■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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