Solid Geometry Advanced – Problem 16: find the radius of the sphere inscribed in the cone (tangent to the base circle and…

A right circular cone is inscribed in a sphere of radius $3$. Among all such cones, the cone volume is maximized.

For this maximum-volume cone, find the radius of the sphere inscribed in the cone (tangent to the base circle and the lateral surface).

(1) Let the sphere radius be $R=3$. Put the sphere center at the origin on the cone axis. Place the cone apex at the top of the sphere (height $z=R$). Let the cone height be $h$, so the base plane is at height $z=R-h$.

(2) The base circle radius is determined by the sphere cross-section: $$r^2=R^2-(R-h)^2=2Rh-h^2.$$ The cone volume is $$V(h)=\frac{1}{3}\pi r^2 h=\frac{1}{3}\pi h(2Rh-h^2)=\frac{1}{3}\pi(2Rh^2-h^3).$$ Differentiate on $(0,2R)$: $$V'(h)=\frac{1}{3}\pi(4Rh-3h^2)=\frac{1}{3}\pi h(4R-3h).$$ Hence the maximum occurs at $h=\frac{4R}{3}=4$.

(3) Then $$r^2=2Rh-h^2=2\cdot 3\cdot 4-16=8\Rightarrow r=2\sqrt2,$$ and the slant height is $$l=\sqrt{r^2+h^2}=\sqrt{8+16}=\sqrt{24}=2\sqrt6.$$ (4) The inscribed sphere radius of the cone equals the inradius of the axial isosceles triangle. Its area is $\Delta=rh$ and its semiperimeter is $s=r+l$, so $$\rho=\frac{\Delta}{s}=\frac{rh}{r+l}.$$ Substitute: $$\rho=\frac{(2\sqrt2)\cdot 4}{2\sqrt2+2\sqrt6}=\frac{4\sqrt2}{\sqrt2+\sqrt6}=\sqrt2(\sqrt6-\sqrt2)=2\sqrt3-2.$$ So $\rho=2(\sqrt3-1)$. \]

For the maximum-volume cone, the inscribed sphere radius is $2(\sqrt3-1)$.