This solution is going to be hard to explain without pictures so bear with me. Call any corner of the dodecahedron a. Call the three corners surrounding a to be b, c, and d. Call q the point at the center of triangle bcd. Call p the point at the center of the dodecahedron.

From problem 123 we know the distance from two non-adjacent corners in a pentagon to be (1+sqr(5))/2. Knowing this and that cos(30)=sqr(3)/2 we can determine that BQ=(sqr(3)+sqr(15))/6.

It is given that AB=1, using the pythagorian formula we can deduce AQ from AB and BQ, which we know. After going through the math we find AQ to be sqr((3-sqr(5))/6).

Lets review what we know so far:

AQ + QP = AP

BP = AP

BQ^{2} + QP^{2} = BP^{2}

BQ = (sqr(3)+sqr(15))/6

AQ = sqr((3-sqr(5))/6)

So:

PQ^{2} + BQ^{2} = BP^{2}, or

PQ^{2} + ((sqr(3)+sqr(15))/6)^{2} = BP^{2}, or

(BP - sqr((3-sqr(5))/6))^{2} + ((sqr(3)+sqr(15))/6)^{2} = BP^{2} (substituting sqr((3-sqr(5))/6) + PQ = BP)

After working through the math we get BP = sqr(6*(3+sqr(5)))/4

From problem 123 we know the distance from the corner of a pentagon to the center of the pentagon is 2*sqr(5+2*sqr(5))/(5+sqr(5)) .

Lets call the center of a pentagon E, where B is one of the corners of the pentagon. We now know the distance from B to P and we know the distance from E to B from problem 123. Next we must work through the pythagorean formula again to solve for EP, which turns out to be sqr((25+11*sqr(5))/40) =~ 1.113516.

The volume of the dodecahedron is composed of 12 pyramids. Each pyramid has base sqr(25+10*sqr(5))/4 and height (25+11*sqr(5))/40. The volume of each of these pyramids is (1/3) * sqr(25+10*sqr(5))/4 * sqr((25+11*sqr(5))/40)

The volume of the entire dodecahedron is 4 * sqr(25+10*sqr(5))/4 * sqr((25+11*sqr(5))/40)

= sqr(25+10*sqr(5)) * sqr((25+11*sqr(5))/40) =

(1/sqr(40)) * sqr(625+275*sqr(5)+250*sqr(5)+550) =

(1/sqr(40)) * sqr(1175+525*sqr(5)) =

(5/sqr(40)) * sqr(47 + 21*sqr(5)) =

(sqr(10)/4) * sqr(47 + 21*sqr(5)) =

(sqr(5)/4) * sqr(94 + 42*sqr(5)) =

(sqr(5)/4) * sqr(49 + 42*sqr(5) + 45) =

(sqr(5)/4) * sqr(((7+3*sqr(5))^{2}) =

(sqr(5)/4) * (7+3*sqr(5)) =

(15+7*sqr(5))/4 =~ 7.663119

Thanks to Mark Hoyle for some help with this problem.

Michael Shackleford, A.S.A.