■シュレーフリの公式と直角三角錐(その48)

3次元版の計算

ユークリッド空間の基本単体では

(sinα)^2(sinγ)^2-(cosβ)^2=0

sinαsinγ-cosβ=0

正四面体ではα=π/3,β=π/3→√3sinγ=1

正八面体ではα=π/3,β=π/4→√3sinγ=√2

正20面体ではα=π/3,β=π/5→√3sinγ=φ

kaleidoscope, p102-103

(-1,0),(0,1),(1,2),(2,3),(3,4)

(-1,1),(0,2),(1,3),(2,4)

(-1,2),(0,3),(1,4)

(-1,3),(0,4)

(-1,4)

a^2(tanγ)^2=1

c^2(tanα)^2=1

b^2=a^2c^2

a=AB,b=BC,c=CD

d=ADとすると

d^2=(tanβ)^2

(secα)^2=(-1,1)(0,2)

(secβ)^2=(0,2)(1,3)

(secγ)^2=(1,3)(2,4)

(tanα)^2=(-1,2)=(2,4)→(s,t)=(t,s+5)が成り立つ

(tanβ)^2=(0,3)

(tanγ)^2=(1,4)

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4次元版の計算に利用

αn: aj=√("2/j(j+1)" )

βn: aj=√("2/j(j+1)" ),an=√("2/n" )

γn: aj=1

{3,5}: a1=1,a2=1/√3,a3=a2・τ2

{5,3}: a1=1,a2=τ √("(τ2+1)/5" ),a3=a2・τ

{3,3,5}: a1=1,a2=1/√3,a3=1/√6,a4=τ3/√2

{5,3,3}: a1=1,a2=τ √("(τ2+1)/5" ),a3=a2・τ,a4=τ4

{3,4,3}: a1=1,a2=√(1/3),a3=√(2/3),a4=√2

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