Solid Geometry Advanced – Problem 13: Find the volume of this circumscribed sphere

A regular triangular pyramid (right pyramid over an equilateral base) has base side length $3$ and all three lateral edges equal to $2$.

All four vertices lie on a common sphere. Find the volume of this circumscribed sphere.

(1) The base is an equilateral triangle of side $3$, so its circumradius is $$R_b=\frac{s}{\sqrt3}=\frac{3}{\sqrt3}=\sqrt3.$$ Let $O$ be the center of the base triangle and let the apex be $A$. In a regular triangular pyramid, $A$ lies on the perpendicular line through $O$.

(2) Let the height be $h=AO_\perp$. Since a lateral edge has length $2$, for any base vertex $B$, $$AB^2=R_b^2+h^2\Rightarrow 4=3+h^2\Rightarrow h=1.$$ (3) By symmetry, the sphere center lies on the axis through $A$ and $O$. Place coordinates so that $O$ is at $z=0$, base vertices have $z=0$, and apex is at $z=1$. Let the sphere center be $(0,0,z_0)$.

Equal distances to a base vertex and to $A$ give $$3+z_0^2=(1-z_0)^2\Rightarrow 3=1-2z_0\Rightarrow z_0=-1.$$ So the sphere radius is $$R=|1-(-1)|=2.$$ (4) Sphere volume: $$V=\frac{4}{3}\pi R^3=\frac{4}{3}\pi\cdot 8=\frac{32\pi}{3}.$$

The circumscribed sphere has radius $2$, so its volume is $\dfrac{32\pi}{3}$.