www.ConcurrentInverse.com
                                      rev A June  4, 2005
                                      rev B June 19, 2005
                                      rev C June 29, 2005
                                      rev D July 11, 2005
                                      (c) copyright 2005 Wm.C.Corwin



http://www.cut-the-knot.org/htdocs/dcforum/DCForumID4/606.shtml

This discussion is maintained at
http://www.issi1.com/corwin/calculator/proof.txt


There will always be a use for large prime numbers, eg 111, when expressing
approximations to irrational numbers.  There are 11*6 feet in a chain,
13 weeks in a quarter, and 2*11/7 = 3.1428 = Pi + 0.0013.

666 - 720 = 54 = 270/5 deg = 3 Pi / 10 radians
  pi/5 is related to the golden ratio
616 - 630 = 14 is not likely to be related to the golden ratio


   5**1/2    = 2.2360679774997896964091736687313
  (5/7)**1/2 = 0.84515425472851657750961832736595
  (5/7)**1/3 = 0.89390353509656765128791642511372

  1/ Phi = Phi -1
  Phi**2 - Phi -1 = 0
  2*Phi = 1 +-(1+4)**0.5 = 1 +- (5)**0.5 = 2* 1.6180339887498948482045868343656



Phi and pentagonal symmetry are discussed in depth at 
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/phi2DGeomTrig.html#trigmore

    1/(5)**1/2 - 0.44721359549995793928183473374626
  pentagonal symmetry  pi/5 = 36deg
    sin36 = 0.58778525229247312916870595463907 = ((5-5**0.5)/8)**0.5
    cos36 = 0.80901699437494742410229341718282 = Phi /2 = (1+(5)**0.5)/4
    tan36 = 0.72654252800536088589546675748062 = (5-2*5**0.5)**0.5

    sin72 = 0.951057 = ((5**0.5+5)/8)**0.5
    cos72 = 0.309017 = (5**0.5-1)/4
                     >< (5**0.5-1))**0.5/2  error in REV A

   Thus in pentagonal symmetry there is a correspondence or relationship of
     Phi, 5**0.5 and Pi/5 . Why or how can this be derived?

      36-54 triangle  hypotenuse  = 4       
                side opposite 36  = 2 Phi = 2.35114100 = 2*1.17557050
                side opposite 54  =       = 3.236067977



    ref:  http://www.mcs.surrey.ac.uk/Personal/r.Knott/Fibonacci/simpleTrig.html#exactrig
            exact trig values
            http://www.rwgrayprojects.com/OswegoOct2001/Presentation/presentationWeb4.html
              Gerald de Jong
    ref:  http://www.mcs.surrey.ac.uk/Personal/r.Knott/Fibonacci/phi2DGeomTrig.html#trigmore


I have observed the phenomenon when I calculated the dodecahedral and
icosahedral angles.

-----------------------------------------------------------------------------
    Proof of simple exact formulas in closed form for the dimensions
          of the dodecahedron and icosahedron.


To illustrate the method of calculating the dihedral angle of an
dodecahedron first consider the square pyramid with equilateral
triangles for four faces i.e. one half of a octahedron.
Alpha is the angle between a unit vector normal to the base and
a vector along the edge of the equilateral triangles.  Consider alpha
as an independent variable. The projection of the edge on the base will be
                  cos(alpha)
If it varies the length of the side of the pyramid base will be
the base of an icosceles triangle with side cos(alpha) and angle pi/3.
              2*cos(alpha)*sin(pi/4)
When the base has sides 1, then the pyramid faces will be equilateral
triagles.
              2*cos(alpha)*sin(pi/4) = 1 
sin(pi/4) = 1/2**0.5, cos(alpha) = 1/2**0.5, and alpha = pi/4 as is
well known.

If the following proofs are difficult to follow without extensive
calculations on your part consider them hints to keep you on the right
path and mileposts to make your errors evident.

THEOREM 1:  The angle, beta, between unit vectors for adjacent faces of the
             icosahedron is given by cos(beta) = 5**0.5/3  

Next consider a pentagonal jello mold with each if the five
sides a cylinder centered at the center of the pentagon base with unit
vectors going from the center of the pentagon and scribing an arc to the
nadir.  At some angle the unit vector points are unity apart; this defines
a pyramid with equilateral triangles for faces.  The sides of the
base are then 2*sin(pi/5) = 1.176 . If the five unit vectors are lowered by
an angle gamma then the sides of the lowered pentagon are
2*cos(gamma)*sin(pi/5) = 1 .  Gamma is the angle between the normal to the
base and unit vector from the peak to one of the five corners along the edge
between two equilateral triangles.

