Question

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 $r_a/r_b?$

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

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

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

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

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

Solution:

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 $CD = \tfrac{12\sqrt{2}}{7}.$

Using $r = \text{area}/s$ with $s$ the semiperimeter, $\dfrac{r_a}{r_b} = \dfrac{[ADC]}{[BCD]} \cdot \dfrac{s_b}{s_a} = \dfrac{3}{4} \cdot \dfrac{4 + \sqrt{2}}{3 + \sqrt{2}},$ where the messy $CD$ terms cancel after factoring.

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

Thus, E is the correct answer.

2008 AMC 12A Problem 20

Solution:

Problem 20 in Other Years