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[几何] 一道正多边形不等式

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hbghlyj posted 2020-1-19 00:23 |Read mode
Last edited by hbghlyj 2020-1-19 00:30圆内接正n边形$A_1A_2\cdots A_n$,P是圆内一点,设$A_iP$再次交圆于$A_i'(i=1,2,\cdots,n)$,求证
(1)$\sum\limits_{i=1}^nA_iP\geq\sum\limits_{i=1}^nA_i'P$
(2)$\sum\limits_{i=1}^nA_iP^2\geq\sum\limits_{i=1}^nA_i'P^2$
均当且仅当P为圆心时取等.
证明:设$A_k$对应n次单位根$e^{\frac{2\pi ik}n}$,由$A_iP+A_i'P\leq2$,知$A_i'P\leq2-A_iP$,
(1)可考虑加强命题$\sum\limits_{i=1}^nA_iP\geq n$.设P对应复数z,则$\sum\limits_{i=1}^nA_iP=\sum\limits_{k=0}^{n-1}\abs{z-e^{\frac{2\pi ik}n}}=\sum\limits_{k=0}^{n-1}\abs{ze^{\frac{-2\pi ik}n}-1}\geq\abs{z\sum\limits_{k=0}^{n-1}e^{\frac{-2\pi ik}n}-n}=n$.
(2)可考虑加强命题$\sum\limits_{i=1}^nA_iP^2\geq n$.所以$\sum\limits_{i=1}^nA_iP^2\geq\abs{\sum\limits_{k=0}^{n-1}\left(ze^{\frac{-2\pi ik}n}-1\right)^2}=\abs{z^2\sum\limits_{k=0}^{n-1}e^{\frac{-4\pi ik}n}-2z\sum\limits_{k=0}^{n-1}e^{\frac{-2\pi ik}n}+n}$,???
改成-1次方、立方、四次方有没有类似关系

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