Now sin(pi/5) = ((5-5**0.5)/8)**0.5 so cos(gamma) = (2/(5-5**0.5))**0.5

Consider three of the above adjacent unit vectors R,S,T with S having no
x component.

       R = ( ((5+5**0.5)/8)**0.5 cos(gamma),((5**0.5-1)/4) cos(gamma),sin(gamma))
       S = (0,cos(gamma),sin(gamma))
       T = (-((5+5**0.5)/8)**0.5 cos(gamma),((5**0.5-1)/4) cos(gamma),sin(gamma))
               
P = R cross S / sin(pi/3) and Q = S cross T / sin(pi/3) are normal unit
vectors to two of the equilateral faces.

 P = ((1-cos(2pi/5)cos(gamma)sin(gamma),sin(2pi/5)cos(gamma)sin(gamma),
                                  sin(2pi/5)cos(gamma)cos(gamma))/sin(pi/3)
 Q = (-(1-cos(2pi/5)cos(gamma)sin(gamma),sin(2pi/5)cos(gamma)sin(gamma),
                                  sin(2pi/5)cos(gamma)cos(gamma))/sin(pi/3)

The angle, beta, between the normal vectors to two icosahedron faces is then
given by

  cos(beta) = P dot Q =

   = -(((1-cos(2pi/5))**2-sin2(2pi/5))cos2(gamma)sin2(gamma)+sin2(2pi/5)cos4(gamma))/(3/4)

Now  sin2(2pi/5) = (5+5**0.5)/8
     cos2(gamma) = (5+5**0.5)/10
     sin2(gamma) = (5-5**0.5)/10
    1-cos(2pi/5) = (5-5**0.5)/4  .

   cos(beta) = (-(((5-5**0.5)/4)**2-((5+5**0.5)/8)(20/100) + ((5+5**0.5)/8)*(30+10*5**0.5)/100))/(3/4)
   cos(beta) = (-(15-5*5**0.5-5-5**0.5)/40) + (20+8*5**0.5)/80)/(3/4)

              cos(beta) = 5**0.5/3      QED

              beta = 41.8103 deg = 0.7297 radians = 0.2322795 pi


The angle of the face normal from the vertical is given by
        cos(delta) = sin(2pi/5)cos(gamma)cos(gamma))/sin(pi/3) .
              = ((5+5**0.5)/8)**0.5 * (2/(5-5**0.5)) / ([3]/2)
              =   (5+5**0.5)**1.5 [2]/(20[3])
                  [5+[5]][30+10[5]] /(10[6])
                  [200+80[5]] /(10[6])     
              =   [5+2[5]] /[15] = 

         delta = arccos([2+[5]]/[15])

?Also it is the inverse cotangent of the circumscribed pentagon over the
?circumscribed sphere
?
?       cos(delta) = [10+2[5]]/2[5] / ([5][3+[5]]/[6*7*8])
?                  = ([6*7*8]/2)  [20-4[5]] / 2
?                  =  [6*7*8] [5-[5]] / 2    ><
?                      7[2][5-[5]]/(5[3])    ?
?Then the angle to the second middle row is
               epsilon = delta + beta .
             cos(epsilon) = cos(delta)cos(beta) - sin(delta)sin(beta)
               = ([5]/3) [2+[5]]/[15] - (2/3) [13-[5]]/[15]
              this may be fruitless

Now the independent variable, now alpha, instead of the angle between an edge
and the horizontal for the icosahedron will be the angle between face normals
for the dodecahedron.
----------------------------------------------------------------------------
THEOREM 2:
Given that cos(pi/5) = (1+(5)**0.5)/4 which is proven at
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/simpleTrig.html#3654
I will show that, alpha, the angle between normals to adjacent faces for the
dodecahedron is given by cos(alpha) = 1/5**0.5 .
PROOF:

Now consider the dodecahedron and three unit vectors normal to the
top face, one of the faces in the upper row of five, and one of the
faces in the lower row adjacent to the second face.

         A = (0,0,1)
         B = (0, sin(alpha), cos(alpha))
         C = (sin(pi/5)sin(alpha),cos(pi/5)sin(alpha),-cos(alpha))

For a regular solid the angles between the normals to adjacent faces must
be the same.

              A dot B = B dot C

Thus cos(alpha) = cos(pi/5)(sin(alpha))**2 - (cos(alpha))**2

Using  cos(pi/5) = (1+5**0.5)/4  and  1+cos(pi/5) = (5+5**0.5)/4

gives the quadratic

(5+(5)**0.5)*(cos(alpha))**2 + 4 cos(alpha) -(1+(5)**0.5) = 0

which gives  cos(alpha) = 1/5**0.5  .


