2008 AMC 12A Problems/Problem 20

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 $r_a/r_b$ ?

$\mathrm{(A)}\ \frac{1}{28}\left(10-\sqrt{2}\right)\qquad\mathrm{(B)}\ \frac{3}{56}\left(10-\sqrt{2}\right)\qquad\mathrm{(C)}\ \frac{1}{14}\left(10-\sqrt{2}\right)\qquad\mathrm{(D)}\ \frac{5}{56}\left(10-\sqrt{2}\right)\\\mathrm{(E)}\ \frac{3}{28}\left(10-\sqrt{2}\right)$

Solution

$[asy] import olympiad; size(300); defaultpen(0.8); pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429); pair O=incenter(A,C,D), P=incenter(B,C,D); picture p = new picture; draw(p,Circle(C,0.2)); draw(p,Circle(B,0.2)); clip(p,B--C--D--cycle); add(p); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); dot(O);dot(P); label("$A$",A,W); label("$B$",B,E); label("$C$",C,W); label("$D$",D,NE); label("$O_A$",O,W); label("$O_B$",P,W); label("$3$",(A+C)/2,W); label("$4$",(B+C)/2,S); label("$\frac{15}{7}$",(A+D)/2,NE); label("$\frac{20}{7}$",(B+D)/2,NE); label("$45^{\circ}$",(.2,.1),E); label("$\sin \theta = \frac{3}{5}$",B-(.2,-.1),W); [/asy]$

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 $s_A = \left({3 + \frac{15}{7} + \frac{12\sqrt{2}}{7}\right)\left/\right.2 = \frac{18+6\sqrt{2}}{7}$ (Error compiling LaTeX. Unknown error_msg) and $s_B = \frac{24+ 6\sqrt{2}}{7}$ . Thus, $\begin{align*} \frac{r_A}{r_B} &= \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A} = \frac{3}{4} \cdot \frac{(24+ 6\sqrt{2})/7}{(18+6\sqrt{2})/7} \\ &= \frac{3(4+\sqrt{2})}{4(3+\sqrt{2})} \cdot \left(\frac{3-\sqrt{2}}{3-\sqrt{2}}\right) = \frac{3}{28}(10-\sqrt{2}) \Rightarrow \mathrm{(E)}\qquad \blacksquare \end{align*}$

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2008 AMC 12A Problems/Problem 20

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