Difference between revisions of "2008 AMC 12A Problems/Problem 20"
m (→Solution 2) |
m (→Solution 2) |
||
Line 111: | Line 111: | ||
</asy></center> | </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> | + | 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>BD = \frac{20}{7}</math>. Using Stewart's Theorem gives us the equation <math>5d^2 + \frac{1500}{49} = \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 of <math>\angle ACD</math>. Similarly, <math>O_b</math> lies on the angle bisector of <math>\angle BCD</math>. Call the point on <math>CD</math> tangent to <math>O_a</math> <math>M</math>, and the point tangent to <math>O_b</math> <math>N</math>. Since <math>\triangle CO_aM</math> and <math>\triangle CO_bN</math> are right, and <math>\angle O_aCM = \angle O_bCN</math>, <math>\triangle CO_aM \sim \triangle CO_bN</math>. Then, <math>\frac{r_a}{r_b} = \frac{CM}{CN}</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 of <math>\angle ACD</math>. Similarly, <math>O_b</math> lies on the angle bisector of <math>\angle BCD</math>. Call the point on <math>CD</math> tangent to <math>O_a</math> <math>M</math>, and the point tangent to <math>O_b</math> <math>N</math>. Since <math>\triangle CO_aM</math> and <math>\triangle CO_bN</math> are right, and <math>\angle O_aCM = \angle O_bCN</math>, <math>\triangle CO_aM \sim \triangle CO_bN</math>. Then, <math>\frac{r_a}{r_b} = \frac{CM}{CN}</math>. |
Revision as of 11:13, 9 October 2016
Contents
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
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
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
(Thanks to above solution writer for the framework of my 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.