The immediate result of the quadratic formula is

      ( -2 + (14 + 6 * 5**0.5)**0.5))  /  (5 + 5**0.5)

It is not immediately obvious that this is 1/5**0.5  .  Nor is
it readily apparent how to prove it.  Numerically it is correct
to as many places as you want to compute but it takes multiple
cross multiplications, conjugate multiplications, and squarings
to prove it.  To the uninitiated it would appear likely that successive
calculations would explode with the number of terms growing geometrically
and one would have to proceed from here only numerically.
However, it is possible to contain the complexity by manipulating square
roots out of the denominator and multiplying out factors.

 ---gore:---------------------------------------------------------------
|             2**0.5 * (7 + 3 * (5)**0.5)**0.5) -2 ) / (5 + (5)**0.5) ? 1/5**0.5
|  1/5**0.5 * 2**0.5 * (7 + 3 * (5)**0.5)**0.5) - 2 *  1/5**0.5 ? 5 + (5)**0.5
|  1/5**0.5 * 2**0.5 * (7 + 3 * (5)**0.5)**0.5) ?  5 + 3 * 5**0.5
|             2**0.5 * (7 + 3 * 5**0.5)**0.5) ?  3 + 5**0.5
|             2**0.5 ?     (3 + 5**0.5) / (7 + 3 * 5**0.5)**0.5)
|             2**0.5 ?    ((3 + 5**0.5) / (7 - 3 * 5**0.5)**0.5))/(49-45)**0.5
|             2**1.5 ?     (3 + 5**0.5) / (7 - 3 * 5**0.5)**0.5)
|             2**1.5 *     (3 - 5**0.5) /4 ? (7 - 3 * 5**0.5)**0.5)
|             (8/4**2) * (14 - 6 * 5**0.5 ) = 7 - 3 * 5**0.5       QED
 ----------------------------------------------------------------------------
         alpha = 1.1071487 radians = 0.35241638 pi


THEOREM 3: The dot product of vector B and vector D which is obtained from B
           by rotating it by -2pi/5 about the z axis is 1/5**0.5 .
           Also C dot D is 1/5**0.5 .

The rotation matrix applied to B is 

           
               |  cos(2pi/5)     sin(2pi/5)    0  |   |    0       |
        D =    | -sin(2pi/5)     cos(2pi/5)    0  | * | (4/5)**0.5 |
               |  0              0             1  |   | (1/5)**0.5 |

               ><?

         B = (0, (4/5)**0.5 , 1/5**0.5)
         D = ( sin(2/5 pi) * sin(alpha) , cos(2/5 pi) * sin(alpha), cos(alpha)
         D = ( ((5**0.5+5)/8)**0.5 * (4/5)**0.5, (5**0.5-1)/4 * (4/5)**0.5, 1/5**0.5)
                                                (3-5**0.5)/8)**0.5

          B dot D = 0 + (4/5) * (5**0.5 - 1)/4 + 1/5
                  = 5**0.5 / 5

                B dot D = 1/5**0.5

Also C dot D = (sin(pi/5)*sin(alpha),cos(pi/5)*sin(alpha),-cos(alpha)) dot D
  = (((5 - 5**0.5)/8)**0.5 * (4/5)**0.5,((1 + 5**0.5)/4) * (4/5)**0.5, -1/5**0.5) dot D
  = (4/5) *  ((5 + 5**0.5) * (5 - 5**0.5))**0.5 /8 + (4/5) * (5**0.5+1)*(5**0.5-1)/16 - 1/5
  = (4/5) *  20**0.5 /8 + (4/5) * 4 /16 - 1/5
  =  1 / 5**0.5

corollary:
The angle between all adjacent faces are all equal by rotating the vectors
B, C, and D and inverting through the x,y plane.  Since B dot D each face in
both the top and bottom row are at the angle alpha; since C dot D = C dot B
= 1/5**0.5 each face in each row is at angle alpha with the faces in the
other row.  Therefore all 12 faces in the the 12 faced figure that we are
at the anlge alpha and the figure that we are studying is a regular solid.


THEOREM 4:  A unit vector to a vertex in the top row of the dodecahedron
         The radius of the superscribed sphere over the radius of the
         inscribed sphere if the dodecahedron is 3**0.5(5-2*5**0.5)**0.5

The faces of the icosahedron are at the vertices of the dodecahedron
so a vector a icosahedron face would be given by the intersection
of three dodecahedral faces.

