Question

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?`

Answer 1

The proportion is `2/sqrt3:1:pi`

Explanation:

Volume of a sphere of radius `r` is given by `4/3pir^3`. As radius of given sphere is unit, its volume will be `(4pi)/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, `3s^2=2^2` or `s=sqrt(4/3)=2/sqrt3` and volume of cube is `(2/sqrt3)^3=8/(3sqrt3)`.

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 `sqrt2/3a^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=2r^2` and `a=rxxsqrt2`.

Hence, volume of tetrahedron in a sphere of unit radius will be `sqrt2/3(sqrt2)^3=4/3`,

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

`8/(3sqrt3):4/3:(4pi)/3`

and multiplying each term by `3/4`, we get

`8/(3sqrt3)xx3/4:4/3xx3/4:(4pi)/3xx3/4`

or `2/sqrt3:1:pi`