AMC12 2008 A

AMC12 2008 A · Q20

AMC12 2008 A · Q20. It mainly tests Angle chasing, Triangles (properties).

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

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

(A) \frac{1}{28}(10 - \sqrt{2}) \frac{1}{28}(10 - \sqrt{2})

(B) \frac{3}{56}(10 - \sqrt{2}) \frac{3}{56}(10 - \sqrt{2})

(D) \frac{5}{56}(10 - \sqrt{2}) \frac{5}{56}(10 - \sqrt{2})

(E) \frac{3}{28}(10 - \sqrt{2}) \frac{3}{28}(10 - \sqrt{2})

Answer

Correct choice: (E)

正确答案：（E）

Solution

By the Angle Bisector Theorem, \[\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7\] By Law of Sines on $\triangle BCD$, \[\frac{BD}{\sin 45^{\circ}} = \frac{CD}{\sin \angle B} \Longrightarrow \frac{20/7}{\sqrt{2}/2} = \frac{CD}{3/5} \Longrightarrow CD=\frac{12\sqrt{2}}{7}\] Since the area of a triangle satisfies $[\triangle]=rs$, where $r =$ the inradius and $s =$ the semiperimeter, we have \[\frac{r_A}{r_B} = \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A}\] Since $\triangle ACD$ and $\triangle BCD$ share the altitude (to $\overline{AB}$), their areas are the ratio of their bases, or \[\frac{[ACD]}{[BCD]} = \frac{AD}{BD} = \frac{3}{4}\] The semiperimeters are $s_A = \left(3 + \frac{15}{7} + \frac{12\sqrt{2}}{7}\right)\left/\right.2 = \frac{18+6\sqrt{2}}{7}$ and $s_B = \frac{24+ 6\sqrt{2}}{7}$. Thus, \begin{align*} \frac{r_A}{r_B} &= \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A} = \frac{3}{4} \cdot \frac{(24+ 6\sqrt{2})/7}{(18+6\sqrt{2})/7} \\ &= \frac{3(4+\sqrt{2})}{4(3+\sqrt{2})} \cdot \left(\frac{3-\sqrt{2}}{3-\sqrt{2}}\right) = \frac{3}{28}(10-\sqrt{2}) \Rightarrow \mathrm{(E)}\qquad \blacksquare \end{align*}

由角平分定理， \[\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7\] 在 $\triangle BCD$ 中用正弦定理， \[\frac{BD}{\sin 45^{\circ}} = \frac{CD}{\sin \angle B} \Longrightarrow \frac{20/7}{\sqrt{2}/2} = \frac{CD}{3/5} \Longrightarrow CD=\frac{12\sqrt{2}}{7}\] 由于三角形面积满足 $[\triangle]=rs$，其中 $r$ 为内切圆半径，$s$ 为半周长，有 \[\frac{r_A}{r_B} = \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A}\] 因为 $\triangle ACD$ 与 $\triangle BCD$ 共享同一高（到底边 $\overline{AB}$），它们面积之比等于底边之比，即 \[\frac{[ACD]}{[BCD]} = \frac{AD}{BD} = \frac{3}{4}\] 半周长为 $s_A = \left(3 + \frac{15}{7} + \frac{12\sqrt{2}}{7}\right)\left/\right.2 = \frac{18+6\sqrt{2}}{7}$，$s_B = \frac{24+ 6\sqrt{2}}{7}$。因此， \begin{align*} \frac{r_A}{r_B} &= \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A} = \frac{3}{4} \cdot \frac{(24+ 6\sqrt{2})/7}{(18+6\sqrt{2})/7} \\ &= \frac{3(4+\sqrt{2})}{4(3+\sqrt{2})} \cdot \left(\frac{3-\sqrt{2}}{3-\sqrt{2}}\right) = \frac{3}{28}(10-\sqrt{2}) \Rightarrow \mathrm{(E)}\qquad \blacksquare \end{align*}

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