         A = (0,0,1)
         B = (0, (4/5)**0.5 , 1/5**0.5)
         D = ( sin(2/5 pi) * sin(alpha) , cos(2/5 pi) * sin(alpha), cos(alpha)
         D = ( ((5**0.5+5)/8)**0.5 * (4/5)**0.5, (5**0.5-1)/4 * (4/5)**0.5, 1/5**0.5)
                                                (3-5**0.5)/8)**0.5

Three simultaneous equations giving a vector, V=(x,y,z) normal to a
icosahedral face are

         V dot A = V dot B = V dot D =  S  ,

where S is the radius to the dodecahedron face.

                           z =  S
                   (4/5)**0.5 * y + 1/5**0.5 * z = S
       (5**0.5+5)/8)**0.5 * (4/5)**0.5 * x + (5**0.5-1)/4 * (4/5)**0.5 * y + 1/5**0.5 * z = S

    ---gore:---------------------------------------------------------------
   |                 y = ((1 - 1/5**0.5) / (4/5)**0.5 ) * S
   |                   =  (5**0.5 - 1)/2
   | (5**0.5+5)**0.5 /2**0.5 * x + (5**0.5-1)/2 * y =  (5**0.5 - 1) * S
   | (5**0.5+5)**0.5 /2**0.5 * x = (5**0.5 -1 -  (5**0.5 - 1)*(5**0.5 - 1)/4) * S
   |                             = (5**0.5 -1 -  (5 + 1 -2*5**0.5)/4) * S
   |                             = (3*5**0.5 -5)/2) * S
   |                           x = (3*5**0.5 -5)/((5**0.5 + 5) *2)**0.5 * S 
    -----------------------------------------------------------------------

              x = (3*5**0.5 -5)/((5**0.5 + 5) *2)**0.5
              y = (5**0.5-1)/2  
              z = 1

To normalize to a unit vector:
    ---gore:---------------------------------------------------------------
   |           N**2 = x**2 + y**2 + z**2
   |       =  (3*5**0.5 - 5)**2 / (2*(5 + 5**0.5)) + (5**0.5 -1)**2/4 + 1
   |       =  (4+5+1-2*5**0.5)/4 + (70 - 30*5**0.5) / (2*(5 + 5**0.5))
   |       =   (5 - 5**0.5) / 2 + 5*(7 - 3*5**0.5)/(5+5**0.5)
   |       =  (20 + 70 - 30*5**0.5)/(2*(5+5**0.5))
   |       N**2 = 15 * (3 - 5**0.5) / (5 + 5**0.5)
   |
   |       N**2 =  3(5-2*5**0.5)
   |
   |       1/N = (5+2*5**0.5)/15**0.5
    -----------------------------------------------------------------------
                            QED

corrolary:
          V/N = (x/N, y/N, z/N)
              =(3*5**0.5 - 5)/(30**0.5(3-5**0.5)**0.5,
               ((5**0.5 - 1)/2) *  ((5**0.5 + 5)/(15**0.5(3-5**0.5)**0.5)),
               (5**0.5 + 5)/(15**0.5(3-5**0.5)**0.5)  )
              =((3-5**0.5)/6)**0.5,
                (5+5**0.5)/30)**0.5,
                (5+2*5**0.5)/15)**0.5)



The cosine of the angle from the center of a face to a vertex is then given
by the dot product of the unit vector to the face, A, and the unit vector
to the vertex V/N.

          cos(FV) = A dot V/N  =  (5+2*5**0.5)/15)**0.5    QED


In order to check the exact equations we will evaluate them and see
if the angles between three faces and the common vertex are equal.

    ---numerical check:---------------------------------------------------------------
   |         A = (0,0,1)                        top
   |         B = (0, 0.8944, 0.4472)            top row at 0
   |         C = (0.5257, 0.7236, -0.4472)  bottom row at pi/5
   |         D = (0.8506, 0.2764, 0.4472)       top row at 2/5 pi from y axis
   |         E = (-0.8506, 0.2764, 0.4472)       top row at -2/5 pi from y axis
   |         y/z = 0.6180
   |         V   = (0.4491, 0.6180. 1.00)
   |                
   |         N   = 1.2584
   |         V/N = (0.3569, 0.4911, 0.7947)
   |
   |
   |         V =  (0.4491, 0.6180, 1.00)
   |         N =  1.2584 
   |         V/N = (0.3568, 0.4911, 0.7946)
   |         W = (?,?,?)
    -----------------------------------------------------------------------
A "great semi-circle" path around the dodecahedron would include segments
of the face center to vertex, then vertex to vertex, then heigth to pentagon,
then pentagon to side to face center which would subtent angles totalling
pi.  A quadrant could include segmants from the center of one edge, across
the heighth of one pentagon through the center to the center of a transverse
edge.  The segments would subtend beta/2, alpha/2, and delta which is the
angle subtended by the segment from the center of a pentagon to a vertex.
Delta is given by
                cos(delta) = A dot V/N = (5+2*5**0.5)/15)**0.5

           beta/2  =  20.905157
           alpha/2 =  31.717474
           delta  =   37.377368
              beta/2 + a;pha/2 + delta = 89.999999

THEOREM 5:  A unit vector to a vertex in the bottom row of the dodecahedron

If we changed the sign on the x componet of D the we would get a vector to
another dodecahedron vertex or icosahedron face.

