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[几何] 求 $(\vv {MB}+\vv{MC})\vv{AD}$ 最大值

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isee Posted 2022-4-1 17:05 |Read mode
融合了知乎提问两问,中线的点积



:在 $\triangle ABC$ 中,已知 $AB=3,$ $AC=2,$ $BC=\sqrt {14}$,若 $D$ 为 $BC$ 边 任意一点, $M$ 为线段 $AD$ 的中点,则 $(\overrightarrow {MB}+\overrightarrow {MC})\cdot \overrightarrow {AD}$ 的最大值是______.


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kuing Posted 2022-4-1 17:12
图与数据严重不符啊

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 Author| isee Posted 2022-4-1 17:13
回复 2# kuing


示意图示意图,反正对你又没影响~

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走走看看 Posted 2022-4-1 22:47
回复 3# isee


是不是等于3?

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 Author| isee Posted 2022-4-1 22:51
回复  isee


是不是等于3?
走走看看 发表于 2022-4-1 22:47
是的是的

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走走看看 Posted 2022-4-2 08:32
Last edited by 走走看看 2022-4-2 20:44写个过程吧:

设BD=x
\begin{align*}
&(\vv{MB}+\vv{MC})\cdot\vv{AD}\\\\
&=(\vv{MD}+\vv{DB}+\vv{MD}+\vv{DC})\cdot\vv{AD}\\\\
&=(\vv{AD}+\vv{DB}+\vv{DC})\cdot\vv{AD}\\\\
&=(\vv{AB}+\vv{DC})\cdot\vv{AD}\\\\
&=\vv{AB}\vv{AD}-\vv{DC}\vv{DA}\\\\
&=\frac{\vv{AB}^2+\vv{AD}^2-\vv{BD}^2}{2}-\frac{\vv{DC}^2+\vv{DA}^2-\vv{AC}^2}{2}\\\\
&=\frac{\vv{AB}^2+\vv{AC}^2-\vv{BD}^2-\vv{DC}^2}{2}\\\\
&=-x^2+\sqrt{14}x-\frac12\\\\
&\le 3
\end{align*}

当D为BC的中点时,取得最大值3。

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 Author| isee Posted 2022-4-2 22:23
回复 6# 走走看看


行距为\\[1em]
1em 表示一个m的距离,所以其实距离类似,任意实数倍
还有距离小一点是\\[1ex],1个x的距离,也是任意实数倍

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