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[几何] 平分线成绩

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天音 posted 2016-9-18 22:06 |Read mode
求三角形的三条内角平分线长的乘积与三边长的乘积之比的最大值

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kuing posted 2016-9-19 02:07
还算简单,由内角平分线长公式及均值,有
\[w_a=\frac{\sqrt{bc(a+b+c)(b+c-a)}}{b+c}\leqslant \frac12\sqrt{(a+b+c)(b+c-a)}=\sqrt{s(s-a)},\]
相乘再由面积公式有
\[\frac{w_aw_bw_c}{abc}\leqslant \frac{s\sqrt{s(s-a)(s-b)(s-c)}}{abc}
=\frac{sS}{abc}=\frac s{4R}=\frac{\sin A+\sin B+\sin C}4\leqslant \frac{3\sqrt3}8,\]
当三角形为正三角形时取等。

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original poster 天音 posted 2016-9-19 16:18
great!

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isee posted 2016-9-20 14:03
还算简单,由内角平分线长公式及均值,有
\[w_a=\frac{\sqrt{bc(a+b+c)(b+c-a)}}{b+c}\leqslant \frac12\sq ...
kuing 发表于 2016-9-19 02:07
竞赛级别,被你秒杀了

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