         A = (0,0,1)
         B = (0, (4/5)**0.5 , 1/5**0.5)
         E = ( - ((5**0.5+5)/8)**0.5 * (4/5)**0.5, (5**0.5-1)/4 * (4/5)**0.5, 1/5**0.5)

To get a vector for the icosahedron face in the middle row of five we need
a vector to one face of the lower row on the dodecahedron.

         B = (0, (4/5)**0.5 , 1/5**0.5)
         D = ( ((5**0.5+5)/8)**0.5 * (4/5)**0.5, (5**0.5-1)/4 * (4/5)**0.5, 1/5**0.5)
         C = (sin(pi/5)sin(alpha),cos(pi/5)sin(alpha),-cos(alpha))
         C = ( ((5-5**0.5)/8)**0.5 * (4/5)**0.5, (1+5**0.5)/4 * (4/5)**0.5, -1/5**0.5 ) 

         W dot B = W dot D = W dot C = S
             W = (x,y,z) 

                2y/5**0.5 + z/5**0.5 = S
  ((5+5**0.5)/8)**0.5 * 2x/5**0.5 + (5**0.5-1)/4 * 2y/5**0.5 + z/5**0.5 = S
  ((5-5**0.5)/8)**0.5 * 2x/5**0.5 + (5**0.5+1)/4 * 2y/5**0.5 1 z/5**0.5 = S

   2**0.5((5+5**0.5)**0.5 + (5-5**0.5)**0.5)x + 2*5**0.5 y = 4*5**0.5 S
   2**0.5((5+5**0.5)**0.5 - (5-5**0.5)**0.5)x - 10 y      = -4*5**0.5 S


   x =  2*10**0.5 * ((5-5**0.5)/(5+5**0.5)**0.5) /
           ( 5(1+((5-5**0.5)/(5+5**0.5))**0.5) + 5**0.5(1-((5-5**0.5)/(5+5**0.5))**0.5) )


  ------------lemma:-------------------------------------------------------
 |
 |   1/( 5(1+((5-5**0.5)/(5+5**0.5))**0.5) + 5**0.5(1-((5-5**0.5)/(5+5**0.5))**0.5) )
 |
 |
 |
 |
 |    = ((7+3*5**0.5)**0.5 - (3-5**0.5)**0.5)/(8*10**0.5)
 |
  --------------------------------------------------------------------------

   y = ((7+3*5**0.5)**0.5 - (3-5**0.5)**0.5)) / 2*2**0.5

   z = 5**0.5 -2

  ------------lemma:-------------------------------------------------------
 |
 |   (7+3*5**0.5)**0.5 - (3-5**0.5)**0.5) = 2*2**0.5
 |
  --------------------------------------------------------------------------

      ...

           x = ((5**0.5+5)**0.5 - (25 - 11*5**0.5)**0.5) / (2*2**0.5)
           y = 1
           z = 5**0.5 -2

           x/N =  ((7+3*5**0.5)**0.5 - (3-5*5**0.5)**0.5)/ (2*6**0.5)2
           y/N =  (20+8*5**0.5)**0.5 / (2*15**0.5)
           z/N =  (20-8*5**0.5)**0.5 / (2*15**0.5)
                                                    QED

The vectors to all twenty faces can be obtained by rotating them by pi/5
about the z axis and then reflecting them through the xy plane by replacing
z with -z.

THEOREM 6:
The angle, beta, between the normal vectors to the icosahedron faces is given by


         cos(beta) = V/N dot W/N = 5**0.5/3          QED

corollaries:
The distance of the dodecahedral vertex from the center of the solid is
             (15 - 6*5**0.5)**0.5  .
The distance of the center of the edge to the center of the solid is
             (5/2)**0.5 * (5**0.2 -1)**0.5
The length of the edge is
             (50 - 22*5**0.5)**0.5  .
The height of the pentagonal face is
             (5/2)**0.5 * (3-5**0.5)**0.5 .
               (20 + 8*5**0.5)/4 * (50-22*5**0.5)**0.5
                 = (1000-880-(440-400)*5**0.5)**0.5/4
                 = 10**0.5 * (3-5**0.5)**0.5 / 2       QED
The distance of the vertex to the center of the face is
             (14 - 6*5**0.5)**0.5  .
The length of the perpendicular from the face center to the edge center is
             ((3 - 5**0.5)/2)**0.5  .

