# Problem 48 Solution

From problem 123 we have the area of a pentagon of side 1 to be sqr(25+10*sqr(5))/4. The critical question to be answered is the distance from the center of a face to the center of the dodecahedron.

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
BQ2 + QP2 = BP2
BQ = (sqr(3)+sqr(15))/6
AQ = sqr((3-sqr(5))/6)

So:

PQ2 + BQ2 = BP2, or

PQ2 + ((sqr(3)+sqr(15))/6)2 = BP2, or

(BP - sqr((3-sqr(5))/6))2 + ((sqr(3)+sqr(15))/6)2 = BP2 (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.