A cube's edge $AB$ is bisected by point $P$, and edge $AC$ is divided by point $Q$ such that $CQ = 2AQ$. Let us draw planes through the line $PQ$ so that they intersect the cube in quadrilaterals. Which of these gives the largest area of intersection?
Let the base of the cube be $ABCD$, and the top face be $EFHG$. If we project the cube onto a plane $S_0$ perpendicular to the line $PQ$, the line $PQ$ is projected as a point, and the intersections of the rotatable plane around the $PQ$ axis are seen as line segments, and we can easily count the intersected edges, which is the number of sides of the intersection.

For any point $X$, we denote its projection as $X'$, where $X=A, B, \ldots, P, Q, \ldots$. The coinciding projections $P'$ and $Q'$ bisect $A'B'$ and trisect $A'C'$, so $C'$ is the reflection of $P'$ over $B'$, and furthermore, $\overrightarrow{B'D'} = \overrightarrow{A'C'}$, and consequently, $A'D'H'E'$ forms a rectangle (the height is compressed in the diagrams). The intersection is a quadrilateral if and only if the projection $e$ of the rotating plane $S_0$ exits the rectangle $A'B'F'E'$ on the segment $E'F'$, including $F'$ but excluding $E'$. If the exit point is on $A'E'$ or exactly at $E'$, the intersection has only 3 sides. However, if the exit point on the segment $F'G'$ (including $G'$) or between $G'$ and $H'$, the intersection has 5, 6, and then again 5 sides, and if it is at $H'$ or on the side $H'D'$. Thus, the additional vertices of our plane sections are points $R$ and $T$ on the edges $EF$ and $EG$, respectively, but such that $RT \parallel PQ$, making the intersection a trapezoid, as the intersection lines with the top and bottom faces are parallel. The height of the trapezoid is seen in its true size in the projection on $S_0$.
![](https://cdn.mathpix.com/cropped/2024_05_02_b5f29510477c29e0ea20g-1.jpg?height=504&width=1820&top_left_y=732&top_left_x=150)

Let the midpoint of $EF$ be $J$. If $R$ is in the interior of the segment $JF$, and the reflection of $R$ over $J$ is $R^*$, then the heights of the trapezoids $PQTR$ and $PQT^*R^*$ are equal: $P'R' = P'R^{*'}$, but the sum of the parallel sides is $PQ + RT > PQ + R^*T^*$, so the area of the first trapezoid is larger. Every allowed plane belongs to such a pair, except for the planes $PQJ$ and $PQF$, so the largest area intersection cannot be given by a plane of type $PQR^*$.

Since as $R$ moves from $J$ to $F$, the height of the trapezoidal intersection and $RT$ also increase strictly monotonically, the sum of the parallel sides also increases strictly monotonically, so the desired intersection is $PQFK$, where $EK = 2 \cdot KG$.

Kovács Zsolt (Szolnok, Verseghy F. Gymnasium, 4th grade)  
Horváth László (Csurgó, Csokonai Vitéz M. Gymnasium, 4th grade)

Note. The area of the found intersection is $AB^2 \sqrt{14} / 4$.