THEOREM 7:  Superscribed sphere to inscribed sphere for icosahedron

The angle subtended by the distance from the center of an equilateral
triangle face to the midpoint of one of the edges is beta/2 where
beta is the angle between the normals to two faces of the icosahedron.

Using the the distance from the center of an equilateral triangle to
a side as 1, the distance from the center of the icosahedron to the
center of one of the edges is
                    1/sin(beta/2)

        cos(beta/2) = (0.5 + 0.5*cos(beta))**0.5
                    = ((3+5**0.5)/6)**0.5
        sin(beta/2) = ((3-5**0.5)/6)**0.5
The face center to solid center is 1/tan(beta/2)
        tan(beta/2) = ((3-5**0.5)/6)**0.5 * ((3+5**0.5)/6)**0.5
                    = (14-6*5**0.5)**0.5 / 2
                    = (7 - 3*5**0.5) / 2**0.5

        1/(sin(beta/2)) =  6**0.5 * (3-5**0.5)**0.5 * (3+5**0.5) /4
                        =  6**0.5 * (3-5**0.5)**0.5 * (14+6**0.5)**0.5 /4
                        =  6**0.5 * (3+5**0.5)**0.5 /2

The distance from the center of a face to the vertex is 3**0.5 so the
superscribed radius R is given by
             R**2 = 3*(3+5**0.5)/2 + 3
                  = 15/2 + (3/2) * 5**0.5
The inscribed radius is
              r = 1/tan(beta/2)
so        R/r = (3/2) * (5+5**0.5)**0.5 * (7-3*5**0.5)**0.5 * (1/2**0.5)
              = (3**0.5/2) * (20-8*5**0.5)**0.5
              = 3**0.5 * (5 - 2*5**0.5)**0.5   QED

It is the same as that for the dodecahedron. Therefore it should be possible
to find simple exact formulas for all the dimensions of the icosahedron as
well.

corollary:
The triangular face edge length is
       2*3**0.5 / ((7 - 3*5**0.5) / 2**0.5) = 6**0.5 * (7-3*5**0.5)**0.5
The triangular face height length is
       2*3**0.5 * 6**0.5 * (7-3*5**0.5)**0.5 = (2/2**0.5) * (7-3*5**0.5)**0.5


APPENDIX A
 trig
   half angle cosine
     cos(A/2) = (1/2 + (1/2)cos(A))**0.5

     cos(2pi/5)   = (5**0.5 - 1)/4
     sin(2pi/5)   = ((5**0.5 + 5)/8)**0.5
     cos(pi/5)    = (5**0.5 + 1)/4
     sin(pi/5)    = ((5 - 5**0.5)/8)**0.5


 equilateral triangle
       height                        3            3**0.5/2
       center to one side            1
       center to vertex              2
       side                          2*3**0.5     1

       side/height                   2/3**0.5
       area                          3*3**0.5     3**0.5/4

 pentagon
       each side subtends                 4 pi/10
       angle at vertex for each triangle  3 pi/10
       right triangle in each triangle
         central angle                    pi / 5
         vertex angle                     3 pi/10
         sin(pi/5)                        ((5-5**0.5)/8)**0.5
         cos(pi/5)                        (5**0.5 + 1)/4

       inscribed radius              (1+5**0.5)/4        1                     (5+2*5**0.5)**0.5/(2*5**0.5)   = 0.68819
       superscribed radius           1                   5**0.5 - 1
       height                        (5+5**0.5)/4        5**0.5                (5+2*5**0.5)/2  
       side                          ((5-5**0.5)/2)**0.5 2*(5-2*5**0.5)**0.5   1
       break                         (3+5**0.5)/4        5**0.5 - 2

       height/side                   (5 + 2*5**0.5)**0.5/2 = 1.53884
       area/5                                                                  (5+2*5**0.5)**0.5/(4*5**0.5)
       area in square sides                                                    (5**0.5 * (5+2*5**0.5)**0.5)/4  
                                                                             
