it is perpendicular to two opposite sides of the hexagon, and the longest when it coincides with one of the hexagon's main diagonals. These two positions correspond to the sections shown in Figures 3a and 3b. In the first case, $P$ and $Q$ are midpoints of edges, since their projections are on the perimeter of the hexagon. In this case, $P^{\prime} Q^{\prime}$ is equal to the face diagonal of the cube, $\sqrt{2}$, so $t=\frac{\sqrt{6}}{2}$. In the second case, $P$ and $Q$ are opposite vertices of the cube, $P^{\prime} Q^{\prime}=\frac{2 \sqrt{6}}{3}$, so $t=\sqrt{2}$.

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Figure 3a

## $1987-10-313-2$. eps

Figure 3b

Thus, the area of the section varies between $\frac{\sqrt{6}}{2} \approx 1.225$ and $\sqrt{2} \approx 1.414$, taking on both extreme values, and since the area of $M$ varies continuously between the two extreme values, it takes on every intermediate value as well.