Question

三角形 $ABC$ 中， $AC = 3$ 、 $BC = 4$ 、 $AB = 5$ 。点 $D$ 在 $AB$ 上，且 $CD$ 平分直角。 $\triangle ADC$ 和 $\triangle BCD$ 的内切圆半径分别为 $r_a$ 和 $r_b$ ，求 $r_a/r_b$ 。

Triangle $ABC$ has $AC = 3,$ $BC = 4,$ and $AB = 5.$ Point $D$ is on $AB,$ and $CD$ bisects the right angle. The inscribed circles of $\triangle ADC$ and $\triangle BCD$ have radii $r_a$ and $r_b,$ respectively. What is $r_a/r_b?$

$\dfrac{1}{28}(10 - \sqrt{2})$

$\dfrac{3}{56}(10 - \sqrt{2})$

$\dfrac{1}{14}(10 - \sqrt{2})$

$\dfrac{5}{56}(10 - \sqrt{2})$

$\dfrac{3}{28}(10 - \sqrt{2})$

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

解答：

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

将 $\triangle ABC$ 沿 $CD$ 分割，该线段与两条直角边都成 $45^\circ$ ，所以得 $CD = \tfrac{12\sqrt{2}}{7}$ 。 $\dfrac{3 \cdot CD}{2\sqrt{2}} + \dfrac{4 \cdot CD}{2\sqrt{2}} = 6,$

用 $r = \text{area}/s$ ，其中为半周长，可得其中复杂的项约去。 $\begin{aligned} s_a &= \dfrac{6}{7}(3 + \sqrt{2}), \\ s_b &= \dfrac{6}{7}(4 + \sqrt{2}). \end{aligned}$ $\begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{[ADC]}{[BCD]} \cdot \dfrac{s_b}{s_a} \\ &= \dfrac{3}{4} \cdot \dfrac{4 + \sqrt{2}}{3 + \sqrt{2}}, \end{aligned}$

有理化得 $\dfrac{4 + \sqrt{2}}{3 + \sqrt{2}} = \dfrac{10 - \sqrt{2}}{7}$ ，因此 $\begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{3}{4} \cdot \dfrac{10 - \sqrt{2}}{7} \\ &= \dfrac{3}{28}(10 - \sqrt{2}). \end{aligned}$

所以正确答案是 E。

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

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

The two semiperimeters are $\begin{aligned} s_a &= \dfrac{6}{7}(3 + \sqrt{2}), \\ s_b &= \dfrac{6}{7}(4 + \sqrt{2}). \end{aligned}$ Using $r = \text{area}/s,$ $\begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{[ADC]}{[BCD]} \cdot \dfrac{s_b}{s_a} \\ &= \dfrac{3}{4} \cdot \dfrac{4 + \sqrt{2}}{3 + \sqrt{2}}, \end{aligned}$

Rationalizing, $\dfrac{4 + \sqrt{2}}{3 + \sqrt{2}} = \dfrac{10 - \sqrt{2}}{7},$ so $\begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{3}{4} \cdot \dfrac{10 - \sqrt{2}}{7} \\ &= \dfrac{3}{28}(10 - \sqrt{2}). \end{aligned}$

Thus, E is the correct answer.

2008 AMC 12A 第 20 题

答案：E

难度评级：2100

解答：

其他年份的第 20 题