■ピザの公平な分け方(その10)

[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/4−α)/R)−arcsin(−sinα/R)

b=arcsin(cosα/R)−arcsin(sin(π/4−α)/R)

c=arcsin(cos(π/4−α)/R)−arcsin(cosα/R)

d=arcsin(sinα/R)−arcsin(cos(π/4−α)/R)

e=arcsin(−sin(π/4−α)/R)−arcsin(sinα/R)

f=arcsin(−cosα/R)−arcsin(−sin(π/4−α)/R)

g=arcsin(−cos(π/4−α)/R)−arcsin(−cosα/R)

h=arcsin(−sinα/R)−arcsin(−cos(π/4−α)/R)

  a+c+e+g=b+d+f+h=0が成り立つ.

  a+e=0,c+g=0,b+f=0,d+h=0

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[まとめ]r^2=cos2(θ−α)+R^2−{4R^2(cos(θ−α))^2−(sin2(θ−α))^2}^1/2

以上より,1/2・πR^2/4・8=πR^2だけが残りことになる.

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