■正三角形の等チェバ線(その9)

  A=Y^2+3/4=9y^2/4(x−1)^2+3/4

={3(x−1)^2+9y^2}/4(x−1)^2

  B=√3Y=−3√3y/2(x−1)

  C=a^2−a/2+5/8=(a−1/4)^2+9/16

  D=−3a/2+3/8=−3/2(a−1/4)

  a=(2√3xy−5x−1)/(2√3y−2x−4)

  a−1/4=(2√3xy−9x/2−√3y/2)/(2√3y−2x−4)

  E=c^2−c/2+5/8=(c−1/4)^2+9/16

  F=−3c/2+3/8=−3/2(c−1/4)

c=(2√3xy+5x+1)/(2√3y+2x+4)

c−1/4=(2√3xy+9x/2−√3y/2)/(2√3y+2x+4)

  E(AD+BC)=F(AC+BD)

  A(DE−CF)=B(DF−CE)

  A/B=(DF−CE)/(DE−CF)=Y/√3+√3/4Y

  A/B=(F/C−E/D)/(E/C−F/D)=Y/√3+√3/4Y

を証明する方がやさしいのではなかろうか?

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F/C=−3/2(c−1/4)/(a^2−a/2+5/8)

E/D=(c^2−c/2+5/8)/(−3/2(a−1/4))

E/C=(c^2−c/2+5/8)/(a^2−a/2+5/8)

F/D=−3/2(c−1/4)/(−3/2(a−1/4))

分子=9/4(a−1/4)(c−1/4)−(a^2−a/2+5/8)(c^2−c/2+5/8)

=9/4(a−1/4)(c−1/4)−(a−1/4)^2(c−1/4)^2−9/16(a−1/4)^2−9/16(c−1/4)^2−(9/16)^2

=−9/16{(a−1/4)^2−2(a−1/4)(c−1/4)+(c−1/4)^2}+9/8(a−1/4)(c−1/4)−(a−1/4)^2(c−1/4)^2−(9/16)^2

=−9/16(a−c)^2−{(a−1/4)(c−1/4)−9/16}^2

分母=−3/2(a−1/4)(c^2−c/2+5/8)+3/2(c−1/4)(a^2−a/2+5/8)

=−3/2(a−1/4)(c−1/4)^2−27/32(a−1/4)

+3/2(c−1/4)(a−1/4)^2+27/32(c−1/4)

=3/2(c−1/4)(a−1/4)(a−c)+27/32(c−a)

=3/2・(a−c){(a−1/4)(c−1/4)−9/16}

分子/分母=−3/8・(a−c)/{(a−1/4)(c−1/4)−9/16}−2/3・{(a−1/4)(c−1/4)−9/16}/(a−c)

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[まとめ]射影幾何的な変形により,かなり簡単になったが,係数比が一致しない.

分子/分母=Y/√3+√3/4Y

1/√3:√3/4=4:3

3/8:2/3=9:16

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