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

m (→‎Solution)
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<math>\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)</math>
 
<math>\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)</math>
  
−
== Solution ==
+
== Solution 1==
 
<center><asy>
 
<center><asy>
 
import olympiad;
 
import olympiad;
Line 49: Line 49:
 
&= \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>
 
&= \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>
  
 +
==Solution 2==
 +
<center><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));
 +
clip(p,P--C--D--cycle);
 +
picture q = new picture;
 +
draw(q, Circle(C, 0.3));
 +
clip(q, O--C--D--cycle);
 +
add(p);
 +
add(q);
 +
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);
 +
label("\(A\)",A,W);
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label("\(B\)",B,E);
 +
label("\(C\)",C,W);
 +
label("\(D\)",D,NE);
 +
label("\(O_a\)",O,W);
 +
label("\(O_b\)",P,NW);
 +
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);
 +
</asy></center>
 +
 +
We start by finding the length of <math>AD</math> and <math>BD</math> as in solution 1. Using the angle bisector theorem, we see that <math>AD = \frac{15}{7}</math> and <math>CD = \frac{20}{7}</math>. Using Stewart's Theorem gives us the equation <math>5d^2 + \frac{1500}{7} = \frac{240}{7} + \frac{180}{7}</math>, where <math>d</math> is the length of <math>CD</math>. Solving gives us <math>d = \frac{12\sqrt{2}}{7}</math>, so <math>CD = \frac{12\sqrt{2}}{7}</math>.
 +
 +
Call the incenters of triangles <math>ACD</math> and <math>BCD</math>  <math>O_a</math> and <math>O_b</math> respectively. Since <math>O_a</math> is an incenter, it lies on the angle bisector
 +
 +
(Thanks to above solution for diagram)
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|num-b=19|num-a=21|ab=A}}
 
{{AMC12 box|year=2008|num-b=19|num-a=21|ab=A}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:12, 16 January 2016

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; 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)); clip(p,P--C--D--cycle); picture q = new picture; draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); add(p); add(q); 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); 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,NW); 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); [/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 $CD = \frac{20}{7}$. Using Stewart's Theorem gives us the equation $5d^2 + \frac{1500}{7} = \frac{240}{7} + \frac{180}{7}$, 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

(Thanks to above solution for diagram)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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