Difference between revisions of "2008 AMC 12A Problems/Problem 20"
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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> bisects the right angle. The inscribed circles of <math>\triangle ADC</math> and <math>\triangle BCD</math> have radii <math>r_a</math> and <math>r_b</math>, respectively. What is <math>r_a/r_b</math>? | 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> bisects the right angle. The inscribed circles of <math>\triangle ADC</math> and <math>\triangle BCD</math> have radii <math>r_a</math> and <math>r_b</math>, respectively. What is <math>r_a/r_b</math>? | ||
| − | <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 == | ||
Revision as of 00:51, 26 April 2008
Problem
Triangle 
has 
, 
, and 
. Point 
is on 
, and 
bisects the right angle. The inscribed circles of 
and 
have radii 
and 
, respectively. What is 
?
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]](http://latex.artofproblemsolving.com/f/0/0/f003e5fe565c49bade1dc1e2b332a46d1a084ca9.png)
By the Angle Bisector Theorem,
![\[\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7\]](http://latex.artofproblemsolving.com/9/c/9/9c9f464a5fad91f50fc120b232c9b27b70d1c93f.png)
By Law of Sines on ,
![\[\frac{BD}{\sin 45^{\circ}} = \frac{CD}{\sin \angle B} \Longrightarrow \frac{20/7}{\sqrt{2}/2} = \frac{CD}{3/5} \Longrightarrow CD=\frac{12\sqrt{2}}{7}\]](http://latex.artofproblemsolving.com/9/d/f/9df704c41b92bf28dfdabff885b9a06d3fcb4441.png)
Since the area of a triangle satisfies ![$[\triangle]=rs$](http://latex.artofproblemsolving.com/6/6/d/66dc3f0f937032d1109f338a8d694db762e4ea41.png)
, where 
the inradius and 
the semiperimeter, we have
![\[\frac{r_A}{r_B} = \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A}\]](http://latex.artofproblemsolving.com/a/f/7/af76ed60fa2fdfb3bfb69329f9c290478535c124.png)
Since 
and share the altitude (to ), their areas are the ratio of their bases, or ![\[\frac{[ACD]}{[BCD]} = \frac{AD}{BD} = \frac{3}{4}\]](http://latex.artofproblemsolving.com/e/f/d/efd8576e6845c12c5b4c017aa1280a1b061fd3a6.png)
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 
. 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*}](http://latex.artofproblemsolving.com/b/4/3/b439f853787f7bda3a9170414daae8b3f8056e69.png)
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 | |
