Difference between revisions of "2008 AMC 12A Problems/Problem 20"
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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; | ||
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&= \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(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | </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
Contents
[hide]Problem
Triangle has
,
, and
. Point
is on
, and
bisects the right angle. The inscribed circles of
and
have radii
and
, respectively. What is
?
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]](http://latex.artofproblemsolving.com/f/0/0/f003e5fe565c49bade1dc1e2b332a46d1a084ca9.png)
By the Angle Bisector Theorem,
By Law of Sines on
,
Since the area of a triangle satisfies
, where
the inradius and
the semiperimeter, we have
Since
and
share the altitude (to
), their areas are the ratio of their bases, or
The semiperimeters are
and
. Thus,
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]](http://latex.artofproblemsolving.com/0/6/4/064bec4b892644507fb15b0dddd147c483f44732.png)
We start by finding the length of and
as in solution 1. Using the angle bisector theorem, we see that
and
. Using Stewart's Theorem gives us the equation
, where
is the length of
. Solving gives us
, so
.
Call the incenters of triangles and
and
respectively. Since
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.