■正多面体の正多角形断面(その168)

以下の図より、d→1/2の初等的証明が可能である。参照点が対角線の交点に載ることを示してみたい。

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7次元の場合

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7次元正単体の8頂点

は超平面:x1+x2+x3+x4+x5+x6+x7+x8=1上にあります。

X=1+2cos(180-180(7-2)/7)=1+2cos(360/7)

一般に X=1+2cos(360/(N+1))

正方形の場合は式が異なり、X=2cos(45)=√2

正五角形の場合はX=1+2cos(360/5)=1+(√5-1)/2=τ

正六角形の場合はX=1+2cos(360/6)=1+1=2

正八角形の場合はX=1+2cos(360/8)=1+√2

また、赤道面

は超平面:x1-x4=X(x2-x3),x2-x5=X(x3-x4),x3-x6=X(x4-x5)

は超平面:x4-x7=X(x5-x6),x5-x8=X(x6-x7),x7=0,x8=0

・・・対角線の長さX=1+√2となるための条件

は超平面:x4=X(x5-x6),x5=X(x6),x7=0,x8=0

x4=X(X-1)x6,

x3==X(x4-x5)=X(X(X-1)-X)+x6=(X^3-2X^2+1)x6

x2=X(x3-x4)+x5=X(X^3-2X^2+1-X^2+X)x6+Xx6=(X^4-3X^3+X^2+X)x6+Xx6=(X^4-3X^3+X^2+2X)x6

x1=X(x2-x3)+x4=X((X^4-3X^3+X^2+2X-X^3+2X^2-1)x6+X(X-1)x6=X(X^4-4X^3+3X^2+2X-1)x6+X(X-1)x6 =(X^5-4X^4+3X^3+3X^2-2X)x6

x1+x2+x3+x4+x5+x6

=X^5-4X^4+3X^3+3X^2-2X

X^4-3X^3+X^2 +2X

X^3-2X^2 +1

X~2-X

X+1

=(X^5-3X^4+X^3+3X^2+2)x6=1

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x6=1/(X^5-3X^4+X^3+3X^2+2)

x5=X/(X^5-3X^4+X^3+3X^2+2)

x4={X^2-X}/(X^5-3X^4+X^3+3X^2+2)

x3={X^3-2X^2+1}/(X^5-3X^4+X^3+3X^2+2)

x2={X^4-3X^3+X^2+2X}/(X^5-3X^4+X^3+3X^2+2)

x1=(X^5-4X^4+3X^3+3X^2-2X)/(X^5-3X^4+X^3+3X^2+2)

この中のいくつかは等しいと思われるが・・・

(X-1)/2=cos(360/8)=cosθ=√2/2

cos(2θ)=2{(X-1)/2}^2-1=(X^2-2X+1)/2-1=(X^2-2X-1)/2=0

cos(3θ)=4{(X-1)/2}^3-3{(X-1)/2}=(X-1)^3/2-(3X-3)/2=(X^3-3X^2+3X-1-3X+3)/2=(X^3-3X^2+2)/2=-cosθ=-(X-1)/2

cos(4θ)=8{(X-1)/2}^4-8{(X-1)/2}^2+1=(X-1)^4/2-4(X-1)^2/2+1=(X^4-4X^3+6X^2-4X+1)/2-4(X^2-2X+1)/2+2/2=(X^4-4X^3+2X^2+4X-1)/2=-1

cos(5θ)=16{(X-1)/2}^5-20{(X-1)/2}^3+5{(X-1)/2}=(X-1)^5/2-5(X-1)^3/2+5{(X-1)/2}=(X^5-5X^4+10X^3-10X^2+5X-1)/2-5(X^3-3X^2+3X-1)/2+5(X-1)/2=(X^5-5X^4+5X^3+5X^2-5X-1)/2)=cos3θ

X^2=2X+1

(X^3-3X^2+2)=-(X-1)

X^3=3X^2-X-1=5X+2

(X^4-4X^3+2X^2+4X+1)=0

X^4=4X^3-2X^2-4X-1=12X+5

(X^5-5X^+5X^3+5X^2-5X-1)=-(X-1)

X^5=5X^4-5X^3-5X^2+4X+2=29X+12

x6=1/(4X+4)

x5=X/(4X+4)

x4={X+1}/(4X+4)

x3={X+1}/(4X+4)

x2={X}/(4X+4)

x1=(1)/(4X+4)

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 8点

  (1,0,0,0,0,0,0,0)

  (0,1,0,0,0,0,0,0)

  (0,0,1,0,0,0,0,0)

  (0,0,0,1,0,0,0,0)

  (0,0,0,0,1,0,0,0)

  (0,0,0,0,0,1,0,0)

  (0,0,0,0,0,0,1,0)

  (0,0,0,0,0,0,0,1)

が,xy平面上の8点

  (cos0π/8,sin0π/8)

  (cos2π/8,sin2π/8)

  (cos4π/8,sin4π/8)

  (cos6π/8,sin6π/8)

  (cos8π/8,sin8π/8)

  (cos10π/8,sin10π/8)

  (cos12π/8,sin12π/8)

  (cos12π/8,sin12π/8)

に投影されるためには,2×8行列

M=[cos0π/8,cos2π/8,・・・,cos14π/8]

  [sin0π/8,sin2π/8,・・・,sin14π/7]

が必要になる.

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正八角形の場合はX=1+2cos(360/8)=1+√2

x6=1/(4X+4)

x5=X/(4X+4)

x4={X+1}/(4X+4)

x3={X+1}/(4X+4)

x2={X}/(4X+4)

x1=(1)/(4X+4)

X0=x1+x2・cosθ+x3・cos2θ+x4・cos3θ+x5・cos4θ+x6・cos5θ

=x1+x2・cosθ+x3・cos2θ+x3・cos3θ+x2・cos4θ+x1・cos3θ

=x1-x2+x2・cosθ+x3・cos3θ+x1・cos3θ

=x1-x2+(x2-x3-x1)√2/2

Y0=x2・sinθ+x3・sin2θ+x4・sin3θ+x5・sin4θ+x6・sin5θ

=x2・sinθ+x3・sin2θ+x3・sin3θ+x2・sin4θ+x1・sin5θ

=x2・sinθ+x3+x3・sin3θ-x1・sin3θ

=x3+x2・sinθ+x3・sin3θ-x1・sin3θ

=x3+(x2+x3-x1)√2/2

X0^2+Y0^2を求めることになる。

=(x1-x2)^2+(x2-x3-x1)^2/2+(x1-x2)(x2-x3-x1)√2

+(x3)^2+(x2+x3-x1)^2/2+x3(x2+x3-x1)√2

=(1-x)^2+(-2)^2/2-2(1-x)√2

+(x+1)^2+(2X)^2/2+2x(x+1)√2・・・/(4x+4)^2

=(2x^2+2)+(4x^2+4)/2+(2x^2+4x-2)√2

={2(2x^2+2)+(2x^2+4x-2)√2}//(4x+4)^2

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