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Last edited by hbghlyj 2024-11-1 00:49
A(1,0),B,D在单位圆上且在上半平面,∠BOD=90°,作B,D关于x轴的对称点E,C,五角星ABCDE的自交点为A',B',C',D',E',求证:$$\frac12<{S_{A'B'C'D'E'}\over S_{A'D'B'E'C'}}<\frac23$$最小值当A,C,D重合时取得,最大值当B,E重合时取得.(五角星的面积按代数规则,即中间的五边形算两次)
设B为$\left(\frac{1-a^2}{a^2+1},\frac{2a}{a^2+1}\right)\;(a>1)$,用这帖的方法计算,$${S_{A'B'C'D'E'}\over S_{A'D'B'E'C'}}={\frac{2\left(a^5-a^4 -2 a^2 +3 a -1\right)}{(a+1)^2\left(a^2+1\right)^2}\over\frac{(a-1)^2\left(3 a^3+a^2 +a -1\right)}{a (a+1)\left(a^2+1\right)^2}}=\frac{2 a \left(a^3+a^2 +a -1\right)}{(a+1)(3 a^3+a^2 +a -1)}=:f(a)$$
$$f'(a)=\frac{2 a \left(a^3+a^2+a-1\right)}{(a+1) \left(3 a^3+a^2+a-1\right)}>0$$
$f(a)$单调增,所以${S_{A'B'C'D'E'}\over S_{A'D'B'E'C'}}>f(1)=\frac12$,${S_{A'B'C'D'E'}\over S_{A'D'B'E'C'}}< f(∞)=\frac23$. |
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