10,11 Find the edge of a cube, one face of which lies in the plane of the base of a regular triangular pyramid, and the four remaining vertices are on its lateral surface, if the sides of the base of the pyramid are equal to $a$, and the height of the pyramid is equal to $h$.

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Let the vertices $A, B, C$ and $D$ of the cube $A B C D A 1 B 1 C 1 D 1$ with edge $x$ lie on the base $K L M$ of a regular triangular pyramid $K L M N$ with vertex $N$, and the vertices $A 1, B 1, C 1$ and $D 1$ lie on the lateral faces of the pyramid. Since the planes $A 1 B 1 C 1 D 1$ and $A B C D$ are parallel, the square $A 1 B 1 C 1 D 1$ is inscribed in the triangle $K 1 L 1 M 1$, obtained by the section of the pyramid $K L M N$ by a plane parallel to its base. Then the points $A 1$ and $B 1$ lie, for example, on the side $K 1 L 1$. The point $C 1$ lies on the side $L 1 M 1$, and the point $D 1$ lies on the side $K 1 M 1$. Since the edge of the cube is $x$, the pyramid $K 1 L 1 M 1 N$ is similar to the pyramid $K L M N$ with a coefficient of $\frac{h-x}{h}$. Therefore, the side of the equilateral triangle $K 1 L 1 M 1$ is $\frac{a(h-x)}{h}$. Further, we have:

$$
\begin{gathered}
B 1 L 1=A 1 K 1=A 1 D 1 \operatorname{ctg} \angle A 1 K 1 D 1=x \operatorname{ctg} 60^{\circ}=\frac{x}{\sqrt{3}} \\
a \cdot \frac{h-x}{h}=K 1 L 1=A 1 K 1+A 1 B 1+B 1 L 1=\frac{x}{\sqrt{3}}+x+\frac{x}{\sqrt{3}}=\frac{2 x}{\sqrt{3}}+x .
\end{gathered}
$$

From the equation

$$
a \cdot \frac{h-x}{h}=\frac{2 x}{\sqrt{3}}+x
$$

we find that

$$
x=\frac{3 a h}{(2 \sqrt{3}+3) h+3 a} .
$$

## Answer

$\frac{3 a h}{(2 \sqrt{3}+3) h+3 a}$