12.417 A sphere is inscribed in a regular quadrilateral pyramid. A tangent plane parallel to the base of the pyramid is drawn to the sphere, dividing the volume of the pyramid in the ratio $m: n$, counting from the vertex. Find the angle between the height of the pyramid and its lateral face.
Solution. Let $A$ and $D$ be the midpoints of the edges $M E$ and $N L$ of the base of the given regular pyramid FEMNL, $B C$ be the line of intersection of $A F D$ with the given plane tangent to the sphere inscribed in the pyramid, $O$ be the projection of the vertex $F$ onto the base.

$N L \perp A D \Rightarrow N L \perp F D$ (Theorem 12) $\Rightarrow N L \perp A F D$ (Theorem 11) $\Rightarrow$

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-297.jpg?height=386&width=417&top_left_y=141&top_left_x=129)

$\Rightarrow A F D \perp F N L$ (Theorem 15$) \Rightarrow$ the angle between $F O$ and the lateral face $F N L$ is precisely $\angle O F D$ (since the projection of point $O$ onto $F N L$ lies on the apothem $F D$ ). $B C \| A D$ (Theorem 13 ) $\Rightarrow A B C D$ is an isosceles trapezoid, into which a circle can be inscribed (the section of the sphere by the plane $A F D$ ). Let $C K$ be the height of the trapezoid $A B C D$.

Then $K D=\frac{1}{2}(A D-B C), C D=\frac{1}{2}(B C+A D)$ (Theorem 2$) \Rightarrow$ $\Rightarrow \sin \angle O F D=\sin \angle K C D=\frac{K D}{C D}=\frac{A D-B C}{A D+B C}=\left(\frac{A D}{B C}-1\right):\left(\frac{A D}{B C}+1\right)$.

But $\frac{A D}{B C}=\sqrt[3]{\frac{m+n}{m}}$, since the volumes of similar pyramids are proportional to the cubes of the corresponding edges $\Rightarrow \sin \angle O F D=\frac{\sqrt[3]{m+n}-\sqrt[3]{m}}{\sqrt[3]{m+n}+\sqrt[3]{m}}$.

Answer: $\arcsin \frac{\sqrt[3]{m+n}-\sqrt[3]{m}}{\sqrt[3]{m+n}+\sqrt[3]{m}}$