5. Given a triangular prism with an equilateral base $m$. The lateral edges are equal to the side of the base. One of the vertices is equidistant from the vertices of the base. Find the largest radius of the sphere that can fit inside the prism.

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# Solution.

Let vertex $A_{1}$ be equidistant from the vertices of the triangle $A, B, C$. Then $A_{1} A B C$ is a tetrahedron with height $A_{1} O=\sqrt{2 / 3} m$, and the apothem $B M=M C=\sqrt{3} / 2 m$. We need to find the radius of the sphere that touches three planes: $A B B_{1}, A C C_{1}, B C B_{1}$. If we draw a section through the center of such a sphere perpendicular to the aforementioned planes, the section will be the inscribed circle of an isosceles triangle. Without loss of generality, we can take the triangle $B M C \left(B M C \perp A A_{1}\right)$ with sides $B M=M C=\sqrt{3} / 2 m$ and $B C=m$. We will find the radius of the inscribed sphere, i.e., the radius of the inscribed circle.

$$
\begin{gathered}
S_{C M B}=p_{C M B} \cdot r=\frac{1}{2} \cdot B C \cdot M P \\
r=(\sqrt{3}-1) /(2 \cdot \sqrt{2}) m
\end{gathered}
$$

The distance between the planes of the bases of the prism $A_{1} O > 2 r$. Indeed,

$$
\begin{aligned}
& \sqrt{2 / 3} m>(\sqrt{3}-1) /(2 \cdot \sqrt{2}) m \Leftrightarrow \\
& 2 / 3>2(4-2 \sqrt{3}) / 4 \Leftrightarrow 27>16
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ada2646b41484dca9e2ag-3.jpg?height=740&width=899&top_left_y=1749&top_left_x=978)

Therefore, the largest radius of the sphere that can fit inside the prism will be

$$
r=\frac{\sqrt{3}-1}{(2 \cdot \sqrt{2}) m}=\frac{\sqrt{6}-\sqrt{2}}{4 m}
$$