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2008 AMC 12A Problems/Problem 20

Revision as of 13:49, 23 November 2018 by Aopsmathabc (talk | contribs) (→‎Solution 3)

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 1

[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 $s_A = \left(3 + \frac{15}{7} + \frac{12\sqrt{2}}{7}\right)\left/\right.2 = \frac{18+6\sqrt{2}}{7}$ 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*}

Solution 2

[asy] import olympiad; import geometry; 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); line cd = line(C, D); picture p = new picture; picture q = new picture; picture r = new picture; picture s = new picture; draw(p,Circle(C,0.2)); clip(p,P--C--D--cycle); draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); line l1 = perpendicular(O, cd); draw(r, l1); clip(r, C--D--O--cycle); line l2 = perpendicular(P, cd); draw(s, l2); clip(s, C--P--D--cycle); add(p); add(q); add(r); add(s); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); draw(C--O); draw(C--P); dot(O); dot(P); point inter1 = intersectionpoint(l1, cd); point inter2 = intersectionpoint(l2, cd); dot(inter1); dot(inter2); 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,E); 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("\(M\)", inter1, 2W); label("\(N\)", inter2, 2E); [/asy]

We start by finding the length of $AD$ and $BD$ as in solution 1. Using the angle bisector theorem, we see that $AD = \frac{15}{7}$ and $BD = \frac{20}{7}$. Using Stewart's Theorem gives us the equation $5d^2 + \frac{1500}{49} = \frac{240}{7} + \frac{180}{7}$, where $d$ is the length of $CD$. Solving gives us $d = \frac{12\sqrt{2}}{7}$, so $CD = \frac{12\sqrt{2}}{7}$.

Call the incenters of triangles $ACD$ and $BCD$ $O_a$ and $O_b$ respectively. Since $O_a$ is an incenter, it lies on the angle bisector of $\angle ACD$. Similarly, $O_b$ lies on the angle bisector of $\angle BCD$. Call the point on $CD$ tangent to $O_a$ $M$, and the point tangent to $O_b$ $N$. Since $\triangle CO_aM$ and $\triangle CO_bN$ are right, and $\angle O_aCM = \angle O_bCN$, $\triangle CO_aM \sim \triangle CO_bN$. Then, $\frac{r_a}{r_b} = \frac{CM}{CN}$.

We now use common tangents to find the length of $CM$ and $CN$. Let $CM = m$, and the length of the other tangents be $n$ and $p$. Since common tangents are equal, we can write that $m + n = \frac{12\sqrt{2}}{7}$, $n + p = \frac{15}{7}$ and $m + p = 3$. Solving gives us that $CM = m = \frac{6\sqrt{2} + 3}{7}$. Similarly, $CN = \frac{6\sqrt{2} + 4}{7}$.

We see now that $\frac{r_a}{r_b} = \frac{\frac{6\sqrt{2} + 3}{7}}{\frac{6\sqrt{2} + 4}{7}} = \frac{6\sqrt{2} + 3}{6\sqrt{2} + 4} = \frac{60-6\sqrt{2}}{56} = \frac{3}{28}(10 - \sqrt{2}) \Rightarrow \boxed{E}$

Solution 3

(Thanks to above solution writer for the framework of my diagram)

[asy] import olympiad; import geometry; 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); line cd = line(C, D); picture p = new picture; picture q = new picture; picture r = new picture; picture s = new picture; draw(p,Circle(C,0.2)); clip(p,P--C--D--cycle); draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); line l1 = perpendicular(O, cd); draw(r, l1); clip(r, C--D--O--cycle); line l2 = perpendicular(P, cd); draw(s, l2); clip(s, C--P--D--cycle); add(p); add(q); add(r); add(s); draw(A--B--C--D--C--cycle); draw(incircle(A,C,D)); draw(incircle(B,C,D)); draw(C--O); draw(C--P); dot(O); dot(P); point inter1 = intersectionpoint(l1, cd); point inter2 = intersectionpoint(l2, cd); dot(inter1); dot(inter2); 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,E); 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("\(M\)", inter1, 2W); label("\(N\)", inter2, 2E); [/asy]

We all know that the inradius of an inscribed triangle is $\frac{\text{Area}}{\text{Semiperimeter}}$ and thus $r

See Also

2008 AMC 12A (Problems • Answer Key • 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
All AMC 12 Problems and Solutions

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