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A cube is inscribed in a right square pyramid such that one edge of the cube lies on the midline of the pyramid's base; the vertices of the cube not belonging to this edge lie on the lateral surface of the pyramid; the center of the cube lies on the height of the pyramid. Find the ratio of the volume of the pyramid to the volume of the cube.
Let $S A B C D$ be a regular quadrilateral pyramid, $M$ and $N$ be the midpoints of the sides $A B$ and $C D$ of the base $A B C D$ (Fig.1), $E$ and $F$ be the vertices of the cube $E F G H E 1 F 1 G 1 H 1$ lying on the segment $M N$, $G$ and $H$ be the vertices lying in the face $B S C$, $E 1$ and $F 1$ be in the face $A S D$, vertex $H 1$ be in the face $A S B$, vertex $G 1$ be in the face $C S D$, $O$ be the center of the cube, and $S T$ be the height of the pyramid. Denote the side of the base $A B C D$ of the pyramid by $a$, and the edge of the cube by $x$. Since the center $O$ of the cube lies on the height $S T$ of the pyramid, the point $T 1$, symmetric to $T$ with respect to $O$, lies on the edge $G 1 H 1$ of the cube and bisects it. Consider the section of the pyramid and the cube by the plane $M S N$ (Fig.2). We obtain an isosceles triangle $M S N$ and a rectangle $E F G 1 H 1$ inscribed in it, with points $G 1$ and $H 1$ lying on the lateral sides $S N$ and $S M$ respectively, and $H 1 G 1=x, E H 1=F G 1=x \sqrt{2}, M N=a$. From the similarity of triangles $G 1 S H 1$ and $N S M$, it follows that $\frac{S T_{1}}{S T^{\prime}}=\frac{G_{1} H_{1}}{M N}=\frac{\boldsymbol{x}}{\mathbf{a}}$. Consider the section of the pyramid and the cube by the plane passing through the vertex $S$ and the midpoints $P$ and $Q$ of the edges $A D$ and $B C$ respectively (Fig.3). We obtain an isosceles triangle $P S Q$ and a square $T P 1 T 1 Q$ inscribed in it, with vertices $P 1$ and $Q 1$ lying on the lateral sides $S P$ and $S Q$ respectively. Then

$$
\frac{x \sqrt{2}}{S I^{\prime}}=\frac{r T_{1}}{S H^{\prime}}=\frac{S T^{\prime}-S T_{1}}{S I^{\prime}}=1-\frac{S T_{1}}{S H^{\prime}}=1-\frac{x}{a}=\frac{a-x}{a},
$$

from which $S T=\frac{a c \sqrt{2}}{a-x}$. Note that

$$
\angle Q T Q 1=90^{\circ}-\angle S T Q 1=90^{\circ}-45^{\circ}=45^{\circ}
$$

Write the area of the right triangle $S T Q$ in two ways:

$$
S_{\triangle S T Q}=\frac{1}{2} T Q \cdot S T, S_{\triangle S T Q}=S_{\Delta S T Q} 1+S_{\Delta Q T Q} 1=
$$

$=\frac{1}{2} S T \cdot T Q 1 \sin 45^{\circ}+\frac{1}{2} T Q \cdot T Q 1 \sin 45^{\circ}=\frac{\sqrt{2}}{4} T Q 1(S T+T Q)$.

From the equality $\frac{1}{2} T Q \cdot S T=\frac{\sqrt{2}}{4} T Q 1(S T+T Q)$, we find that $T Q 1=\frac{\sqrt{2} T Q \cdot S T^{\prime}}{T Q+S T^{\prime}}$, or

![](https://cdn.mathpix.com/cropped/2024_05_06_a98533106cc029c284d7g-22.jpg?height=128&width=760&top_left_y=925&top_left_x=649)

Then

$$
\begin{gathered}
S T=\frac{a x \sqrt{2}}{a-x}=\frac{x^{2} \sqrt{2} \cdot(2 \sqrt{2}-1)}{x(2 \sqrt{2}-1)-x}=\frac{x \sqrt{2}(3+\sqrt{2})}{2}, \\
V_{\text {pyramid }}=\frac{1}{3} a 2 \cdot S T=\frac{1}{3} \cdot x 2(2 \sqrt{2}-1) 2 \cdot \frac{x \sqrt{2}(3+\sqrt{2})}{2}=\frac{x^{3}(12 \sqrt{2}-6)}{6} .
\end{gathered}
$$

Therefore,

$$
\frac{v_{\text {pyramid }}}{v_{\text {cube }}}=\frac{x^{3}(19 \sqrt{2}-6)}{6 x^{3}}=\frac{19 \sqrt{2}-6}{6} .
$$

## Answer

$\frac{19 \sqrt{2}-6}{6}$