A pyramid with a parallelogram base has base edges of lengths $a$ and $2a$, and side edges of length $a \sqrt{3}$. The pyramid is intersected by a plane that cuts the base along one of its diagonals and is parallel to one of the side edges. What is the area of the plane section?
The vertices $A, B, C, D$ of the base lie on the opposite vertex $E$ (tetrahedral) on a sphere with radius $a \sqrt{3}$. This sphere intersects the base plane in a circle, so the base quadrilateral is a parallelogram circumscribed by a circle: a rectangle, and the center of the circle circumscribed around it is the intersection point $O$ of the diagonals, and $EO$ is the height of the pyramid. Thus, the pyramid is symmetric with respect to the perpendicular bisector planes of the base edges $AB$ and $AD$, and a $180^{\circ}$ rotation around $EO$ as an axis also maps the body onto itself.
![](https://cdn.mathpix.com/cropped/2024_05_02_9f442f187bf4359a3e5bg-1.jpg?height=538&width=1204&top_left_y=346&top_left_x=450)

According to this, the two diagonals of the base are equal, so it is sufficient to take the desired intersecting plane $S$ through the diagonal $BD$. To determine $S$, we can essentially choose the side edge in two ways: $S$ is either parallel to $EA$ or $EC$, or it is parallel to $EB$ and in this case, it also passes through $EB$ and $ED$, and the intersection is an isosceles triangle $EBD$. (In fact, we only consider the examination of the first possibility as our real task. The second interpretation is somewhat forced, but the problem statement allows it - hence we will address it immediately: the length of the diagonal $BD$ is $a \sqrt{5}$ from the right triangle $EBO$, the height $EO = a \sqrt{7} / 2$, and the area of the triangle $EBD$ is $a^{2} \sqrt{35} / 4$.)

The rotational symmetry mentioned remains even after the selection of the diagonal $BD$, so we can choose that $S$ is parallel to $EA$. Denoting the intersection point of $S$ with $EC$ by $P$, the intersection line of $S$ and the plane $EAC$ is clearly $PO$, and this is the midline of the triangle $EAC$, so $PO = EA / 2$.

Let the perpendicular projections of $P$ and $C$ onto $BD$ be $P_{1}$ and $C_{1}$, respectively, then $OP_{1} = P_{1}C_{1}$, since these segments are the distances between adjacent pairs of the perpendicular planes erected from $E$, $P$, and $C$ to $BD$, and these distances are equal because $EP = PC$ (in the right part of the diagram, $P_{2}$ is the projection of $P$ onto the base plane).

Now, from consecutive right triangles:

$$
\begin{gathered}
BC_{1} = \frac{BC^2}{BD} = \frac{4a}{\sqrt{5}}, \quad OC_{1} = \frac{3a}{2\sqrt{5}}, \quad OP_{1} = \frac{3a}{4\sqrt{5}} \\
PP_{1}^2 = \sqrt{OP^2 - OP_{1}^2} = \frac{\sqrt{51}a}{4\sqrt{5}}
\end{gathered}
$$

and thus the area of the triangle $PBD$ is $a^{2} \sqrt{51} / 8$.