Wm.C.Corwin,Ph.D.,P.E.
                               www.ConcurrentInverse.com
                               www.byeless.com
                               billc@issi1.com
                               2005 July 22
                               (C) copyright 2005 Wm.C.Corwin


TITLE:  PROOF OF EXACT FORMULAS FOT PENTAGON

pentagon
inscribed in circle or radius R
each side subtends 72 from center
each side subtends 36 from any place on perimeter
two sides subtend 72 from any place on perimeter

two equations and two unknowns:
half angle formula
cos(36) = [1/2 + 1/2 cos(72)]
law of cosines
diag = [1 + diag**2 -2 diag cos(72)]
1    = [diag**2 + diag**2 -2 diag**2 cos(36)]

1    = [diag**2 + diag**2 -2 diag**2 [1/2 + 1/2 cos(72)]

1-1/(2d**2)= [1/2+1/2cos(72)]
1/d = 2cos(72)

1-2cos(72)**2 = [1/2+1/2cos(72)]
1-4cos(72)**2+4cos(72)**4 = 1/2+1/2cos(72)
  4cos(72)**4-4cos(72)**2-(1/2)cos(7)+1/2 = 0
the length of the diagonals is given by
1/(4d**4)-1/d**2-1/4d+1/2 = 0
2d**4-d**3-4d**2+1 = 0
(2d**2+d-1)(d**2-d-1) = 0
(2d-1)(d+1)(d+phi)(d-Phi) = 0
d = -1,-0.6180339887,0.5,1.6180339887

d = ([5]+1)/2
cos(72) = ([5]-1)/4
cos(36) = [1/2 + (1/2)(([5]-1)/4)] = [([5]+3)/8] = ([5]+1)/4
cos(18) = [1/2 + (1/2)(([5]+1)/4)] = [([5]+5)/8]    

1/cos(36) = [5]-1

sin(72) = [1-(6-2[5])/16] = [(10+2[5])/16] = [5+[5]]/(2[2])
sin(36) = [5-[5]]/(2[2])
sin(18) = [1-(5+[5])/8] = [(3-[5])/8] = ([5]-1)/4

tan(72) = [2][[5+5[5]]([5]+1)/4 = [5+[5]] [6+2[5]]/2[2] =  [40+16[5]]/2[2] = [5+2[5]]
tan(36) = [2][5-[5]] ([5]-1) / 4 =  [5-[5]] [6-2[5]]/2[2] = [40-16[5]]/2[2] = [5-2[5]]
tan(18) = [8/([5]+5)] ([5]-1)/4 = [6-2[5]][5-[5]]/[2][20] = [5-2[5]]/[5] = [5-2[5]]/[5]

R = (1/2) / sin(36) = [5+[5]]/[10]
r = R cos(36) = [5+[5]][6+2[5]]/4[10] = [40+16[5]/4[2][5] = [5+2[5]]/2[5]



 1+cos(36) = (5+[5])/4
 (1+cos(36))/cos(36) = (5+[5])/(1+[5]) = 4[5]/4 =[5]
  1+1/cos(36) = 1+4([5]-1)/4 = [5]
h = R(1+cos(36)) = (([5]+5)/4) R = [5+2[5]] / 2

the radius of the circle inscribed in a star
instar = ([5]+1)/(4tan(72)) = [6+2[5]] [5-2[5]]/4[5] = [10-2[5]/4[5] = [5-[5]]/2[10]
        = cos(72)R = [5+[5]]([5]-1)/4[10] = [5+[5]][3-[5]]/4[5] = [5-[5]]/2[10]
        =
        = 0.262865556



Since the pentagon side is 1 the base of the iscosoles point is
 base = instar/R = cos(72) = ([5]-1)/4
      =

A = 5r/2 = ([5][5+2[5]]) / 4


APPENDIX A: SUMMARY

side          1                         2[5-2[5]]      [5-[5]]/[2]    2[5-2[5]]/[5]     [2][5+[5]]
diagonal      ([5]+1)/2 =Phi  =1.618034  [10-2[5]]      [5+[5]]/[2]    [2/5][5-[5]]            
circumscribed [5+[5]]/[10]    =0.850651 [5]-1          1              1-1/[5]           (5+[5])/[5]
inscribed     [5+2[5]] /(2[5])=0.688191 1              (1+[5])/4      1/[5]             (5+3[5])/(2[5])
height        [5+2[5]] / 2    =1.538841 [5]            [5](1+[5])/4      1              (5+3[5])/2

point         [5-[5]]/2[2]    =0.587785  (3[5]-5)/2    (5-[5])/4        (3-[5])/2        [5]           
base          [7-3[5]]/[2] chk=0.381966  [2][65-29[5]] [25-11[5]]/[2] [2/5][65-29[5]]    [20-8[5]]            
instar        [5-[5]]/2[10]   =0.262866 [7-3[5]]/[2]   ([5]-1)/4       [7-3[5]/[10]      1
                                                        

area          ([5][5+2[5]]) /4          5[5-2[5]]     5[20+4[5]]/(8[2])  [5-2[5]]       5[50+22[5]]/2                //5[35+19[5]]/2 ??   

R/r           [5]-1

These factors have been used at http://www.issi1.com/corwin/calculator/pentagon.html
and found to be consistent.

APPENDIX B: IDENTITIES

    1+[5] = [6+2[5]]
    [5]-1 = [6-2[5]]
    5-[5] = [30-10[5]
    5+[5] = [30+10[5]
    3[5]-5 = [70-30[5]]
    5+3[5] = [70+30[5]]
    3-5[5] = [134-30[5]]





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ref

http://www.issi1.com/corwin/calculator/icosahedron.jpg
http://www.issi1.com/corwin/calculator/proof.txt
http://www.issi1.com/corwin/calculator/possible.txt
http://www.issi1.com/corwin/calculator/unit_v.txt
http://www.issi1.com/corwin/calculator/platonic.txt

http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/simpleTrig.html