已知△ $ABC$ 面积为 $S$,内切圆半径和三个旁切圆半径分别为 $r,{{r}_{a}},{{r}_{b}},{{r}_{c}}$,求证:
$S=\sqrt{r{{r}_{a}}{{r}_{b}}{{r}_{c}}}$.
$S=\sqrt{r{{r}_{a}}{{r}_{b}}{{r}_{c}}}$.
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【答案】
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【解析】
$S=\dfrac{1}{2}ab\sin C=2{{R}^{2}}\sin A\sin B\sin C$,
其中 $r\left(\cot \dfrac{A}{2}+\cot \dfrac{B}{2} \right)=c=2R\sin C$,
即 $r=\dfrac{2R\sin C\sin \dfrac{A}{2}\sin \dfrac{B}{2}}{\sin \left( \dfrac{A+B}{2} \right)}=4R\sin\dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}$,
如图 ${{r}_{a}}\left(\cot \dfrac{A+B}{2}+\cot \dfrac{A+C}{2} \right)=a$,
得到 ${{r}_{a}}=\dfrac{a}{\tan\dfrac{C}{2}+\tan \dfrac{B}{2}}=\dfrac{a\cos \dfrac{B}{2}\cos \dfrac{C}{2}}{\sin\dfrac{B+C}{2}}=\dfrac{a\cos \dfrac{B}{2}\cos \dfrac{C}{2}}{\cos \dfrac{A}{2}}=4R\sin\dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2}$
同理可得 ${{r}_{b}}=4R\cos\dfrac{A}{2}\sin \dfrac{B}{2}\cos \dfrac{C}{2}$,${{r}_{c}}=4R\cos \dfrac{A}{2}\cos\dfrac{B}{2}\sin \dfrac{C}{2}$,
$r{{r}_{a}}{{r}_{b}}{{r}_{c}}={{4}^{4}}{{R}^{4}}{{\sin}^{2}}\dfrac{A}{2}{{\sin }^{2}}\dfrac{B}{2}{{\sin }^{2}}\dfrac{C}{2}{{\cos}^{2}}\dfrac{A}{2}{{\cos }^{2}}\dfrac{B}{2}{{\cos }^{2}}\dfrac{C}{2}=4{{R}^{4}}{{\sin}^{2}}A{{\sin }^{2}}B{{\sin }^{2}}C={{S}^{2}}$.
其中 $r\left(\cot \dfrac{A}{2}+\cot \dfrac{B}{2} \right)=c=2R\sin C$,
即 $r=\dfrac{2R\sin C\sin \dfrac{A}{2}\sin \dfrac{B}{2}}{\sin \left( \dfrac{A+B}{2} \right)}=4R\sin\dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}$,
如图 ${{r}_{a}}\left(\cot \dfrac{A+B}{2}+\cot \dfrac{A+C}{2} \right)=a$,
得到 ${{r}_{a}}=\dfrac{a}{\tan\dfrac{C}{2}+\tan \dfrac{B}{2}}=\dfrac{a\cos \dfrac{B}{2}\cos \dfrac{C}{2}}{\sin\dfrac{B+C}{2}}=\dfrac{a\cos \dfrac{B}{2}\cos \dfrac{C}{2}}{\cos \dfrac{A}{2}}=4R\sin\dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2}$
同理可得 ${{r}_{b}}=4R\cos\dfrac{A}{2}\sin \dfrac{B}{2}\cos \dfrac{C}{2}$,${{r}_{c}}=4R\cos \dfrac{A}{2}\cos\dfrac{B}{2}\sin \dfrac{C}{2}$,
$r{{r}_{a}}{{r}_{b}}{{r}_{c}}={{4}^{4}}{{R}^{4}}{{\sin}^{2}}\dfrac{A}{2}{{\sin }^{2}}\dfrac{B}{2}{{\sin }^{2}}\dfrac{C}{2}{{\cos}^{2}}\dfrac{A}{2}{{\cos }^{2}}\dfrac{B}{2}{{\cos }^{2}}\dfrac{C}{2}=4{{R}^{4}}{{\sin}^{2}}A{{\sin }^{2}}B{{\sin }^{2}}C={{S}^{2}}$.
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