AMC12 2004 B

AMC12 2004 B · Q24

AMC12 2004 B · Q24. It mainly tests Triangles (properties), Trigonometry (basic).

In $\triangle ABC$, $AB = BC$, and $\overline{BD}$ is an altitude. Point $E$ is on the extension of $\overline{AC}$ such that $BE = 10$. The values of $\tan \angle CBE$, $\tan \angle DBE$, and $\tan \angle ABE$ form a geometric progression, and the values of $\cot \angle DBE,$ $\cot \angle CBE,$ $\cot \angle DBC$ form an arithmetic progression. What is the area of $\triangle ABC$?

在 $\triangle ABC$ 中，$AB = BC$，且 $\overline{BD}$ 是一条高。点 $E$ 在 $\overline{AC}$ 的延长线上，且 $BE = 10$。$\tan \angle CBE$、$\tan \angle DBE$、$\tan \angle ABE$ 的值成等比数列，而 $\cot \angle DBE,$ $\cot \angle CBE,$ $\cot \angle DBC$ 的值成等差数列。求 $\triangle ABC$ 的面积。

(A) 16 16

(B) \frac{50}{3} \frac{50}{3}

(D) 8\sqrt{5} 8\sqrt{5}

(E) 18 18

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let $\alpha = DBC$. Then the first condition tells us that \[\tan^2 DBE = \tan(DBE - \alpha)\tan(DBE + \alpha) = \frac {\tan^2 DBE - \tan^2 \alpha}{1 - \tan ^2 DBE \tan^2 \alpha},\] and multiplying out gives us $(\tan^4 DBE - 1) \tan^2 \alpha = 0$. Since $\tan\alpha \neq 0$, we have $\tan^4 DBE = 1 \Longrightarrow \angle DBE = 45^{\circ}$. The second condition tells us that $2\cot (45 - \alpha) = 1 + \cot \alpha$. Expanding, we have $1 + \cot \alpha = 2\left[\frac {\cot \alpha + 1}{\cot \alpha - 1}\right] \Longrightarrow (\cot \alpha - 3)(\cot \alpha + 1) = 0$. Evidently $\cot \alpha \neq - 1$, so we get $\cot \alpha = 3$. Now $BD = 5\sqrt {2}$ and $AC = \frac {2BD} {\cot \alpha} = \frac {10\sqrt {2}}{3}$. Thus, $[ABC] = \frac {1}{2} \cdot 5\sqrt {2} \cdot \frac {10\sqrt {2}}{3} = \frac {50}{3}\ \mathrm{(B)}$.

令 $\alpha = DBC$。第一个条件告诉我们 \[\tan^2 DBE = \tan(DBE - \alpha)\tan(DBE + \alpha) = \frac {\tan^2 DBE - \tan^2 \alpha}{1 - \tan ^2 DBE \tan^2 \alpha},\] 两边同乘并整理得 $(\tan^4 DBE - 1) \tan^2 \alpha = 0$。由于 $\tan\alpha \neq 0$，所以 $\tan^4 DBE = 1 \Longrightarrow \angle DBE = 45^{\circ}$。第二个条件告诉我们 $2\cot (45 - \alpha) = 1 + \cot \alpha$。展开得 $1 + \cot \alpha = 2\left[\frac {\cot \alpha + 1}{\cot \alpha - 1}\right] \Longrightarrow (\cot \alpha - 3)(\cot \alpha + 1) = 0$。显然 $\cot \alpha \neq - 1$，因此 $\cot \alpha = 3$。于是 $BD = 5\sqrt {2}$ 且 $AC = \frac {2BD} {\cot \alpha} = \frac {10\sqrt {2}}{3}$。因此 $[ABC] = \frac {1}{2} \cdot 5\sqrt {2} \cdot \frac {10\sqrt {2}}{3} = \frac {50}{3}\ \mathrm{(B)}$。

Topics