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Difference between revisions of "2008 AMC 12A Problems/Problem 20"

(New page: ==Problem== Triangle <math>ABC</math> has <math>AC=3</math>, <math>BC=4</math>, and <math>AB=5</math>. Point <math>D</math> is on <math>\overline{AB}</math>, and <math>\overline{CD}</math>...)
 
(moved from #21)
Line 3: Line 3:
  
 
<math>\textbf{(A)}\ \frac{1}{28}\left(10-\sqrt{2}\right) \qquad \textbf{(B)}\ \frac{3}{56}\left(10-\sqrt{2}\right) \qquad \textbf{(C)}\ \frac{1}{14}\left(10-\sqrt{2}\right) \qquad \textbf{(D)}\ \frac{5}{56}\left(10-\sqrt{2}\right) \\ \textbf{(E)}\ \frac{3}{28}\left(10-\sqrt{2}\right)</math>
 
<math>\textbf{(A)}\ \frac{1}{28}\left(10-\sqrt{2}\right) \qquad \textbf{(B)}\ \frac{3}{56}\left(10-\sqrt{2}\right) \qquad \textbf{(C)}\ \frac{1}{14}\left(10-\sqrt{2}\right) \qquad \textbf{(D)}\ \frac{5}{56}\left(10-\sqrt{2}\right) \\ \textbf{(E)}\ \frac{3}{28}\left(10-\sqrt{2}\right)</math>
 +
 +
== Solution ==
 +
<center><asy>
 +
import olympiad;
 +
size(300);
 +
defaultpen(0.8);
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pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429);
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pair O=incenter(A,C,D), P=incenter(B,C,D);
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picture p = new picture;
 +
draw(p,Circle(C,0.2)); draw(p,Circle(B,0.2));
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clip(p,B--C--D--cycle);
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add(p);
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draw(A--B--C--D--C--cycle);
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draw(incircle(A,C,D));
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draw(incircle(B,C,D));
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dot(O);dot(P);
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label("\(A\)",A,W);
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label("\(B\)",B,E);
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label("\(C\)",C,W);
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label("\(D\)",D,NE);
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label("\(O_A\)",O,W);
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label("\(O_B\)",P,W);
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label("\(3\)",(A+C)/2,W);
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label("\(4\)",(B+C)/2,S);
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label("\(\frac{15}{7}\)",(A+D)/2,NE);
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label("\(\frac{20}{7}\)",(B+D)/2,NE);
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label("\(45^{\circ}\)",(.2,.1),E);
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label("\(\sin \theta = \frac{3}{5}\)",B-(.2,-.1),W);
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</asy></center>
 +
 +
By the [[Angle Bisector Theorem]],
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<cmath>\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7</cmath>
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By [[Law of Sines]] on <math>\triangle BCD</math>,
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<cmath>\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}</cmath>
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Since the area of a triangle satisfies <math>[\triangle]=rs</math>, where <math>r = </math> the [[inradius]] and <math>s =</math> the [[semiperimeter]], we have
 +
<cmath>\frac{r_A}{r_B} = \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A}</cmath>
 +
<!--Using any of various formulas for triangle area, we find the area <math>[BCD]</math> to be
 +
<cmath>[BCD] = \frac{1}{2} (\sin \angle CBD) \cdot (BD) \cdot (CD) = \frac 12 \cdot \frac 35 \cdot \frac{20}{7} \cdot 4 = \frac{24}{7}</cmath>
 +
and
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<cmath>[ACD] = [ABC] - [BCD] = \frac 12 (3)(4) - \frac{24}{7} = \frac{18}{7}</cmath>-->
 +
Since <math>\triangle ACD</math> and <math>\triangle BCD</math> share the [[altitude]] (to <math>\overline{AB}</math>), their areas are the ratio of their bases, or <cmath>\frac{[ACD]}{[BCD]} = \frac{AD}{BD} = \frac{3}{4}</cmath>
 +
The semiperimeters are <math>s_A = \left({3 + \frac{15}{7} + \frac{12\sqrt{2}}{7}\right)\left/\right.2 = \frac{18+6\sqrt{2}}{7}</math> and <math>s_B = \frac{24+ 6\sqrt{2}}{7}</math>. Thus,
 +
<cmath>\begin{align*}
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\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*}</cmath>
 +
 +
==See Also==
 +
{{AMC12 box|year=2008|num-b=19|num-a=21|ab=A}}

Revision as of 21:41, 19 February 2008

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 $r_a/r_b$?

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

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*}

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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