A prism with an equilateral triangle in its base, whose height is two times bigger than base-side, is given.
Find out the ratio of volumes of inscribed and circumscribed spheres.
I tried to solve this, but I got stuck. I have a few questions which I cannot answer myself:
- Where do the centroids of these spheres lie ?
- What criteria does a sphere have to meet to be inscribed or circumscribed in such a prism ?
Best Answer
The centers of the inscribed & circumscribed spheres lie at the mid-point of the axis of the prism. The inscribed sphere must touch the nearest faces of the prism while the the surface of the circumscribed sphere must pass through all six vertices of the prism.
Thus the maximum possible value of radius $r$ of inscribed sphere is equal to the distance of its center from the lateral face of prism i.e. $$\color{blue}{r=\frac{a}{2\sqrt3}}$$
Drop a perpendicular from the center of circumscribed sphere to the base of prism & join the center of equilateral triangular base to one vertex to obtain a right triangle & apply Pythagorean theorem $$R^2=a^2+\left(\frac{a}{\sqrt 3}\right)^2$$ $$\color{blue}{R=\frac{2a}{\sqrt3}}$$
Hence, the ratio of the volumes of inscribed to the circumscribed spheres is given as $$=\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3}=\left(\frac{r}{R}\right)^3=\left(\frac{\frac{a}{2\sqrt 3}}{\frac{2a}{\sqrt 3}}\right)^3=\left(\frac{1}{4}\right)^3=\frac1{64}$$ or $$\color{red}{1:64}$$