Difference between revisions of "2008 AMC 12A Problems/Problem 20"
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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 | + | ==Solution 2== |
<center><asy> | <center><asy> | ||
import olympiad; | import olympiad; | ||
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label("\(\frac{15}{7}\)",(A+D)/2,NE); | label("\(\frac{15}{7}\)",(A+D)/2,NE); | ||
label("\(\frac{20}{7}\)",(B+D)/2,NE); | label("\(\frac{20}{7}\)",(B+D)/2,NE); | ||
− | label("\(M\)", inter1, | + | label("\(M\)", inter1, 2W); |
− | label("\(N\)", inter2, | + | label("\(N\)", inter2, 2E); |
</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>. | ||
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We now use common tangents to find the length of <math>CM</math> and <math>CN</math>. Let <math>CM = m</math>, and the length of the other tangents be <math>n</math> and <math>p</math>. Since common tangents are equal, we can write that <math>m + n = \frac{12\sqrt{2}}{7}</math>, <math>n + p = \frac{15}{7}</math> and <math>m + p = 3</math>. Solving gives us that <math>CM = m = \frac{6\sqrt{2} + 3}{7}</math>. Similarly, <math>CN = \frac{6\sqrt{2} + 4}{7}</math>. | We now use common tangents to find the length of <math>CM</math> and <math>CN</math>. Let <math>CM = m</math>, and the length of the other tangents be <math>n</math> and <math>p</math>. Since common tangents are equal, we can write that <math>m + n = \frac{12\sqrt{2}}{7}</math>, <math>n + p = \frac{15}{7}</math> and <math>m + p = 3</math>. Solving gives us that <math>CM = m = \frac{6\sqrt{2} + 3}{7}</math>. Similarly, <math>CN = \frac{6\sqrt{2} + 4}{7}</math>. | ||
− | We see now that <math>\frac{r_a}{r_b} = \frac{\frac{6\sqrt{2} + 3}{7}}{\frac{6\sqrt{2} + 4}{7}} = \frac{6\sqrt{2} + 3}{6\sqrt{2} + 4} = \frac{60-6\sqrt{2}}{56} = \frac{3}{28}(10 - \sqrt{2}) \Rightarrow \boxed{ | + | We see now that <math>\frac{r_a}{r_b} = \frac{\frac{6\sqrt{2} + 3}{7}}{\frac{6\sqrt{2} + 4}{7}} = \frac{6\sqrt{2} + 3}{6\sqrt{2} + 4} = \frac{60-6\sqrt{2}}{56} = \frac{3}{28}(10 - \sqrt{2}) \Rightarrow \boxed{E}</math> |
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==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}} |
Latest revision as of 15:31, 2 August 2019
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
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.