■ピザの公平な分け方(その14)

 問題となるとすれば,22.5°16ピースの場合である.

  a=1/2・∫(0,π/8)r^2dθ

  b=1/2・∫(π/8,π/4)r^2dθ

  c=1/2・∫(π/4,3π/8)r^2dθ

  d=1/2・∫(3π/8,π/2)r^2dθ

  e=1/2・∫(π/2,5π/8)r^2dθ

  f=1/2・∫(5π/8,3π/4)r^2dθ

  g=1/2・∫(3π/4,7π/8)r^2dθ

  h=1/2・∫(7π/8,π)r^2dθ

  i=1/2・∫(π,9π/8)r^2dθ

  j=1/2・∫(9π/8,5π/4)r^2dθ

  k=1/2・∫(5π/4,11π/8)r^2dθ

  l=1/2・∫(11π/8,3π/2)r^2dθ

  m=1/2・∫(3π/2,13π/8)r^2dθ

  n=1/2・∫(13π/8,7π/4)r^2dθ

  o=1/2・∫(7π/4,15π/8)r^2dθ

  p=1/2・∫(15π/8,2π)r^2dθ

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[3]∫cos(θ−α){R^2−(sin(θ−α))^2}^1/2dθ

=R/2・sin(θ−α){1−(sin(θ−α)/R)^2}^1/2+2R^2・arcsin(sin(θ−α)/R)の

arcsin(sin(θ−α)/R)

a=arcsin(sin(π/8−α)/R)−arcsin(−sinα/R)

b=arcsin(sin(π/4−α)−arcsin(sin(π/8−α)/R)

c=arcsin(sin(3π/8−α)/R)−arcsin(sin(π/4−α)/R)

d=arcsin(sin(π/2−α)/R)−arcsin(sin(3π/8−α)/R)

e=arcsin(sin(5π/8−α)/R)−arcsin(sin(π/2−α)/R)

f=arcsin(sin(3π/4−α)/R)−arcsin(sin(5π/8−α)/R)

g=arcsin(sin(7π/8−α)/R)−arcsin(sin(3π/4−α)/R)

h=arcsin(sin(π−α)/R)−arcsin(sin(7π/8−α)/R)

i=arcsin(sin(9π/8−α)/R)−arcsin(sin(π−α)/R)

j=arcsin(sin(5π/4−α)/R)−arcsin(sin(9π/8−α)/R)

k=arcsin(sin(11π/8−α)/R)−arcsin(sin(5π/4−α)/R)

l=arcsin(sin(3π/2−α)/R)−arcsin(sin(11π/8−α)/R)

m=arcsin(sin(13π/8−α)/R)−arcsin(sin(3π/2−α)/R)

n=arcsin(sin(7π/4−α)/R)−arcsin(sin(13π/8−α)/R)

o=arcsin(sin(15π/8−α)/R)−arcsin(sin(7π/4−α)/R)

p=arcsin(sin(2π−α)/R)−arcsin(sin(15π/8−α)/R)

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