In a regular quadrilateral prism $A B C D A_1 B_1 C_1 D_1$, the lateral edge is $\sqrt{14}$, and the side of the base $A B C D$ of the prism is 6. The base circle of the cone is inscribed in the triangle $B C_1 D$, and the vertex of the cone lies in the plane $A B C_1$. Find the volume of the cone.
Let $O$ be the center of the base of the cone, i.e., the center of the circle inscribed in the isosceles triangle $B C 1 D$ with

$$
\begin{gathered}
\text { sides } \\
D C 1 = B C 1 = \sqrt{C B^{2} + C C_{1}^{2}} = \sqrt{36 + 14} = \sqrt{50} = 5 \sqrt{2}, \\
B D = A B \sqrt{2} = 6 \sqrt{2}, \\
Q - \text { the center of the square } A B C D, \text { i.e., the midpoint of } B D. \text { Then } \\
C 1 O = \sqrt{C C_{1}^{2} + C Q^{2}} = \sqrt{14 + (3 \sqrt{2})^{2}} = \sqrt{14 + 18} = \sqrt{32} = 4 \sqrt{2} .
\end{gathered}
$$

If $r$ is the radius of the circle, $p$ is the semiperimeter of triangle $B C 1 D$, and $S$ is its area, then

$$
r = \frac{S}{\rho} = \frac{1}{B} B D C_{1} + B Q = \frac{3 \sqrt{2} \cdot 4 \sqrt{2}}{5 \sqrt{2} + 3 \sqrt{2}} = \frac{3 \sqrt{2}}{2}
$$

Since $A$ and $C 1$ are common points of the planes $A B C 1$ and $B D C 1$, these planes intersect along the line $A C 1$. In the plane $B C 1 D$, construct a perpendicular to $C 1 Q$ from point $O$. Let $P$ be the point of intersection of this perpendicular with the line $A C 1$. The line $B D$ is perpendicular to the intersecting lines $A C$ and $C C 1$ in the plane $A C C 1$, so $O P \perp B Q$, and since $O P \perp Q C 1$, the line $O P$ is perpendicular to the plane $B C 1 D$, i.e., $P O$ is the height of the cone. Consider the triangle $A C C 1$. Let $\angle A C 1 C = \alpha, \angle Q C 1 C = \beta$. Then

$$
\operatorname{tg} \alpha = \frac{A C}{C C_{1}} = \frac{6 \sqrt{2}}{\sqrt{14}} = \frac{6}{\sqrt{7}}, \operatorname{tg} \beta = \frac{Q C}{C C_{1}} = \frac{3 \sqrt{2}}{\sqrt{14}} = \frac{3}{\sqrt{7}} .
$$

Then

![](https://cdn.mathpix.com/cropped/2024_05_06_3af6d4d9a845a74ce61dg-20.jpg?height=151&width=858&top_left_y=2017&top_left_x=606)

From the right triangle $O P C 1$, we find that

$$
\begin{gathered}
P O = O C 1 \operatorname{tg} \angle O P C 1 = (C 1 Q - O Q) \operatorname{tg}(\alpha - \beta) = \left(4 \sqrt{2} - \frac{3 \sqrt{2}}{2}\right) \cdot \frac{3 \sqrt{7}}{25} = \\
= \frac{5 \sqrt{2}}{2} \cdot \frac{3 \sqrt{7}}{25} = \frac{3 \sqrt{14}}{10} . \\
\text { Therefore, if } V \text { is the volume of the cone, then } \\
V = \frac{1}{3} \pi r 2 \cdot O P = \frac{1}{3} \pi \left(\frac{3 \sqrt{2}}{2}\right) 2 \cdot \frac{3 \sqrt{14}}{10} = \frac{7 \sqrt{14}}{20} \pi .
\end{gathered}
$$

## Answer

The base of the cylinder is inscribed in the lateral face $S A B$ of a regular quadrilateral pyramid $S A B C D$ ( $S$ is the vertex), and the center of the other base of the cylinder lies in the plane $S B C$. Find the volume of the cylinder if $A B = 6, S B = 5$.

## Answer

In a regular quadrilateral prism $A B C D A 1 B 1 C 1 D 1$, the side of the base $A B C D$ is 2, and the lateral edge is $\sqrt{14}$. The base of the cone is inscribed in the triangle $A B 1 D 1$, and the vertex of the cone lies in the plane $A B 1 C 1$. Find the volume of the cone.

## Answer

$$
\frac{\sqrt{7}}{9} \pi \text {. }
$$

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