Points $A, B, C, D, E$ and $F$ are the vertices of the lower base of a regular hexagonal prism, points $M, N, P, Q$, $R$ and $S$ are the midpoints of the sides of the upper base, points $O$ and $O_1$ are the centers of the lower and upper bases, respectively. Find the volume of the common part of the pyramids $O_1 A B C D E F$ and $O M N P Q R S$, if the volume of the prism is $V$.
Let point $P$ lie on edge $C_1D_1$ of the prism, $T$ be the midpoint of edge $CD$. Then $O_1P = OT$. If the diagonals $O_1T$ and $OP$ of rectangle $OO_1PT$ intersect at point $G$, then $G$ is the midpoint of the apothem $O_1T$ of the regular pyramid $O_1ABCDEF$. Consequently, the lateral edge $OP$ of the regular pyramid $OMNPQRS$ intersects the lateral face $CO_1D$ of the regular pyramid $O_1ABCDEF$ at the midpoint of the apothem lying in this face. Similarly for the other lateral edges of the pyramid $OMNPQRS$. Let point $N$ lie on edge $B_1C_1$ of the prism, $H$ be the intersection point of segment $O_1C_1$ with side $NP$ of the base of the pyramid $OMNPQRS$. Then $H$ is the midpoint of $NP$ and $O_1H = \frac{3}{4} O_1C_1 = \frac{3}{4} OC$. Consider the plane of rectangle $OO_1C_1C$. If segments $OH$ and $O_1C$ intersect at point $X$, then triangles $HXO_1$ and $OHC$ are similar with a ratio of $\frac{O_1H}{OC} = \frac{3}{4}$, meaning the lateral edge $O_1C$ of the regular pyramid $O_1ABCDEF$ intersects the lateral face $ONP$ of the regular pyramid $OMNPQRS$ at a point lying on its apothem and divides this edge in the ratio $3:4$, counting from vertex $O_1$. Similarly for the other lateral edges of the pyramid $O_1ABCDEF$. Let $Y$ be the intersection point of the lateral edge $O_1D$ of the pyramid $O_1ABCDEF$ with the apothem lying in the face $OPQ$ of the pyramid $OMNPQRS$. Consider the quadrilateral pyramid $OO_1XGY$ with vertex $O$. Its height is the perpendicular $OZ$ dropped from point $O$ to the plane $CO_1D$, i.e., the height of the triangular pyramid $OO_1CD$. Denote $OZ = h, S_{\Delta CO_1D} = s$. Then $V_{OO_1CD} = \frac{1}{3}sh$. On the other hand,

$$
V_{OO_1CD} = \frac{1}{6} V_{O_1ABCDEF} = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{3} V = \frac{1}{18} V
$$

and since

$$
S_{O_1XGY} = 2 S_{\Delta O_1GX} = 2 \cdot \frac{O_1X}{O_1C} \cdot \frac{O_1G}{O_1T'} \cdot S_{\Delta O_1CT} = 2 \cdot \frac{3}{7} \cdot \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{14 S}
$$

Then

$$
V_{OO_1XGY} = \frac{1}{3} S_{O_1XGY} \cdot h = \frac{13}{3} 14 S h = \frac{3}{14} V_{OO_1CD} = \frac{3}{14} \cdot \frac{1}{18} V = \frac{1}{146} V
$$

The common part of the pyramids $O_1ABCDEF$ and $OMNPQRS$ can be represented as the union of six quadrilateral pyramids equal to the pyramid $OO_1XGY$, so the volume of the common part is $6 \cdot \frac{1}{146} V = \frac{1}{14} V$.

## [ Volume Ratio ] Problem 111417 Topics: [ Fourier Series ] [ Regular Pyramid ]

The height $SO$ of a regular quadrilateral pyramid $SABCD$ forms an angle $\alpha$ with a lateral edge, and the volume of this pyramid is $V$. The vertex of a second regular quadrilateral pyramid is at point $S$, the center of the base is at point $C$, and one of the vertices of the base lies on the line $SO$. Find the volume of the common part of these pyramids.

## Answer

$\frac{V}{2(1+\cos(2))}$.