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2004 AMC 12B 第 24 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

24.

在 △ABC\triangle ABC 中，AB=BCAB = BC，且 BD‾\overline{BD} 是高。点 EE 在 AC‾\overline{AC} 的延长线上，且 BE=10BE = 10。tan⁡∠CBE\tan \angle CBE、tan⁡∠DBE\tan \angle DBE、tan⁡∠ABE\tan \angle ABE 的值构成等比数列，而 cot⁡∠DBE\cot \angle DBE、cot⁡∠CBE\cot \angle CBE、cot⁡∠DBC\cot \angle DBC 的值构成等差数列。△ABC\triangle ABC 的面积是多少？

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

1616

503\dfrac{50}{3}

10310\sqrt{3}

858\sqrt{5}

1818

答案：B
知识点：三角恒等式等比数列等腰三角形
难度评级：2390
小提示：

令 ∠DBE=α\angle DBE = \alpha、∠DBC=β\angle DBC = \beta；则 ∠CBE=α−β\angle CBE = \alpha - \beta，∠ABE=α+β\angle ABE = \alpha + \beta。

Let ∠DBE=α\angle DBE = \alpha and ∠DBC=β;\angle DBC = \beta; then ∠CBE=α−β\angle CBE = \alpha - \beta and ∠ABE=α+β\angle ABE = \alpha + \beta

大提示：

等比数列给出 tan⁡(α−β)tan⁡(α+β)=tan⁡2α\tan(\alpha - \beta)\tan(\alpha + \beta) = \tan^2\alpha，这会推出 α=45∘\alpha = 45^\circ。

The geometric progression gives tan⁡(α−β)tan⁡(α+β)=tan⁡2α,\tan(\alpha - \beta)\tan(\alpha + \beta) = \tan^2\alpha, which forces α=45∘\alpha = 45^\circ

解答：

令 ∠DBE=α\angle DBE = \alpha、∠DBC=β\angle DBC = \beta。由于 BD‾\overline{BD} 是等腰三角形的高，∠CBE=α−β\angle CBE = \alpha - \beta，且 ∠ABE=α+β\angle ABE = \alpha + \beta。等比数列给出 tan⁡(α−β)tan⁡(α+β)=tan⁡2α\tan(\alpha - \beta)\tan(\alpha + \beta) = \tan^2\alpha，化简为 tan⁡2β(tan⁡4α−1)=0\tan^2\beta(\tan^4\alpha - 1) = 0，所以 tan⁡α=1\tan\alpha = 1，α=45∘\alpha = 45^\circ。

设 DC=aDC = a、BD=bBD = b，等差数列 cot⁡∠DBE\cot\angle DBE、cot⁡∠CBE\cot\angle CBE、cot⁡∠DBC\cot\angle DBC 变为 1,b+ab−a,ba1, \dfrac{b + a}{b - a}, \dfrac{b}{a}，从而推出 b=3ab = 3a。又 BE=10BE = 10，且 ∠DBE=45∘\angle DBE = 45^\circ，所以 b=BE2=52b = \dfrac{BE}{\sqrt2} = 5\sqrt2，于是 a=523a = \dfrac{5\sqrt2}{3}。

△ABC\triangle ABC 的面积为 12(AC)(BD)=ab\tfrac12 (AC)(BD) = ab =52⋅523= 5\sqrt2 \cdot \dfrac{5\sqrt2}{3} =503= \dfrac{50}{3}。

因此正确答案是 B。

Let ∠DBE=α\angle DBE = \alpha and ∠DBC=β.\angle DBC = \beta. Since BD‾\overline{BD} is the altitude of the isosceles triangle, ∠CBE=α−β\angle CBE = \alpha - \beta and ∠ABE=α+β.\angle ABE = \alpha + \beta. The geometric progression gives tan⁡(α−β)tan⁡(α+β)=tan⁡2α,\tan(\alpha - \beta)\tan(\alpha + \beta) = \tan^2\alpha, which simplifies to tan⁡2β(tan⁡4α−1)=0,\tan^2\beta(\tan^4\alpha - 1) = 0, so tan⁡α=1\tan\alpha = 1 and α=45∘.\alpha = 45^\circ.

Writing DC=aDC = a and BD=b,BD = b, the arithmetic progression cot⁡∠DBE,\cot\angle DBE, cot⁡∠CBE,\cot\angle CBE, cot⁡∠DBC\cot\angle DBC becomes 1,b+ab−a,ba,1, \dfrac{b + a}{b - a}, \dfrac{b}{a}, forcing b=3a.b = 3a. With BE=10BE = 10 and ∠DBE=45∘,\angle DBE = 45^\circ, we get b=BE2=52,b = \dfrac{BE}{\sqrt2} = 5\sqrt2, so a=523.a = \dfrac{5\sqrt2}{3}.

The area of △ABC\triangle ABC is 12(AC)(BD)=ab\tfrac12 (AC)(BD) = ab =52⋅523= 5\sqrt2 \cdot \dfrac{5\sqrt2}{3} =503.= \dfrac{50}{3}.

Thus, the correct answer is B.

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