Before going further with the dodecahedron some pentagonal
expressions must be derived.


The radius of the circumscribed circle for the pentagon is

1/(2cos(3pi/10)) = [2]/[5-[5]] = [5+[5]]/[10]

The inscribed circle radius of the altitude of one of the
five triangles making up the pentagon is

   (1/2)tan((3pi/10) =

   (1/2)([3+[5]]/[5-[5]]) = (1/2)([20+8[5]]/[20]) = (1/2)([5+2[5]]/[5])

The inscribed ot circumscribed ratio is

   (1/2)([5+2[5]]/[5]) / [5+[5]]/[10] = 1/[2] [15+5[5]]/[20]
       = [3+[5]]/(2[2]) = (1+[5])/4

The circumscribed to inscribed ratio is

   [5+[5]]/[10]  2[5]/[5+2[5]] = [2][15-5[5]]/[5] = [2][3-[5]] = [5]-1

The area of the pentagon is then

  5 (1/2) (1/2)([5+2[5]]/[5]) = [5][5+2[5]]/4



Now going back to the expression in proof_new.txt
for the dodecahedron inscribed sphere radius in terms of an edge.


  1/([2][25-11[5]]) = 1/[2]  [25+11[5]]/[625-605] = [25+11[5]]/[40]


Using the inscribed sphere radius of the dodecahedron the volume of the
dodecahedron is then


  12/3 [25+11[5]]/[40]  [5][5+2[5]]/4   = 5[235+105[5]]/[200]

    = [5][47+21[5]]/[8]


To determin whether an expression of the form [a+b[5]] is a perfect square
of the form c+d[5] consider if
             [a+b[5]] = c+d[5]
             = [c^2 + 5d^2 +2cd[5]]
then    a = c^2 + 5d^2 and b = 2cd

or      a = c^2 + 5/4 b^2/c^2

            c^4 - a c^2 + 5/4 b^2 = 0

has a real solution if  a^2 - 5b^2 = 47^2 - 5 21^2> 0


Now 47^2 - 5 21^2 = 1600+560+49-2205 = 4 > 0 so it may be possible to
simplify the expression for the volume of the dodecahedron.
c^2 = (a +- 2)/2 = 45/2 , 49/2      d =  21/[90]  , 21/[94]

multiplying by 5/8  c^2 = 225/16 c=15/4    d= 21(5/8) 2/15 = 7/4            


So the volume of the dodecahedron can also be expressed as

           (15+7[5])/4


These expressions are rendered in MathML at pentagon.xml


ref:
  /wiki/Nth_root
  /wiki/Periodic_continued_fraction#Reduced_surds
  /wiki/Continued_fraction#Continued_fraction_expansions_of_.CF.80
  /wiki/Galois_theory

(c)2009 Wm.C.Corwin  billc(a)issi1(dot)com
-----------------------------------------------------------------------

superfulous:

21 has factors 3 and 7 and what c and d could have cd/2 / c^2+5d^2 = 21/47

    470 = c^2 + 5 d^2  and 210 = 2cd

    235 = 5 47  ;      105 = 2cd =  3 35 = 3 5 7
    470 = 2 5 47 ;     210 = 2 3 5 7
    = 225 + 245

        5 3^2 + 7^2 = 3 15 + 7^2 = 94 = 2 47 
    = 15^2 + 5 7^2     210 = 2 7 15
      c = 15  ;  d = 7

          (15 + 7[5])/4  = [225+245 + 210[5]]/4 = [5/8][47+21[5]]



(c) 2009 Wm.C.Corwin  www.ConcurrentInverse.com