$\left[\begin{array}{c}\text { Quadrilateral Pyramid }\end{array}\right]$

In the pyramid $S A B C D$, the base is a quadrilateral $A B C D$, where sides $A D$ and $B C$ are parallel, side $A B$ is 4, side $B C$ is 8, and angle $A B C$ is $60^{\circ}$. The edge $S B$ is $8 \sqrt{2}$. Find the volume of the pyramid, given that two planes can be drawn through the lines $A D$ and $B C$, not coinciding with the base of the pyramid, and intersecting the pyramid in equal quadrilaterals.
Let the plane passing through the side $AD$ of the base $ABCD$ of the pyramid $SABCD$ intersect the lateral edges $BS$ and $CS$ at points $M$ and $N$, respectively, and the plane passing through the side $BC$ intersect the lateral edges $AS$ and $DS$ at points $P$ and $Q$, respectively. The planes $ASD$ and $BPQC$ pass through the parallel lines $AD$ and $BC$ and intersect along the line $PQ$. Therefore, $PQ \| BC$. Similarly, $MN \| AD$. Suppose that $AM \| DN$. Then $BP \| CQ$. In this case, two intersecting lines of the plane $ASB$ are respectively parallel to two intersecting lines of the plane $CSD$, which means these planes are parallel, which is impossible. Therefore, the given quadrilaterals are trapezoids. Moreover, $PQ < AD$ and $MN < BC$, so in the equal trapezoids $BPQC$ and $AMND$, the bases $BC$ and $AD$ and the bases $PQ$ and $MN$ are respectively equal. In the quadrilateral $ABCD$, the opposite sides $AD$ and $BC$ are equal and parallel, so $ABCD$ is a parallelogram and

$$
S_{ABCD} = AD \cdot AB \sin 60^\circ = 16 \sqrt{3}
$$

Since $PQ \| AD$ and $MN \| BC$, then

$$
\frac{SP}{AS} = \frac{PQ}{AD} = \frac{MN}{BC} = \frac{SM}{BS}
$$

Therefore, $PM \| AB$. Similarly, $QN \| CD$, so $PM \| QN$, and since $PQ \| MN$, $PMNQ$ is a parallelogram. Thus, $PM = NQ$. Let the segments $AM$ and $BP$ intersect at point $E$, and the segments $CQ$ and $DN$ at point $F$. Suppose that $AM = CQ$ and $BP = DN$. Then the triangles $PEM$ and $NFQ$ are equal by three sides, so $\angle AMP = \angle CQN$. Therefore, the triangles $APM$ and $CQN$ are equal by two sides and the angle between them. Then $AP = CN$, and since $AP / AS = DQ / DS$, we have $AS = DS$. Similarly, $BS = CS$. Let $O$ be the orthogonal projection of the vertex $S$ onto the plane of the base $ABCD$. Then $OA = OD$ and $OB = OC$ as the orthogonal projections of equal oblique lines. Therefore, the point $O$ lies on the perpendicular bisectors of the opposite sides $AD$ and $BC$ of the parallelogram $ABCD$. Since the parallelogram $ABCD$ is not a rectangle, the perpendicular bisectors of its two opposite sides are parallel. Thus, the assumption that $AM = DN$ and $BP = CQ$ leads to a contradiction. The only remaining case is when $AM = BP$ and $CQ = DN$. Reasoning similarly, we get that $AS = CS$ and $BS = DS$. Then the point $O$ belongs to the perpendicular bisectors of the diagonals $AC$ and $BD$ of the parallelogram $ABCD$, i.e., coincides with the center of the parallelogram $ABCD$. Further, we find:

$$
\begin{gathered}
BD = \sqrt{AD^2 + AB^2 - 2AD \cdot AB \cos 120^\circ} = \sqrt{64 + 16 + 3} \\
SO = \sqrt{BS^2 - OB^2} = \sqrt{(8\sqrt{2})^2 - (2\sqrt{7})^2} = 10, \\
V_{SABCD} = \frac{1}{3} S_{ABCD} = \frac{1}{3} \cdot 16 \sqrt{3} \cdot 10 = \frac{160 \sqrt{3}}{3}.
\end{gathered}
$$

## Answer

$160 \sqrt{3}$