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32 lines
1.2 KiB
32 lines
1.2 KiB
2 years ago
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有公共端点的三条线段，方法就是找出两条线段所在三角形做$60^{\circ}$旋转。
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如下图:
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$\because \triangle AB'P' \cong \triangle ABP$
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$\therefore \angle PAP'=60 ^{\circ},AP=AP'$
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$\therefore \triangle AP'P$是等边三角形
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$\therefore AP=PP'$
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问题中要求解的$AP+BP+PC$也就成功转化为
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$$PP'+B'P'+PC$$
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通过图形可知，图中$B'C$就是三点共线时的最小值！
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下面开始求解$B'C$长度。
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求边形一般用勾股定理，构造直角三角形：从$B'$向$BC$边引垂线，交$BC$延长线于$H$。
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$\because \triangle AB'B$是等边三角形
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$AB=B'B=2$ 可以理解为旋转了$60^{\circ}$
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$\because \angle ABB'=60 ^{\circ}$
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$\therefore \angle B'BH=30^{\circ}$
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$\therefore B'H=1,BH=\sqrt{3}$
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$\therefore B'C=\sqrt{(\sqrt{3}+2)^2+1^2}=\sqrt{8+4\sqrt{3}}$
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复合二次根式 需要用配方法化简：
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$=\sqrt{8+2\sqrt{12}}=\sqrt{\sqrt{6}^2+2\sqrt{12}+\sqrt{2}^2}=\sqrt{(\sqrt{6}+\sqrt{2})^2}=\sqrt{6}+\sqrt{2}$
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