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2008 AMC 12A 第 20 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。

20.

三角形 ABCABC 中，AC=3AC = 3、BC=4BC = 4、AB=5AB = 5。点 DD 在 ABAB 上，且 CDCD 平分直角。△ADC\triangle ADC 和 △BCD\triangle BCD 的内切圆半径分别为 rar_a 和 rbr_b，求 rarb\frac{r_a}{r_b}。

Triangle ABCABC has AC=3,AC = 3, BC=4,BC = 4, and AB=5.AB = 5. Point DD is on AB,AB, and CDCD bisects the right angle. The inscribed circles of △ADC\triangle ADC and △BCD\triangle BCD have radii rar_a and rb,r_b, respectively. What is rarb?\frac{r_a}{r_b}?

128(10−2)\dfrac{1}{28}(10 - \sqrt{2})

356(10−2)\dfrac{3}{56}(10 - \sqrt{2})

114(10−2)\dfrac{1}{14}(10 - \sqrt{2})

556(10−2)\dfrac{5}{56}(10 - \sqrt{2})

328(10−2)\dfrac{3}{28}(10 - \sqrt{2})

答案：E
知识点：角平分线定理内切圆、内心与内切圆半径面积比
难度评级：2100
小提示：

由角平分线定理，AD:DB=CA:CB=3:4AD:DB = CA:CB = 3:4。

By the Angle Bisector Theorem, AD:DB=CA:CB=3:4AD:DB = CA:CB = 3:4

大提示：

对每个小三角形使用 r=面积sr = \frac{\text{面积}}{s}；两个三角形共享底边 CDCD。

For each small triangle r=areas;r = \frac{\text{area}}{s}; the two triangles share the base CDCD

解答：

由角平分线定理，AD:DB=CA:CB=3:4AD:DB = CA:CB = 3:4，所以 AD=157AD = \tfrac{15}{7}，BD=207BD = \tfrac{20}{7}。两个小三角形 △ADC\triangle ADC 和 △BCD\triangle BCD 共用底边 CDCD，面积比为 3:43:4，面积分别为 187\tfrac{18}{7} 和 247\tfrac{24}{7}。

将 △ABC\triangle ABC 沿 CDCD 分割，该线段与两条直角边都成 45∘45^\circ，于是 3⋅CD22+4⋅CD22=6， \dfrac{3 \cdot CD}{2\sqrt{2}} + \dfrac{4 \cdot CD}{2\sqrt{2}} = 6\text{，} 所以 CD=1227CD = \tfrac{12\sqrt{2}}{7}。

两个三角形的半周长分别为 sa=67(3+2),sb=67(4+2)。 \begin{aligned} s_a &= \dfrac{6}{7}(3 + \sqrt{2}), \\ s_b &= \dfrac{6}{7}(4 + \sqrt{2}) \end{aligned}\text{。} 利用 r=面积sr = \frac{\text{面积}}{s}，可得 rarb=[ADC][BCD]⋅sbsa=34⋅4+23+2， \begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{[ADC]}{[BCD]} \cdot \dfrac{s_b}{s_a} \\ &= \dfrac{3}{4} \cdot \dfrac{4 + \sqrt{2}}{3 + \sqrt{2}} \end{aligned}\text{，}

有理化得 4+23+2=10−27\dfrac{4 + \sqrt{2}}{3 + \sqrt{2}} = \dfrac{10 - \sqrt{2}}{7}，因此 rarb=34⋅10−27=328(10−2)。 \begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{3}{4} \cdot \dfrac{10 - \sqrt{2}}{7} \\ &= \dfrac{3}{28}(10 - \sqrt{2}) \end{aligned}\text{。}

所以正确答案是 E。

By the Angle Bisector Theorem, AD:DB=CA:CB=3:4,AD:DB = CA:CB = 3:4, so AD=157AD = \tfrac{15}{7} and BD=207.BD = \tfrac{20}{7}. The areas of △ADC\triangle ADC and △BCD\triangle BCD share base CD,CD, so they are in ratio 3:4,3:4, namely 187\tfrac{18}{7} and 247.\tfrac{24}{7}.

Splitting △ABC\triangle ABC along CD,CD, which meets each leg at 45∘,45^\circ, gives 3⋅CD22+4⋅CD22=6, \dfrac{3 \cdot CD}{2\sqrt{2}} + \dfrac{4 \cdot CD}{2\sqrt{2}} = 6, so CD=1227.CD = \tfrac{12\sqrt{2}}{7}.

The two semiperimeters are sa=67(3+2),sb=67(4+2). \begin{aligned} s_a &= \dfrac{6}{7}(3 + \sqrt{2}), \\ s_b &= \dfrac{6}{7}(4 + \sqrt{2}). \end{aligned} Using r=areas,r = \frac{\text{area}}{s}, rarb=[ADC][BCD]⋅sbsa=34⋅4+23+2, \begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{[ADC]}{[BCD]} \cdot \dfrac{s_b}{s_a} \\ &= \dfrac{3}{4} \cdot \dfrac{4 + \sqrt{2}}{3 + \sqrt{2}}, \end{aligned}

Rationalizing, 4+23+2=10−27,\dfrac{4 + \sqrt{2}}{3 + \sqrt{2}} = \dfrac{10 - \sqrt{2}}{7}, so rarb=34⋅10−27=328(10−2). \begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{3}{4} \cdot \dfrac{10 - \sqrt{2}}{7} \\ &= \dfrac{3}{28}(10 - \sqrt{2}). \end{aligned}

Thus, E is the correct answer.

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