A square-based pyramid has every edge 1 unit long. What is the radius of the sphere inscribed in the pyramid?
Solution. Let the base of the pyramid be the quadrilateral $ABCD$, its fifth vertex be $E$, and the foot of the perpendicular from $E$ to the base be $T$. Since every edge of the pyramid is 1 unit long, the quadrilateral $ABCD$ is a rhombus. The segment $ET$ is perpendicular to the plane of $ABCD$, so by the Pythagorean theorem,

\[
\begin{array}{ll}
A T^{2}=A E^{2}-E T^{2}=1-E T^{2}, & B T^{2}=B E^{2}-E T^{2}=1-E T^{2} \\
C T^{2}=C E^{2}-E T^{2}=1-E T^{2}, & D T^{2}=D E^{2}-E T^{2}=1-E T^{2}
\end{array}
\]

Thus, $A T = B T = C T = D T$. Therefore, a circle can be inscribed around the rhombus $ABCD$ with center $T$, which means the rhombus is a square with $T$ as its center.

![](https://cdn.mathpix.com/cropped/2024_05_02_f5bdb2741a0a4f4c296fg-1.jpg?height=415&width=588&top_left_y=584&top_left_x=758)

The pyramid $ABCDE$ is thus a square-based right pyramid. Let the radius of the inscribed sphere be $r$, the surface area be $F$, and the volume be $V$. It is known that $V = F \cdot \frac{r}{3}$. The radius of the inscribed sphere can be determined if we first calculate the surface area and volume of the pyramid.

Since $ABCD$ is a unit square, $A T = \frac{\sqrt{2}}{2}$, and by the Pythagorean theorem,

\[
E T = \sqrt{A E^{2} - A T^{2}} = \sqrt{1 - \frac{1}{2}} = \frac{\sqrt{2}}{2}
\]

This is the height of the pyramid corresponding to the base $ABCD$ with unit area, so the volume of the pyramid is

\[
V = \frac{1 \cdot \frac{\sqrt{2}}{2}}{3} = \frac{\sqrt{2}}{6}
\]

The lateral faces of the pyramid are equilateral triangles with unit side length, so the surface area is

\[
F = 1 + 4 \cdot \frac{\sqrt{3}}{4} = 1 + \sqrt{3}
\]

Therefore, the radius of the inscribed sphere is

\[
r = \frac{3 V}{F} = \frac{\sqrt{2}}{2(1 + \sqrt{3})} = \frac{\sqrt{6} - \sqrt{2}}{4}
\]