How to calculate the volume of a regular convex icosahedron?

Question

I know that the volume of an icosahedron is $\frac {5(3+\sqrt{5})} {12}x^3$, where x is the length of any given side. However, I am not sure how to prove it. I have looked into a few methods, yet none of my research has yielded a solid method to find this formula. Any help would be appreciated

Answer 1

Here is how I would go about doing this calculation.

Begin by identifying a ring of edges looping around any vertex and making five plane cuts, one through each ring edge and the center of the icosahedron. This sets off a pentagonal bipyramid with one apical vertex being the vertex around which the ring was originally drawn and the other apical vertex at what was the center.

The icosahedron is then a superposition of twelve of these bipyramids, which overlap three at a time throughout the volume of the icosahedron (because each face of the icosahedron is triangular). So the volume of the icosahedron must be four times the volume of the bipyramid, from which we may then render

$V=4[(1/3)(A)(l/2)]=(2/3)Al,$

where $V$ is the volume, $A$ is the area of the regular pentagonal "equator", and $l$ is the length of a diagonal through the center of the icosahedron.

The area of the pentagon is half the perimeter times its inradius, thus

$A=(5x/2)(x\tan54°/2)=(5x^2/4)(\phi/\sqrt{3-\phi})=(5x^2/4)(\phi^2/\sqrt{1+\phi^2}),$

where $\phi=(1+\sqrt5)/2$ and we have used the fact that $\phi+(1/\phi^2)=2$.

The length of the diagonal is a bit simpler to figure out if you recognize the embedded golden rectangles in a regular icosahedron. Then the required diagonal is the hypotenuse of a right triangle whose legs are an edge and the diagonal of one of the regular pentagonal rings described above, thus:

$l=x\sqrt{1+\phi^2}.$

We therefore have

$V=(2/3)[(5x^2/4)][(\phi^2/\sqrt{1+\phi^2})][x\sqrt{1+\phi^2}]=(5x^3/6)\phi^2,$

from which the claimed result emerges by putting in $\phi=(1+\sqrt5)/2$.

Answer 2

Consider one of the pentagonal caps in the icosahedron (seen from inside in figure below). To compute the volume, you need distance $OG$ between the centre $O$ of the icosahedron and the centre $G$ of a face. If you consider two neighbouring faces and the midpoint $H$ of their common edge, line $OH$ bisects at $M$ segment $AC$, joining the non-common vertices of the triangles.

But triangles $AHM$ and $OHG$ are similar. If the edges have unit length, then $AH=\sqrt3/2$, $HG=\sqrt3/6$, $AC=(1+\sqrt5)/2$ (diagonal of a regular pentagon with unit sides) and $MH=(\sqrt5-1)/4$ (by Pythagoras theorem). From: $$ OG:AH=HG:MH $$ one can then find $$ OG={\sqrt3(\sqrt5+3)\over12}. $$ You can now easily compute the volume of a pyramid with vertex $O$ and a face as base: multiply that by $20$ and you'll get the volume of the icosahedron.