2008 AMC 12A Problem 20

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 $r = \frac{\text{area}}{s},$ $$\begin{aligned} \dfrac{r_a}{r_b} &= \dfrac{[ADC]}{[BCD]} \cdot \dfrac{s_b}{s_a} \\ &= \dfrac{3}{4} \cdot \dfrac{4 + \sqrt{2}}{3 + \sqrt{2}}, \end{aligned}$$ 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.