■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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