■正三角形の等チェバ線(その7)

  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(CE−DF)

DE=−3/2(a−1/4){(c−1/4)^2+9/16}

CF=−3/2(c−1/4){(a−1/4)^2+9/16}

DE−CF=−3/2{(a−1/4)(c−1/4)^2+9(a−1/4)/16−(a−1/4)^2(c−1/4)−9(c−1/4)/16}

=−3/2{(a−1/4)(c−1/4)(c−a)+9(a−c)/16}

=−3/2(c−a){(a−1/4)(c−1/4)−9/16}

CE={(a−1/4)^2+9/16}{(c−1/4)^2+9/16}

DF=9/4・(a−1/4)(c−1/4)

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

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

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a−c={(2√3xy−5x−1)(2√3y+2x+4)−(2√3xy+5x+1)(2√3y−2x−4)}/(2√3y−2x−4)(2√3y+2x+4)

={4√3xy(2x+4)−2√3y(5x+1)}/{(2√3y)^2−(2x+4)^2}

=2√3y(4x^2−x−2)/(12y^2−4x^2−16x−16)

 これ以上計算しないが,方針だけは示せたと思う.

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