10,11

A cube is inscribed in a regular quadrilateral pyramid $S A B C D$. All four vertices of one face of the cube lie on the base $A B C D$ of the pyramid. The vertices of the opposite face of the cube lie on the lateral edges of the pyramid. It is known that $S A=A B=a$, i.e., the lateral edge of the pyramid is equal to $a$ and is equal to the side of its base. What is the volume of the cube?
Let the vertices $K, L, M$ and $N$ of the cube $K L M N K_1 L_1 M_1 N_1$ lie respectively on the lateral edges $A S, B S, C S$ and $D S$ of a regular quadrilateral pyramid $S A B C D$ with vertex $S$, and the vertices $K_1, L_1, M_1$ and $N_1$ lie on the base of the pyramid (Fig.1). Since the pyramid is regular, the orthogonal projections of its lateral edges onto the plane of the base are the segments $O A, O B, O C$ and $O D$, where $O$ is the point of intersection of the diagonals of the square $A B C D$. Therefore, the vertices $K_1, L_1, M_1$ and $N_1$ of the cube lie on the diagonals of the base of the pyramid. Consider the section of the pyramid by the plane passing through the points $A, S$ and $C$ (Fig.2). We obtain an isosceles triangle $A S C$ and a rectangle $K K_1 M_1 M$ inscribed in it, the vertices $K$ and $M$ of which lie on the lateral sides $A S$ and $C S$, and the vertices $K_1$ and $M_1$ lie on the base $A C$. Let the edge of the cube be denoted by $x$. Then

$$
K K_1 = M M_1 = x, \quad K M = K_1 M_1 = x \sqrt{2}.
$$

The sides of the triangle $A S C$ are $a$, $a$ and $a \sqrt{2}$. Therefore, the triangle $A S C$ is a right triangle, so

$$
\angle S A C = \angle S C A = 45^\circ, \quad A K_1 = K K_1 = x, \quad C M_1 = M M_1 = x,
$$

and since $A C = A K_1 + K_1 M_1 + C M_1$, we have

$$
a \sqrt{2} = x + x \sqrt{2} + x = x(2 + \sqrt{2}),
$$

from which $x = \frac{a \sqrt{2}}{2 + \sqrt{2}} = \frac{a}{\sqrt{2} + 1}$. Therefore,

$$
V_{K L M N K_1 L_1 M_1 N_1} = x^3 = \frac{\frac{a^3}{(\sqrt{2} + 1)^3}}{}.
$$

## Answer