POW: Cube and Octahedron

[AuSmith]

(1) A cube and a regular octahedron are inscribed in a sphere. Which has the larger volume? Guess first, then prove or disprove your guess.

(2) A regular dodecahedron and a regular icosahedron are inscribed in a sphere. Which has the larger volume?

(3) What if we change from inscribing to circumscribing?

(4) Exactly what are the volumes of these four solids in terms of their inradii and circumradii?

[bradp]

1) Cube: circumradius = R = half the longest diagonal = (1/2)sqrt(3)(side)
the side then equals 2Rsqrt(3)/3
the volume is then 8*sqrt3*R^3 / 9

Octahedron: circumradius = half the diagonal of the "base" = (1/2)sqrt2(side)

so the side = Rsqrt(2)
the volume is twice the volume of one of the pyramids --> 2(1/3)(height)(area of base) = (2/3)(R)(side)^2 = (4/3)R^3

[8sqrt(3)/9] > [4/3]

i hope this is right
someone teach me LaTeX

[Kurt]

I'm not going to do the problems (don't really have time right now) :(.

However, as a general note, a sphere can be thought of as a polyhedron with an infinite number of faces/vertices. Therefore, inscribed/circumscribed polyhedra that have more faces/vertices will have a closer volume (and surface area) to a sphere with the same "radius".

When inscribing, the "radius" applies to the vertices, so the one with the closer volume would be the one with more vertices - i.e., a cube. This would therefore be the larger one. When circumscribing, the "radius" applies to the middle of each face, so the one with the larger number of faces would have a closer volume to the sphere - which is the octahedron. So therefore the cube should have a larger volume again.

So does the cube have a larger volume in both cases? Tell me if there is a flaw in my logic. Applying this to the other problem, I get that a dodecahedron would have a larger volume both inscribed and circumscribed.

[Kurt]

[AuSmith]