Inside a regular triangular pyramid, a straight prism is placed, with a rhombus lying at its base. One of the faces of the prism belongs to the base of the pyramid, and another face - to the lateral face of the pyramid. What is the maximum volume that the prism can have if the edge of the base of the pyramid is 2, and the height of the pyramid is 2 $\sqrt{2}$?
The lateral face of a regular pyramid is neither parallel nor perpendicular to the base plane, so the adjacent lateral faces of the prism lie in the lateral face and the base of the pyramid. Then one of the lateral edges, for example, $K K 1$, of the prism $K L M N K 1 L 1 M 1 N 1$ lies on the edge $A B$ of the base $A B C$ of the regular pyramid $D A B C$, the lateral face $K K 1 N 1 N$ of the prism - on the base $A B C$ of the pyramid, and the lateral face $K K 1 L 1 L-$ on the lateral face $A B D$ of the pyramid. Since the opposite lateral faces of the prism are pairwise parallel, the lateral face $L L 1 M 1 M$ lies in the section of the pyramid by a plane parallel to the base plane. Let $A 1 B 1 C 1$ be this section. Then among all such prisms, the one with the largest volume has vertices $M$ and $M 1$ lying on the sides $A 1 C 1$ and $B 1 C 1$ of the section $A 1 B 1 C 1$, respectively. Let the height $D H$ of the pyramid intersect the plane $A 1 B 1 C 1$ at point $O, P$ - the midpoint of side $A B, F$ - the midpoint of $A 1 B 1, E$ - the projection of point $F$ onto the plane $A B C D$. Denote $K L=K N=x, \frac{D A_{1}}{D A}=y, \angle D P H=\beta$. Then

$$
\begin{aligned}
& P H=\frac{1}{3} C H=\frac{1}{3} \cdot \frac{2 \sqrt{3}}{2}=\frac{\sqrt{3}}{3}, \operatorname{ctg} \angle L K N=\operatorname{ctg} \beta=\frac{p^{3} H}{D H}=\frac{\frac{\sqrt{3}}{3}}{2 \sqrt{2}}=\frac{\sqrt{6}}{12}, \\
& \sin \beta=\frac{1}{\sqrt{1+\alpha t g^{2} \beta}}=\frac{1}{\sqrt{1+\frac{1}{2 M}}}=\frac{2 \sqrt{6}}{5}, F E=O H=(1-y) D H=2 \sqrt{2}(1-y)
\end{aligned}
$$

$$
\begin{aligned}
& L L 1=A 1 B 1-2 L B 1=y \cdot A B-2 \cdot \frac{M L}{\sqrt{3}}=2 y-\frac{2 x}{\sqrt{3}}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_a98533106cc029c284d7g-07.jpg?height=112&width=774&top_left_y=2202&top_left_x=642)

from which we find that $y=\frac{5-x \sqrt{3}}{5}$. Then

$$
\begin{gathered}
L L 1=2 y-\sqrt{3}=\frac{2(5-x \sqrt{3})}{5}-\sqrt{3}=\frac{2(5 \sqrt{3}-8 x)}{5 \sqrt{3}}, \\
V_{K L M N K} 1 L 1 M 1 N 1=V(x)=K L \cdot K N \sin \beta L L 1=x 2 \cdot \frac{2 \sqrt{6}}{5} \cdot \frac{2(5 \sqrt{3}-8 x)}{5 \sqrt{3}}=
\end{gathered}
$$

$$
=\frac{4 \sqrt{2}}{25} \cdot x 2(5 \sqrt{3}-8 x)=\frac{4 \sqrt{2}}{25 \cdot 16} \cdot 4 x \cdot 4 x(5 \sqrt{3}-8 x) \leqslant
$$

$$
\left.\leqslant \frac{\sqrt{2}}{1000} \frac{4 x+4 x+(5 \sqrt{3}-8 x)}{3}\right) 3=\frac{\sqrt{2}}{1000}\left(\frac{5 \sqrt{3})}{3}\right) 3=\frac{5 \sqrt{6}}{36},
$$

with equality if $4 x=5 \sqrt{3}-8 x, x=\frac{5 \sqrt{3}}{12}$, i.e., when

$$
\frac{D A_{1}}{D A}=y=\frac{5-x \sqrt{3}}{5}=\frac{5-\frac{5 \sqrt{5}}{12}-\sqrt{3}}{5}=\frac{3}{4}
$$

Therefore, the sought maximum volume is $\frac{5 \sqrt{6}}{36}$.

## Answer

$5 \sqrt{6}$.

36.