APPENDIX B    SUMMMARY

 dodecahedron
             face normal angle      arccos(1/5**0.5)                 63.4349488 deg

             edge length            (50 - 22*5**0.5)**0.5           = 0.89805595
             inscribed radius       1
             center of edge radius  ((5 - 5**0.5)/2)**0.5           = 1.17557
             superscribed radius    3**0.5 * (5 - 2*5**0.5)**0.5    = 1.25840857
             pentagon height        (5/2)**0.5 * ( 3 - 5**0.5)**0.5 = 1.381966

             surface area           30 * 2**0.5 * (65 - 29*5**0.5)**0.5
             volume                 10 * 2**0.5 * (65 - 29*5**0.5)**0.5  = 5.55029  
                                    4pi/3                                = 4.1887

 icosahedron
             face normal angle      arccos(5**0.5/3)

             edge length            6**0.5 * (7 - 3*5**0.5)**0.5       = 1.32317  
             inscribed radius       1
             center of edge radius  (3/2)**0.5 * (3 - 5**0.5)**0.5
             superscribed radius    3**0.5 * (5 - 2*5**0.5)**0.5
             triangle heigth        (3/2**0.5) * (7 - 3*5**0.5)**0.5   = 1.145898

             surface area           30 * 3**0.5 * (7 - 3*5**0.5)      
             volume                 10 * 3**0.5 * (7 - 3*5**0.5)       = 5.05406
                                                                       = 5.05405
APPENDIX C  IDENTITIES  simplification by removing radicals in the denominator

  Phi = (1+[5])/2 = 2/(1-[5]) = 1/phi = phi + 1


 (a+b*5**0.5)**0.5/(c+d*5**0.5)**0.5 = ((ac-bd*5) +(bc-ad)*5**0.5))**0.5 / (c**2 - 5*d**2)**0.5

    examples; let % = 5**0.5 and [x] = x**0.5
  [a+b%] / [c+d%] = [ac-5bd + (bc-ad)%]/ [c**2 - 5d%]

   [5-%]/[3-%]  = [10+2%] / 2 = [5+%] / [2]
   [3-%]/[5+%]  = [20-8%] /[20] = [5-2%]/[5] 
   [5-%]/[5+%]  = [30-10%] / [20] = [3-%] / [2] ; [15-5%]/[10] = [3-%]/[2]

   [5+%]/[3-%]  = [20+8%] / 2 = [5+2%]
   [3-%]/[7+3%] = [36-16%]/[4] = [9-4%]
   [5+%]/[7+3%] = [20-8%]/2 = [5-2%]

   [3-%]/[3+%]  = [14-6%]/[4] = [7-3%]/[2]              
   [3+%]/[5+%]  = [10+2%]/[20] = [5+%]/[10] 
   [3-%]/[5+%]  = [5-2%]/[5]   ; [20-8%]/[20] = [5-2%]/[5]

   [7-3%]/(5+%) =   [7-3%][6-2%]/4[5] = [72-32%]/4[5] = [9-4%]/[10]  
   (5+%)/(3-5%) =  (40+28%)  / -116 = (10+7%)/-29 = [345+140%]/-29 = [5][69+28%]/-29 
   [7-3%]/(3-5%) =  [7-3%][134+30%] / -116 = [488-192%]/-116 = [61-24%] / -29[2] ; [61-24%] / -29[2]

   [3+%]/[5+%]  = [10+2%]/[20] = [5+%]/[10] 
   [5+%]/(1+%)  = [5+%][6-2%]/-4 = [20-4%]/-4 = [5-%]/-2 
   [3+%]/(1+%)   =  [3+%](1-%) / -4 = -[3+%][6-2%] / 4  = -[8]/4 = -1/[2] ; -1/[2]

           (5+%) = [10][3+1%]
           (3+%) = [2][7+3%]
           (7+3%) = [94+42%] 
   [7+3%]/(5+%) =   [7+3%][30-10%]/20  = [7+3%][3-1%]/2[10] = [3+1%]/[20]
   (3+%)/[7+3%] =  [14+6%][7-3%] /[4]  =  [49-45]/[2] =[2]           
   (3+%)/(5+%)  = (5+%)/10 = [3+%]/[10]

   (3-%)/[5+%] = [14-6%][5-%]/[20]=[7-3%][5-%]/[10]=[50-22%]/[10]=[25-11%]/[5]
   (5-%)/(3-%) = (10+2%)/4 = (5+%)/2 [30+10%]/2 
   (5-%)/[5+%]  =  [30-10%][5-%] /[20] = [20-8%]/[2] = [10-4%] ;  [200-80%]/2[5] = [10-4%]

    [3+%]/[5-%] = [20+8%]/[20] = [5+2%]/[5]
    (5+%)/[3+%] = [10]
    (5+%)/[5-%]  =  [3+%][5+%] /[2] = [20+8%]/[2] = [10+4%]

One way to check these would be to find two with the same numerator or
denominator, make an identity with the denominator or numerator cancelling
that of the first and replacing it with that for the last; then multiplying
the first with the new yields the last result.

