2008 AMC 12A Problem 20

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

\frac{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})

By the Angle Bisector Theorem,

AD:DB = CA:CB = 3:4

For each small triangle

r = \frac{\text{area}}{s};

the two triangles share the base

CD

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.

Problem 20 in other years