87. In a regular quadrilateral pyramid $S A B C D$ ( $A B C D$ - base), the side of the base is equal to $a$, and the angle between the lateral edge and the plane of the base is $\alpha$. A plane, parallel to the diagonal of the base $A C$ and the lateral edge $B S$, intersects the pyramid such that a circle can be inscribed in the section. Determine the radius of this circle.
87. If a plane intersects the edges $A D$ and $C D$, then the section will be a triangle, and the radius of the inscribed circle will vary from 0 to $\frac{a}{\sqrt{2}\left(2 \cos \alpha+\sqrt{4 \cos ^{2} \alpha+1}\right)}$.

Now let the plane intersect the edges $A B$ and $B C$ at points $P$ and $N$, $S A$ and $S C$ at points $Q$ and $R$, $S D$ at point $K$, and the extensions of $A D$ and $C D$ at points $L$ and $M$ (Fig. 8). Since the lines $P Q$ and $N R$ are parallel and tangent to the circle inscribed in our section, $P N$ is the diameter of this circle. Let $|P N|=2 r$,

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-061.jpg?height=501&width=477&top_left_y=291&top_left_x=121)

Fig. 8. We will have

$$
\begin{gathered}
|M L|=2 a \sqrt{2}-2 r \\
|K L|=\frac{a \sqrt{2}-r}{2 \cos \alpha} \sqrt{4 \cos ^{2} \alpha+1} \\
S_{M K L}=\frac{(a \sqrt{2}-r)^{2}}{2 \cos \alpha}
\end{gathered}
$$

Thus,

$$
r=\frac{a \sqrt{2}-r}{2 \cos \alpha+\sqrt{4 \cos ^{2} \alpha+1}}
$$

from which

$r=\frac{a \sqrt{2}}{1+2 \cos \alpha+\sqrt{4 \cos ^{2} \alpha+1}}$

Answer:

$$
\begin{gathered}
0<r \leqslant \frac{a}{\sqrt{2}\left(2 \cos \alpha+\sqrt{4 \cos ^{2} \alpha+1}\right)} \\
r=\frac{a \sqrt{2}}{1+2 \cos \alpha+\sqrt{4 \cos ^{2} \alpha+1}}
\end{gathered}
$$