■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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