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2008 AMC 12A Problems/Problem 20

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Problem

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 $r_a/r_b$ ?

$\mathrm{(A)}\ \frac{1}{28}\left(10-\sqrt{2}\right)\qquad\mathrm{(B)}\ \frac{3}{56}\left(10-\sqrt{2}\right)\qquad\mathrm{(C)}\...$

Solution

import olympiad;size(300);defaultpen(0.8);pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429);pair O=incenter(A,C,D), P=incent...

By the Angle Bisector Theorem, $\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7$ By Law of Sines on $\triangle BCD$ , $\frac{BD}{\sin 45^{\circ}} = \frac{CD}{\sin \angle B} \Longrightarrow \frac{20/7}{\sqrt{2}/2} = \frac{CD}{3/5} \Longrightarro...$ 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, $\begin{align*}\frac{r_A}{r_B} &= \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A} = \frac{3}{4} \cdot \frac{(24+ 6\sqrt{2})/7}{(18...$

See Also

2008 AMC 12A (Problems • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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