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
![[asy] import olympiad; import geometry; 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); line cd = line(C, D); picture p = new picture; picture q = new picture; picture r = new picture; picture s = new picture; draw(p,Circle(C,0.2)); clip(p,P--C--D--cycle); draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); line l1 = perpendicular(O, cd); draw(r, l1); clip(r, C--D--O--cycle); line l2 = perpendicular(P, cd); draw(s, l2); clip(s, C--P--D--cycle); add(p); add(q); add(r); add(s); 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); point inter1 = intersectionpoint(l1, cd); point inter2 = intersectionpoint(l2, cd); dot(inter1); dot(inter2); 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,E); 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("\(M\)", inter1, 2W); label("\(N\)", inter2, 2E); [/asy]](http://latex.artofproblemsolving.com/c/4/b/c4b25343784f830ea4d327aef305fb52a67bbc4a.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 of
. Similarly,
lies on the angle bisector of
. Call the point on
tangent to
, and the point tangent to
. Since
and
are right, and
,
. Then,
.
We now use common tangents to find the length of and
. Let
, and the length of the other tangents be
and
. Since common tangents are equal, we can write that
,
and
. Solving gives us that
. Similarly,
.
We see now that
Solution 3
(Thanks to above solution writer for the framework of my diagram)
![[asy] import olympiad; import geometry; 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); line cd = line(C, D); picture p = new picture; picture q = new picture; picture r = new picture; picture s = new picture; draw(p,Circle(C,0.2)); clip(p,P--C--D--cycle); draw(q, Circle(C, 0.3)); clip(q, O--C--D--cycle); line l1 = perpendicular(O, cd); draw(r, l1); clip(r, C--D--O--cycle); line l2 = perpendicular(P, cd); draw(s, l2); clip(s, C--P--D--cycle); add(p); add(q); add(r); add(s); 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); point inter1 = intersectionpoint(l1, cd); point inter2 = intersectionpoint(l2, cd); dot(inter1); dot(inter2); 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,E); 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("\(M\)", inter1, 2W); label("\(N\)", inter2, 2E); [/asy]](http://latex.artofproblemsolving.com/c/4/b/c4b25343784f830ea4d327aef305fb52a67bbc4a.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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.