In the quadrilateral pyramid $S A B C D$, the base $A B C D$ has its axis of symmetry as the diagonal $A C$, which is equal to 9, and the point $E$ of intersection of the diagonals of the quadrilateral $A B C D$ divides the segment $A C$ such that the segment $A E$ is smaller than the segment $E C$. A plane is drawn through the midpoint of the lateral edge of the pyramid $S A B C D$, parallel to the base and intersecting the edges $S A, S B, S C, S D$ at points $A 1, B 1, C 1, D 1$ respectively.

The polyhedron $A B C D A 1 B 1 C 1 D 1$, which is part of the pyramid $S A B C D$, intersects the plane $\alpha$ in a regular hexagon with a side length of 2. Find the area of the triangle $A B D$, if the plane $\alpha$ intersects the segments $B B 1$ and $D D 1$.
From the condition, it follows that the diagonals $A C$ and $B D$ of the quadrilateral $A B C D$ are perpendicular, and the triangles $A B D$ and $C B D$ are isosceles. Since the truncated pyramid $A B C D A 1 B 1 C 1 D 1$ has 6 faces, the plane $\alpha$ intersects each face along a segment, the ends of which do not coincide with the vertices of the polyhedron. Let the vertices of the hexagon lying on the sides of the base $A B C D$ be denoted by $K$ and $M$, on the sides of the quadrilateral $A 1 B 1 C 1 D 1$ by $K 1$ and $M 1$, and on the lateral edges $D D 1$ and $B B 1$ by $P$ and $T$ respectively (for definiteness, we assume that the vertex $P$ is adjacent to the vertices $K$ and $K1$). Since $K M T M 1 K 1 P$ is a regular hexagon, the line $P T$ is parallel to the lines $K M$ and $K 1 M 1$ and, consequently, parallel to the planes $A B C D$ and $A 1 B 1 C 1 D 1$ and equidistant from them. Therefore, the points $P$ and $T$ are the midpoints of the edges $D D 1$ and $B B 1$ respectively, and $P T$ is the midline of the trapezoid $B B 1 D 1 D$, so the lines $K M, K 1 M 1, P T, D 1 B 1$, and $D B$ are parallel. Let $E 1$ be the point of intersection of the diagonals of the quadrilateral $A 1 B 1 D 1 C 1$, $H$ be the point of intersection of the side $K M$ with the diagonal $A C$, and $H 1$ be the point of intersection of the side $K 1 M 1$ with the diagonal $A 1 C 1$. There are two possible cases for the position of the plane $\alpha$: in the first case, the point $H$ lies between $A$ and $E$, and in the second case, between $E$ and $C$. We will prove that only the first case is possible. Let $O$ be the point of intersection of the plane $A A 1 C 1 C$ with the diagonal $P T$. The triangles $H E O$ and $H 1 E 1 O$ are equal by two sides and the angle between them. Therefore, $H E = H 1 E 1$. Thus, the trapezoids $B D K M$ and $B 1 D 1 K 1 M 1$ have equal heights $H E$ and $H 1 E 1$ and smaller bases $K M$ and $K 1 M 1$. Since these trapezoids are isosceles (they are symmetric with respect to the diagonals $A C$ and $A 1 C 1$ respectively) and $B D > B 1 D 1$, then $\angle K D B < \angle K 1 D 1 B 1$. Therefore, the second case is impossible (otherwise, $\angle C D B < \angle A D B$, which contradicts the condition $A E < C E$). Let $\frac{\sigma E}{A E} = m$. Since the triangle $B C D$ is similar to the triangle $K 1 C 1 M 1$, and the triangle $B A D$ is similar to the triangle $K A M$, then

$$
\frac{C_{1} H_{1}}{G E} = \frac{K_{1} M M_{1}}{B D} = \frac{K M}{B D} = \frac{A B}{A E}
$$

Therefore, $\frac{C_{1} H_{1}}{A B} = m$. Then

$$
A H + H E = A E = \frac{\square E}{m} = \frac{2 C_{l} H_{1}}{m} = \frac{2\left(C_{l} H_{l} + H_{l} E_{l}\right)}{m} = 2 A H + \frac{2 E H}{m}.
$$

From this, it follows that $A H = \frac{m-2}{2 m-2} \cdot A E \cdot$ Therefore,

$$
\frac{K M}{B D} = \frac{A H}{A E} = \frac{m-2}{2 m-2}
$$

On the other hand,

$$
K M = \frac{1}{2} P T = \frac{1}{1}(B 1 D 1 + B D) = \frac{3}{\frac{3}{8}} B D
$$

Therefore, $\frac{m-2}{2 m-2} = \frac{3}{3}$, from which $m = 5$, and $A E = \frac{1}{6} A C = \frac{3}{2}$. Therefore,

$$
S_{\triangle A B D} = \frac{1}{2} A E \cdot B D = \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{16}{3} = 4
$$

## Answer

4.00