A sphere is inscribed in a quadrilateral pyramid $S A B C D$, whose base is a trapezoid $A B C D$, and is also inscribed in a regular tetrahedron, one of whose faces coincides with a lateral face of the pyramid $S A B C D$.

Find the radius of the sphere, if the volume of the pyramid $S A B C D$ is 64.
Let $AD$ and $BC$ be the bases of the trapezoid ($AD > BC$), and the lines $AB$ and $CD$ intersect at point $E$. Suppose the lateral face $SBC$ of the pyramid is also a face of a regular tetrahedron. Through a line that has no common points with the sphere, exactly two planes can be drawn that are tangent to the sphere, so the planes of the other faces of the regular tetrahedron coincide with the planes $SCD$, $SAB$, and $ABCD$. Then $SBCD$ is a regular tetrahedron, which is impossible, because the point of tangency of the sphere with the plane $ABCD$ lies outside the triangle $BCE$. If the lateral face $SCD$ is a face of the regular tetrahedron, then the planes of its other faces coincide with the planes $SAD$, $SBC$, and $ABCD$, which is also impossible, because the line $AD$ of intersection of the planes $SAD$ and $ABCD$ and the line $BC$ of intersection of the planes $SBC$ and $ABCD$ are parallel, and therefore, the edges of the tetrahedron cannot lie on these lines. Therefore, the triangle $AED$ is a face of the regular tetrahedron mentioned in the problem, and $S$ is the vertex of this tetrahedron. Then $ADE$ is an equilateral triangle. Let the edges of the regular tetrahedron $SADE$ be $a$. Consider the section of the pyramid $SABCD$ and the tetrahedron by the plane passing through the points $S$, $E$, and the midpoint $M$ of the edge $AD$. This plane also passes through the midpoint $N$ of the edge $BC$ and the points $P$ and $Q$ of tangency of the sphere with the faces $BSC$ and $ABCD$ respectively. From the equality of the right triangles $SQM$ and $SQN$

$$
\text{it follows that } NQ = QM = \frac{1}{2} QE \text{. Therefore, } N \text{ is the midpoint of the segment } QE \text{ and } \frac{BC}{AD} = \frac{EN}{EM} = \frac{1}{3} \text{. Therefore,}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_3af6d4d9a845a74ce61dg-39.jpg?height=103&width=1208&top_left_y=2710&top_left_x=425)

$$
V_{SABCD} = \frac{1}{3} S_{ABCD} \cdot SQ = \frac{1}{3} \cdot \frac{2a^2 \sqrt{3}}{9} \cdot a \sqrt{\frac{2}{3}} = \frac{2a^3 \sqrt{2}}{27} = 64
$$

from which we find that $a = 6 \sqrt{2}$. The center of the sphere inscribed in a regular tetrahedron coincides with the center of the tetrahedron, and since the center of the tetrahedron divides each of its heights (medians) in the ratio 1:3, counting from the vertex, the radius of the inscribed sphere is one-fourth of the height of the regular tetrahedron, so if $r$ is the radius of the sphere, then

$$
r = \frac{1}{4} SQ = \frac{1}{4} a \sqrt{\frac{2}{3}} = \frac{1}{4} \cdot 6 \sqrt{2} \cdot \sqrt{\frac{2}{3}} = \sqrt{3}
$$

## Answer

$\sqrt{3}$.

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