■シュレーフリの公式と直角三角錐(その175)

以下の数列を横方向に伸ばして、フリーズを形成することは可能だろうか?

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正8胞体では

cosδ= 0→1/2,1

1   1   1   1   1   1

  1   2   2   2   1・・・(sec)^2

    1   3   3   1・・・(tan)^2

      1   4   1

        1   1

          0

{{(tanα)^2(tanβ)^2(tanγ)^2-(tanα)^2-(tanγ)^2 -1}{{(tanδ)^2(tanβ)^2(tanγ)^2-(tanδ)^2-(tanβ)^2 -1}=(secβ)^2(secγ)^2

(9-5)(9-5)=(secβ)^2(secγ)^2=16・・・OK

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正16胞体では

cosδ= -1/2→1/4,3

1   1   1   1   1   1

  2   2   2   1   4・・・(sec)^2

    3   3   1   3・・・(tan)^2

      4   1   2

        1   1 

          0

{{(tanα)^2(tanβ)^2(tanγ)^2-(tanα)^2-(tanγ)^2 -1}{{(tanδ)^2(tanβ)^2(tanγ)^2-(tanδ)^2-(tanβ)^2 -1}=(secβ)^2(secγ)^2

(9-5)(9-7)=(secβ)^2(secγ)^2=8・・・OK

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正24胞体では

cosδ= -1/2→1/4,3

1   1   1   1   1   1

  2   2   1   4   1

    3   1   3   3

      1   2   2

        1   1

          0

{{(tanα)^2(tanβ)^2(tanγ)^2-(tanα)^2-(tanγ)^2 -1}{{(tanδ)^2(tanβ)^2(tanγ)^2-(tanδ)^2-(tanβ)^2 -1}=(secβ)^2(secγ)^2

(9-7)(9-5)=(secβ)^2(secγ)^2=8・・・OK

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正八面体では

1   1   1   1   1

  2   2   1   3

    3   1   2

      1   1

        0

とおけて

(tanα)^2=3

(tanβ)^2=1

(tanγ)^2=2

cosδ= -1/3→1/3,2 

(tanα)^2(tanγ)^2=6

(secα)^2(secγ)^2/(secβ)^2=4・3/2=6

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立方体では

1   1   1   1   1

  1   2   2   1

    1   3   1

      1   1

        0

とおけて

(tanα)^2=1

(tanβ)^2=3

(tanγ)^2=1

cosδ= 0→1/2,1

α=π/4,β=π/3→一致  

(tanα)^2(tanγ)^2=1

(secα)^2(secγ)^2/(secβ)^2=2・2/4=1

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