3. Given a triangular pyramid $DABC$ whose base $ABC$ is an equilateral triangle, and the side $DBC$ is an isosceles triangle with leg $b$ and is perpendicular to the base. Find the area of the square that is obtained as the intersection of the pyramid with a plane parallel to the edges $DA$ and $BC$.
Solution. Edge $D A$ is normal to edge $B C$, so any plane parallel to these two edges cuts the pyramid into a rectangle. We are asked whether, at some position of the plane, this rectangle can be a square. If this is possible, we need to find its side.

Let us assume that this is possible and let it be the square $E F G H$ (see drawing) with side $x$. If $O$ is the midpoint of edge $B C$, then we have

$$
\overline{D O}^{2}=b^{2}-\frac{a^{2}}{4}, \overline{D A}^{2}=b^{2}+\frac{a^{2}}{2}
$$

If $M$ and $N$ are the intersection points of the plane with the lines $A O$ and $D O$ respectively, then from the fact that triangle $A E F$ is equilateral, we get $\overline{A M}=\frac{x \sqrt{3}}{2}$, i.e.

![](https://cdn.mathpix.com/cropped/2024_06_05_faf4f26986d5dd6dd724g-7.jpg?height=461&width=512&top_left_y=255&top_left_x=955)

$$
\overline{O M}=\overline{O A}-\overline{A M}=\frac{\sqrt{3}}{2}(a-x)
$$

From the similarity of triangles $O A D$ and $O M N$, it follows that

$$
\overline{D A}: x=\overline{O A}: \overline{O M}
$$

from which we get

$$
x=a \sqrt{2 b^{2}+a^{2}}:\left(a \sqrt{2}+\sqrt{2 b^{2}+a^{2}}\right) .
$$

Thus, the rectangle can be a square with side $x$ given by (1), and area $P=x^{2}$