■葉序らせん(その136)

 ねじれ角に相当するのは,∠AODである.

cos(∠AOD)=((x−ξ)^2−b^2s^2)/((x−ξ)^2+b^2s^2

===================================

 (その2)の正四面体

  A(1/2√2,0,−1/2)

  D(1/2√2,0,1/2)

  C(−1/2√2,1/2,0)

  B(−1/2√2,−1/2,0)

 これをx軸周りにθだけ回転させて,5点が同一円周上にあるような投影方向を求めなければならない.c=cosθ,s=sinθ

  A(1/2√2,s/2,−c/2)

  D(1/2√2,−s/2,c/2)

  C(−1/2√2,c/2,s/2)

  B(−1/2√2,−c/2,−s/2)

  E(−5/6√2,−5s/6,5c/6)

  O(x,0,0)

 s^2=1/10,c^2=9/10

  A(1/2√2,1/2√10,−3/2√10)

  D(1/2√2,−1/2√10,3/2√10)

  C(−1/2√2,3/2√10,1/2√10)

  B(−1/2√2,−3/2√10,−1/2√10)

===================================

 (x−1/2√2)^2+s^2/4=(x+1/2√2)^2+c^2/4=(x+5/6√2)^2+s^2・(5/6)^2

 x^2−x/√2+1/8+s^2/4

=x^2+x/√2+1/8+1/4−s^2/4

=x^2+5x/3√2+25/72+25s^2/36

2x/√2+1/4−s^2/2=0

8x/3√2+2/9+4s^2/9=0

8x/3√2+1/3−2s^2/3=0

1/9−10s^2/9=0,s^2=1/10,c^2=9/10

x√2+1/4−1/20=0→x=−1/5√2

===================================