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?#

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#