每日一题[857]举一反三

设$\triangle ABC$的三边$a,b,c$ 上的高分别为$h_a,h_b,h_c$ ,满足$3\cdot\dfrac{a}{{{h_a}}} - \dfrac{b}{{{h_b}}} + 6\cdot\dfrac{c}{{{h_c}}} = 6$.

(1)若$\triangle ABC$的面积为$S$,试证$S = \dfrac{1}{{12}}(3{a^2} - {b^2} + 6{c^2})$;

(2)用$b,c$ 表示$\sin \left( {A + \dfrac{\pi}{4}} \right)$,并求$A$的大小;

(3)根据上述解题过程所得到的$\triangle ABC$结论,请你设计一个与三角形有关的类似结论,并证明你所给的结论.


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分析与解 (1)由 $S = \dfrac{1}{2}a{h_a} = \dfrac{1}{2}b{h_b} = \dfrac{1}{2}c{h_c}$,于是$$\dfrac{1}{{{h_a}}} = \dfrac{a}{{2S}}\;,\;\dfrac{1}{{{h_b}}} = \dfrac{b}{{2S}}\;,\;\dfrac{1}{{{h_c}}} = \dfrac{c}{{2S}},$$
因此,$\dfrac{{3{a^2}}}{{2S}} - \dfrac{{{b^2}}}{{2S}} + \dfrac{{6{c^2}}}{{2S}} = 6$,整理得$$S = \dfrac{1}{{12}}\left( {3{a^2} - {b^2} + 6{c^2}} \right).$$

(2)由面积公式 $$S = \dfrac{1}{2}bc\sin A = \dfrac{1}{{12}}\left( {3{a^2} - {b^2} + 6{c^2}} \right)\qquad\cdots\cdots①$$
再根据余弦定理,$${a^2} = {b^2} + {c^2} - 2bc\cos A\qquad\cdots\cdots②$$
将②代入①,有$$bc\sin A = \dfrac{1}{6}\left( {2{b^2} + 9{c^2} - 6bc\cos A} \right),$$整理得$\sin A + \cos A = \dfrac{{2{b^2} + 9{c^2}}}{{6bc}}$,于是$$\sin \left( {A + \dfrac{\pi}{4}} \right) = \dfrac{{2{b^2} + 9{c^2}}}{{6\sqrt 2 bc}}.$$
考虑到$2{b^2} + 9{c^2} \geqslant 6\sqrt 2 bc$,于是$\sin \left( {A + \dfrac{\pi}{4}} \right) = 1$,从而$A = \dfrac{\pi}{4}$.

(3)结论一:若$x \cdot \dfrac{a}{{{h_a}}} + y \cdot \dfrac{b}{{{h_b}}} + z \cdot \dfrac{c}{{{h_c}}} = 1$,则$S = \dfrac{1}{2}\left( {x{a^2} + y{b^2} + z{c^2}} \right)$;

结论二:$\sin \left( {A + \arctan 2x} \right) = \dfrac{{\left( {x + y} \right){b^2} + \left( {x + z} \right){c^2}}}{{bc\sqrt {1 + 4{x^2}} }}$;

结论三:当$xy + yz + zx = \dfrac{1}{4}$时,$A = \arctan \dfrac{1}{{2x}}$.

证明过程与(1)(2)完全类似,略去.

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