2008 AMC 12A Problems/Problem 20
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 (Not Finished, will do later)
![[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 |
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