   1/[3+%] = [3-%] / 2
   1/[3-%] = [3+%] / 2
   1/[5-2%] = [5+2%] / [5]
   1/[5+2%] = [5-2%] / [5]

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

Above the three things taken two at a time give six relationships; I
used three in order to check the work.  There may be sets closed under
multiplication and division.  These resemble complex numbers; maybe
there are similar relationships.

All of the relationships can be expressed in expressions consisting of
two terms and a factor.  There is one term of an integer and a second
term of an integer times the square root of five.  The factor is a ratio
of two integers.  Thus each 'form factor' can be a vector or array of
four components each of which is a vector or array of a sign and the
number of times each prime is in the factor.  The array of primes needs
to go to no higher than 29.  A calculator for multiplying these 'form
factors' is in the realm of possibility and could be a first step towards
the computer manipulation of equations.
If the simplification did not exist the expressions would easily become
unmanageably unweildy.

-----------------------------------------------------------------------------

We have levels of complexity
        Pi/2         2**0.5, Pythagoras, square root
        Pi/6         3**0.5
        Pi/5         Phi, golden, quadratic equation root
        dodecahedral quadratic equation root
        Pi/7         cubic equation root
        ...

My discussion of possible and impossible things is at
  http://www.issi1.com/corwin/calculator/possible.txt
Ray tracing is at
  http://www.issi1.com/corwin/calculator/icosahedron.jpg
Other details are at
  http://www.issi1.com/corwin/calculator/pentagon.txt
  http://www.issi1.com/corwin/calculator/platonic.txt
  http://www.issi1.com/corwin/calculator/unit_v.txt
  http://www.issi1.com/corwin/calculator/addition.txt
This page is on line at
  http://www.issi1.com/corwin/calculator/proof.txt
and is summarized at
  http://www.issi1.com/corwin/calculator/summary.txt

cc:golden numerology
     cosmos2000@iquebec.com
            www.666myth.co.nr/
     cosmos2000@iquebec.com?subject=666 Bivalent Myth
       www.chez.com/cosmos2000/Numbers/IsomorphousTriplets.html

After doing these computations I told whoever I thought may be interested
and George Hart replied to me that it was well known and in the books:

Regular Polytropes  by H.S.M.Coxeter Prof. Mathematics University of Toronto
                    isbn 0486614808 Dover
Zome Geometry       by George Hart and Henri Picciotto KeyCurriculumPress 2001
                    isbn 1559533854 KeyCurriculumPress
                    sum and ratio patterns; coordinates; fractals;
                    fourth dimension

13 other references that look like they may be of particular use for
computations are among the 187 listed at:
ref: http://www.georgehart.com/virtual-polyhedra/references.html
       mathematical text focusing on combinatorial issues
         Branko Grunbaum, Convex Polytropes, Interscience, 1967
         Branko Grunmaum. Regular Polyhedra---Old and New"
           Aequationes Mathematicae Vol 15 pp 118-120
           more regular polyhedra
       five fold symmetry
         Istvan Hargittai editor, Fivefold Symmetry, World Scientific, 1992
       computational method for locating vertex coordinates, with exact formulas for angles
         Andrew Hume, Exact Descriptions of Regular and Semi-Regular Polyhedra and their Duals
         Computing Science Technical Report #130, AT&T Bell Laboratories, Murray Hill, 1986
       non-rigorous computations of strut lengths in tensegrity structures and geodesic domes
         Hugh Kenner, Geodesic Math and How to Use It, Univ.Cal. Pr. 1976
       symmetry groups of polyhedra ... translation of 1884 German
         Felix Klein, The Icosahedron and the Solution of Equations of the Fifth Degree
         Dover 1956
       exact formulas for constructing all 77 kinds of uniform polyhedra
         Peter W. Messer, Closed Form Expressions for Uniform Polyhedra and Their Duals,
         Discrete and Computational Geometry 27 pp 353-375, 2002  
       reassemblies
         David Peterson, Two Dissections in 3-D, Journal of Recreational Mathematics,
         Vol 20 pp 257-270, 1988
       Coexeter enumeration complete
         J. Skilling, The Complete Set of Uniform Polyhedra, Philosophical Transactions of the Royal Society
         Ser A, 278 pp 111-135, 1975
       tables
         Eric W. Weisstein, The CRC Concise Encyclopedia of Mathematics, CRC Press, 1998
       references
         David Wells, The Penguin Dictionary of Curious and Interesting Geometry,
         Penguin, 1991
       design principles
         Robert Williams, Natural Structure: Toward a Form Language, Eudaemon Pr.,
         1972
         The Geometrical Foundations of Natural Structure: A Source Book of Design,
         Dover, 1979