A prism has a base that is a regular triangle with unit side length. A plane intersects the prism's lateral surface forming an isosceles right triangle. What are the lengths of the sides of the intersecting triangle?
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Translate the text into English:

Move the intersecting plane parallel so that the vertex $A$ of the intersection triangle falls into one of the vertices of the base triangle (1st figure).

Due to axial symmetry, the intersecting plane cuts the other two edges of the prism at the same height: $B E=C D$.

This implies that $D E \# C B=1$. In the right-angled triangle $A E B$, $A E>A B=1$. Since the triangle $A D E$ is also right-angled, $D E=1>A E>1$, which is a contradiction.

Therefore, $D$ and $E$ cannot be at the same distance from the plane of $A B C$.

Let $C D=x, B E=y$, where $y>x$ (2nd figure).

From the triangle $A D C$,

\[
\begin{aligned}
& A D=\sqrt{1+x^{2}}, \text { similarly } \\
& A E=\sqrt{1+y^{2}}
\end{aligned}
\]

Since $y>x$, $A E>A D$; thus, the right angle is at vertex $D$, and therefore

\[
A D=A E=\frac{1}{\sqrt{2}} A E
\]

From the triangle $D E F$:

\[
E D=\sqrt{1+(y-x)^{2}} \text {, so } \sqrt{1+(y-x)^{2}}=\sqrt{1+x^{2}}
\]

Thus,

\[
(y-x)^{2}=x^{2}, \text { and since } y-x>0, \text { then }
\]

Substituting (1) and (2) into the equation,

\[
\frac{1}{\sqrt{2}} \sqrt{1+4 x^{2}}=\sqrt{1+x^{2}}
\]

From this, $x^{2}=\frac{1}{2}, A D=\sqrt{\frac{3}{2}}$ and $A E=\sqrt{3}$.
![](https://cdn.mathpix.com/cropped/2024_05_02_0551bc44ba4c4bd54b1ag-1.jpg?height=862&width=424&top_left_y=1317&top_left_x=838)

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