Point $D$ is the midpoint of the lateral edge $B B 1$ of a regular triangular prism $A B C A 1 B 1 C 1$. On the lateral face $A A 1 C 1 C$, a point $E$ is taken, and on the base $A B C$, a point $F$ is taken such that the lines $E B 1$ and $F D$ are parallel. What is the maximum volume that the prism $A B C A 1 B 1 C 1$ can have if $E B 1=1, F D=\frac{3}{4}, E F=\frac{1}{2 \sqrt{3}}$?
Consider the section of the prism by a plane passing through the parallel lines $B 1 E$ and $D F$ (Fig.1). This plane contains the lateral edge $B B 1$ and intersects the edges $A C$ and $A 1 C 1$ at some points $L$ and $K$ respectively. Then $B L K B 1$ is a rectangle. Denote $\angle K B 1 E=\angle B F D=\alpha$. From the right triangles $K B 1 E$ and $B F D$ we find that

$$
\begin{aligned}
& K B 1=B 1 E \cos \alpha=\cos \alpha, K E=B 1 E \sin \alpha=\sin \alpha \\
& B F=D F \cos \alpha=\frac{3}{4} \cos \alpha, B D=D F \sin \alpha=\frac{3}{4} \sin \alpha
\end{aligned}
$$

Then

$$
\begin{gathered}
K L=B B 1=2 B D=\frac{3}{2} \sin \alpha, E L=K L-K E=\frac{3}{2} \sin \alpha-\sin \alpha=\frac{1}{2} \sin \alpha, \\
L F=B L-B F=K B 1-B F=\cos \alpha-\frac{3}{4} \cos \alpha=\frac{1}{4} \cos \alpha .
\end{gathered}
$$

By the Pythagorean theorem $L F 2+E L 2=E F 2$, or $\frac{1}{16} \cos 2 \alpha+\frac{1}{4} \sin 2 \alpha=\frac{1}{12}$, from which we find that $\sin \alpha=\frac{1}{3}$. Then

$$
\cos \alpha=\frac{2 \sqrt{2}}{3}, B 1 K=B L=\cos \alpha=\frac{2 \sqrt{2}}{3}
$$

Let $a-$ be the side of the base of the prism. Since the height of the prism is constant and equal to $K L=\frac{3}{2} \sin \alpha=\frac{3}{2} \cdot \frac{1}{3}=\frac{1}{2}$, the volume will be the largest when $a$ has the largest possible value. Consider all possible equilateral triangles with vertex $B 1$ and point $K$ lying on the opposite side (Fig.2). From their vertices lying on one side of the line $B 1 K$, the segment $B 1 K$ is seen at the same angle, equal to $60^{\circ}$. Therefore, these vertices lie on the same circle with points $B 1$ and $K$. The largest side has the equilateral triangle whose side is the diameter of this circle, so the largest side has the triangle $A 1 B 1 C 1$, for which $B 1 K$ is the height, and since $B 1 K=\frac{a \sqrt{3}}{2}$, then

$$
a=\frac{2 B_{1} K}{\sqrt{3}}=\frac{2 \cdot \frac{2 \sqrt{2}}{2}}{\sqrt{3}}=\frac{4 \sqrt{2}}{3 \sqrt{3}}
$$

Therefore,

$$
V_{\text {prism }}=\frac{\underline{a}^{2} \sqrt{3}}{4} \cdot K L=\frac{\frac{32}{2} \cdot \sqrt{3}}{4} \cdot \frac{1}{2}=\frac{4 \sqrt{3}}{27}
$$

## Answer