■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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