Question

三角形 $ABC$ 中， $\angle C = 60^\circ$ ， $BC = 4$ 。点 $D$ 是 $BC$ 的中点。求 $\tan(\angle BAD)$ 的最大可能值。

Triangle $ABC$ has $\angle C = 60^\circ$ and $BC = 4.$ Point $D$ is the midpoint of $BC.$ What is the largest possible value of $\tan(\angle BAD)?$

$\dfrac{\sqrt{3}}{6}$

$\dfrac{\sqrt{3}}{3}$

$\dfrac{\sqrt{3}}{2\sqrt{2}}$

$\dfrac{\sqrt{3}}{4\sqrt{2} - 3}$

$1$

答案：D

知识点：坐标几何三角恒等式最优化

难度评级：2380

解答：

取 $C = (0, 0)$ ， $B = (2, 2\sqrt{3})$ ，使 $\angle C = 60^\circ$ 且 $BC = 4$ ，令 $A = (x, 0)$ ，其中 $x \gt 0$ 。则 $D = (1, \sqrt{3})$ 是 $BC$ 的中点。

直线 $\overrightarrow{AB} = (2-x, 2\sqrt{3})$ 和 $\overrightarrow{AD} = (1-x, \sqrt{3})$ 的斜率分别为 $\sqrt{3}\,x$ 与 $x^2 - 3x + 8$ 。用正切差公式化简得 $\tan(\angle BAD) = \dfrac{\sqrt{3}\,x}{x^2 - 3x + 8}.$

求导可知最大值在 $8-x^2$ ，即 $x = 2\sqrt{2}$ 处取得。代入得 $\begin{aligned} \tan(\angle BAD) &= \dfrac{2\sqrt{6}}{16 - 6\sqrt{2}} \\ &= \dfrac{\sqrt{6}}{8 - 3\sqrt{2}} \\ &= \dfrac{\sqrt{3}}{4\sqrt{2} - 3}. \end{aligned}$

所以正确答案是 D。

Place $C = (0, 0),$ $B = (2, 2\sqrt{3})$ so that $\angle C = 60^\circ$ and $BC = 4,$ and let $A = (x, 0)$ with $x \gt 0.$ Then $D = (1, \sqrt{3})$ is the midpoint of $BC.$

The vectors $\overrightarrow{AB} = (2-x, 2\sqrt{3})$ and $\overrightarrow{AD} = (1-x, \sqrt{3})$ have cross-product magnitude $\sqrt{3}\,x$ and dot product $x^2 - 3x + 8,$ which is always positive. Hence $\tan(\angle BAD) = \dfrac{\sqrt{3}\,x}{x^2 - 3x + 8}.$

The derivative has the sign of $8-x^2,$ so the unique maximum occurs at $x = 2\sqrt{2}.$ Substituting, $\begin{aligned} \tan(\angle BAD) &= \dfrac{2\sqrt{6}}{16 - 6\sqrt{2}} \\ &= \dfrac{\sqrt{6}}{8 - 3\sqrt{2}} \\ &= \dfrac{\sqrt{3}}{4\sqrt{2} - 3}. \end{aligned}$

Thus, D is the correct answer.

2008 AMC 12A 第 24 题

答案：D

难度评级：2380

解答：

其他年份的第 24 题