AMC12 2008 A

AMC12 2008 A · Q24

AMC12 2008 A · Q24. It mainly tests Manipulating equations, Triangles (properties).

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

三角形 $ABC$ 满足 $\angle C = 60^{\circ}$ 且 $BC = 4$。点 $D$ 是 $BC$ 的中点。$\tan{\angle BAD}$ 的最大可能值是多少？

(A) \frac{\sqrt{3}}{6} \frac{\sqrt{3}}{6}

(B) \frac{\sqrt{3}}{3} \frac{\sqrt{3}}{3}

(D) \frac{\sqrt{3}}{4\sqrt{2} - 3} \frac{\sqrt{3}}{4\sqrt{2} - 3}

(E) 1 1

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $x = CA$. Then $\tan\theta = \tan(\angle BAF - \angle DAE)$, and since $\tan\angle BAF = \frac{2\sqrt{3}}{x-2}$ and $\tan\angle DAE = \frac{\sqrt{3}}{x-1}$, we have \[\tan\theta = \frac{\frac{2\sqrt{3}}{x-2} - \frac{\sqrt{3}}{x-1}}{1 + \frac{2\sqrt{3}}{x-2}\cdot\frac{\sqrt{3}}{x-1}}= \frac{x\sqrt{3}}{x^2-3x+8}\] With calculus, taking the derivative and setting equal to zero will give the maximum value of $\tan \theta$. Otherwise, we can apply AM-GM: \begin{align*} \frac{x^2 - 3x + 8}{x} = \left(x + \frac{8}{x}\right) -3 &\geq 2\sqrt{x \cdot \frac 8x} - 3 = 4\sqrt{2} - 3\\ \frac{x}{x^2 - 3x + 8} &\leq \frac{1}{4\sqrt{2}-3}\\ \frac{x\sqrt{3}}{x^2 - 3x + 8} = \tan \theta &\leq \frac{\sqrt{3}}{4\sqrt{2}-3}\end{align*} Thus, the maximum is at $\frac{\sqrt{3}}{4\sqrt{2}-3} \Rightarrow \mathbf{(D)}$.

设 $x = CA$。则 $\tan\theta = \tan(\angle BAF - \angle DAE)$，且由于 $\tan\angle BAF = \frac{2\sqrt{3}}{x-2}$、$\tan\angle DAE = \frac{\sqrt{3}}{x-1}$，有 \[\tan\theta = \frac{\frac{2\sqrt{3}}{x-2} - \frac{\sqrt{3}}{x-1}}{1 + \frac{2\sqrt{3}}{x-2}\cdot\frac{\sqrt{3}}{x-1}}= \frac{x\sqrt{3}}{x^2-3x+8}\] 用微积分对其求导并令导数为零可得 $\tan \theta$ 的最大值。否则可用 AM-GM： \begin{align*} \frac{x^2 - 3x + 8}{x} = \left(x + \frac{8}{x}\right) -3 &\geq 2\sqrt{x \cdot \frac 8x} - 3 = 4\sqrt{2} - 3\\ \frac{x}{x^2 - 3x + 8} &\leq \frac{1}{4\sqrt{2}-3}\\ \frac{x\sqrt{3}}{x^2 - 3x + 8} = \tan \theta &\leq \frac{\sqrt{3}}{4\sqrt{2}-3}\end{align*} 因此最大值为 $\frac{\sqrt{3}}{4\sqrt{2}-3} \Rightarrow \mathbf{(D)}$。

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