10,11

The height of the cylinder is $h$. In each base, an equilateral triangle with side $a$ is inscribed, and one of these triangles is rotated relative to the other by an angle of $60^{\circ}$. Find the volume of the polyhedron whose vertices are all the vertices of these triangles.
Let $O$ and $O_1$ be the centers of triangles $ABC$ and $A_1B_1C_1$, and $B'$ be the orthogonal projection of vertex $B$ onto the plane $A_1B_1C_1$. Then $B'$ is the midpoint of the arc $A_1B_1$ of the circumcircle of triangle $A_1B_1C_1$. Therefore, $C_1B'$ is the diameter of this circle, and $\angle C_1B_1B' = 90^\circ$. Since $B'B_1 \perp C_1B_1$, by the theorem of three perpendiculars, $BB_1 \perp C_1B_1$. Consequently, quadrilateral $ABB_1C_1$ is a rectangle, in which $AB = B_1C_1 = a$. Let $BB_1 = AC_1 = b$. The polyhedron $ABC A_1B_1C_1$ consists of two equal quadrilateral pyramids with a common base $ABB_1C_1$, the area of which is $ab$. Let $CP$ be the height of the quadrilateral pyramid $CABB_1C_1$. Since $CB = CA$, then $PB = PA$. Therefore, point $P$ lies on the perpendicular bisector of sides $AB$ and $B_1C_1$ of the rectangle $ABB_1C_1$. Let $M$ and $M_1$ be the midpoints of edges $AB$ and $B_1C_1$ respectively, and $C'$ be the orthogonal projection of vertex $C$ onto the plane $A_1B_1C_1$. Then $C'$ is the midpoint of the arc $B_1C_1$ of the circumcircle of triangle $A_1B_1C_1$, $MC \parallel M_1C'$, and point $P$ lies on $MM_1$. Consider the right trapezoid $CMM_1C'$. In it,

$$
CM = \frac{a \sqrt{3}}{2}, \quad M_1C' = \frac{a \sqrt{3}}{6}, \quad CC' = h, \quad MM_1 = b
$$

Let $\angle MCP = \alpha$. Drop a perpendicular $M_1Q$ from point $M_1$ to $CM$. Then

$$
\angle MM_1Q = \angle MCP = \alpha, \quad \cos \alpha = \cos \angle MM_1Q, \quad \frac{M_1Q}{MM_1} = \frac{h}{b}
$$

Therefore,

$$
CP = CM \cos \alpha = \frac{a \sqrt{3}}{2} \cdot \frac{h}{b}
$$

Consequently,

$$
\begin{aligned}
V_{ABC A_1B_1C_1} & = 2 V_{CABB_1C_1} = 2 \cdot \frac{1}{3} S_{ABB_1C_1} \cdot CP = \\
& = \frac{2}{3} ab \cdot \frac{a \sqrt{3}}{2} \cdot \frac{h}{b} = \frac{1}{3} a^2 h \sqrt{3}
\end{aligned}
$$

## Answer

$\frac{1}{3} a^2 h \sqrt{3}$.