# A cube and a regular octahedron are carved out of of unit-radius wooden spheres . If the vertices are on the spheres, how do you prove that their volumes compare with that of the sphere, in the proportions 3/sqrt 2 : 2 : pi?

Nov 16, 2016

The proportion is $\frac{2}{\sqrt{3}} : 1 : \pi$

#### Explanation:

Volume of a sphere of radius $r$ is given by $\frac{4}{3} \pi {r}^{3}$. As radius of given sphere is unit, its volume will be $\frac{4 \pi}{3}$.

Now let us consider a cube carved in unit sphere. It should appear as follows:

As the diameter of sphere is the longest diagonal of sphere, which is $2$ here, $3 {s}^{2} = {2}^{2}$ or $s = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$ and volume of cube is ${\left(\frac{2}{\sqrt{3}}\right)}^{3} = \frac{8}{3 \sqrt{3}}$.

A regular octahedron is a solid object made of eight equilateral triangles and appears as shown below. It is made of two tetrahedrons and volume of an octahedron of side $a$ is given by $\frac{\sqrt{2}}{3} {a}^{3}$.

Let us consider an octahedron in a sphere, so that when a sphere is divided into eight equal parts each part contains an equilateral triangle. Using Pythagoras theorem, the side of a tetrahedron will be given by ${a}^{2} = {r}^{2} + {r}^{2} = 2 {r}^{2}$ and $a = r \times \sqrt{2}$.

Hence, volume of tetrahedron in a sphere of unit radius will be $\frac{\sqrt{2}}{3} {\left(\sqrt{2}\right)}^{3} = \frac{4}{3}$,

Now we have to find ratio of volume of such cube, octahedron and sphere and it is

$\frac{8}{3 \sqrt{3}} : \frac{4}{3} : \frac{4 \pi}{3}$

and multiplying each term by $\frac{3}{4}$, we get

$\frac{8}{3 \sqrt{3}} \times \frac{3}{4} : \frac{4}{3} \times \frac{3}{4} : \frac{4 \pi}{3} \times \frac{3}{4}$

or $\frac{2}{\sqrt{3}} : 1 : \pi$