963

在等边三角形 ${\displaystyle ABC}$ 的三边上各取一点 ${\displaystyle D,E,F}$, 满足 ${\displaystyle DE=3}$, ${\displaystyle DF=2\sqrt{3}}$, ${\displaystyle \mathrm{\angle }DEF={90}^{\circ }}$, 求三角形 ${\displaystyle ABC}$ 的面积的最大值.

作出如图所示的图形. 设 ${\displaystyle \alpha =\mathrm{\angle }FEC}$. 根据正弦定理，$$\begin{array}{r}\frac{\sqrt{3}}{\sin {\displaystyle \frac{\pi }{3}}}=\frac{CE}{\sin ({\displaystyle \frac{\pi }{3}}+\alpha )},\frac{3}{\sin {\displaystyle \frac{\pi }{3}}}=\frac{BE}{\sin ({\displaystyle \frac{\pi }{3}}+{\displaystyle \frac{\pi }{2}}-\alpha )},\end{array}$$
整理得$$\begin{array}{r}BC=BE+CE=2\sqrt{3}\cos \alpha +4\sin \alpha \le 2\sqrt{7},\end{array}$$
从而 ${\displaystyle S_{ABC}={\displaystyle \frac{\sqrt{3}}{4}B{C}^2\le 7\sqrt{3}}}$.