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。

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

7次元の場合

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

この計算を一般のｎに拡張することは難しそうであると思われたのであるが

対角線の交点を求めることは可能である。

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

ｎを奇数とする。nn=n+1

xi=(cos(2πi/nn),sin(2πi/nn))

x0(1,0)とx(nn-3)を結んだ線を基準に考える

x0(1,0)とx(nn/2-1)=(cos((nn-2)π/nn),sin((nn-2)π/nn))を結ぶ。

x(nn-3)=(cos((nn-3)2π/nn),sin((nn-3)2π/nn))とx((nn/2-2)=(cos((nn-4)π/nn),sin((nn-4)π/nn))を結ぶ

y=(sin((nn-2)π/nn)-sin0)/(cos((nn-2)π/nn)-cos0)(x-1)

y-sin(2nn-6)π/nn)=sin((nn-4)π/nn)-sin(2nn-6)π/nn))/(cos((nn-4)π/nn)-cos(2nn-6)π/nn))(x-cos(2nn-6)π/nn))

y=-cot((nn/2-1)π/nn)(x-1)

y-sin(2nn-6)π/nn)=-cot((3nn/2-5)π/nn)(x-cos(2nn-6)π/nn))

-cot((nn/2-1)π/nn)(x-1)-sin(2nn-6)π/nn)=-cot((3nn/2-5)π/nn)(x-cos(2nn-6)π/nn))

{cot((3nn/2-5)π/nn)-cot((nn/2-1)π/nn)}x=cot((3nn/2-5)π/nn)cos(2nn-6)π/nn)-cot((nn/2-1)π/nn)+sin(2nn-6)π/nn))

-{sin(nn-4)π/nn)/sin((3nn/2-5)π/nn)sin((nn/2-1)π/nn))}x=cot((3nn/2-5)π/nn)cos(2nn-6)π/nn)-cot((nn/2-1)π/nn)+sin(2nn-6)π/nn))

-{sin(nn-4)π/nn}x=cos((3nn/2-5)π/nn)sin((nn/2-1)π/nn)cos(2nn-6)π/nn)-sin((3nn/2-5)π/nn)cos((nn/2-1)π/nn)+sin((3nn/2-5)π/nn)sin((nn/2-1)π/nn)sin(2nn-6)π/nn))

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

NN=8を代入して検算

-x=(√2/2)・(√2/2)・0+(√2/2)-(√2/2)・(√2/2) ＝（√2-1）/2 ・・・一致

NN=6を代入して検算

-x√3/2=+1/2・√3/2+√3/2・1/2-√3/2・√3/2・0＝0 ・・・一致

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

nn/2の項をなくす

-{sin(nn-4)π/nn}x=cos((3nn/2-5)π/nn)sin((nn/2-1)π/nn)cos(2nn-6)π/nn)-sin((3nn/2-5)π/nn)cos((nn/2-1)π/nn)+sin((3nn/2-5)π/nn)sin((nn/2-1)π/nn)sin(2nn-6)π/nn))

-{sin(nn-4)π/nn}x=1/2{sin((2nn-6)π/nn)-sin((nn-4)π/nn)}cos(2nn-6)π/nn)-1/2{sin((2nn-6)π/nn)+sin((nn-4)π/nn)}-1/2{cos((2nn-6)π/nn)-cos((nn-4)π/nn)}sin(2nn-6)π/nn))

-{sin(nn-4)π/nn}x=1/2{-sin((nn-4)π/nn)}cos(2nn-6)π/nn)-1/2{sin((2nn-6)π/nn)+sin((nn-4)π/nn)}-1/2{-cos((nn-4)π/nn)}sin(2nn-6)π/nn))

-{sin(nn-4)π/nn}x=1/2{sin((nn-2)π/nn)}-1/2{sin((2nn-6)π/nn)+sin((nn-4)π/nn)}

NN=8を代入して検算

-x=1/2｛(√2/2)｝-1/2｛-√2/2+1｝ ＝（√2-1）/2 ・・・一致

NN=6を代入して検算

-x√3/2=+1/2・√3/2-√3/2・1/2-√3/2・√3/2・0＝0 ・・・一致

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

y=(sin((nn-2)π/nn))/(cos((nn-2)π/nn)-1)(x-1)

-{sin(nn-4)π/nn}(x-1)=1/2{sin((nn-2)π/nn)}-1/2{sin((2nn-6)π/nn)-sin((nn-4)π/nn)}

(cos((nn-2)π/nn)-1)y=(sin((nn-2)π/nn))(x-1)

=-1/2{sin((nn-2)π/nn)}+1/2{sin((2nn-6)π/nn)-sin((nn-4)π/nn)}・(sin((nn-2)π/nn))/{sin(nn-4)π/nn}

(-cos(2π/nn)-1)y=(sin(2π/nn))(x-1)

=-1/2{sin(2π/nn)}+1/2{-sin(6π/nn)-sin(4π/nn)}・(sin((2π/nn))/{sin(4π/nn)

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

-{sin(4π/nn)}x=-sin(5π/nn)cos(π/nn)cos(6π/nn)+cos((5π/nn)sin(π/nn)+cos(5π/nn)cos(π/nn)sin(6π/nn))

-{sin(4π/nn)}x=(cos(5π/nn)sin(6π/nn)-sin(5π/nn)cos(6π/nn))cos(π/nn)+cos((5π/nn)sin(π/nn)

-{sin(4π/nn)}x=(sin(π/nn)cos(π/nn)+cos((5π/nn)sin(π/nn)

-{sin(4π/nn)}x=(sin(π/nn)2cos(3π/nn)cos(2π/nn)

これが最も簡単な形である・・・正しいことが確認された

(-cos(2π/nn)-1)y=(sin(2π/nn))(x-1)

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

NN=8を代入して検算

-x=(√2/4)-(2-√2)/4) ＝（√2-1）/2 ・・・一致

-x=√(2-√2）√(2-√2）/2・√2)/2 ＝（√2-1）/2 ・・・一致

NN=6を代入して検算

-x√3/2=√3/4-√3/2・1/2＝0 ・・・一致

-x√3/2=0・・・一致

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

-{2sin(2π/nn)cos(2π/nn)}x=(sin(π/nn)2cos(3π/nn)cos(2π/nn)

-{sin(2π/nn)}x=(sin(π/nn)cos(3π/nn)

-{2sin(π/nn)cos(π/nn)}x=(sin(π/nn)cos(3π/nn)

-{2cos(π/nn)}x=cos(3π/nn)

x=-cos(3π/nn)/{2cos(π/nn)}

x={-4cos^2(π/nn)+3}/2→-1/2

(-cos(2π/nn)-1)y=(sin(2π/nn))(x-1)

(-2cos^2(π/nn))y=(2sin(π/nn)cos(π/nn))(x-1)

(-cos(π/nn))y=(sin(π/nn))(x-1)

y=-tan(π/nn))(x-1)

y=-tan(π/nn)){-4cos^2(π/nn)+1}/2

>y={2sin(2π/nn)-tan（π/nn)}/2→0

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