Difference between revisions of "2008 AMC 12A Problems/Problem 20"
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<cmath>[ACD] = [ABC] - [BCD] = \frac 12 (3)(4) - \frac{24}{7} = \frac{18}{7}</cmath>--> | <cmath>[ACD] = [ABC] - [BCD] = \frac 12 (3)(4) - \frac{24}{7} = \frac{18}{7}</cmath>--> | ||
Since <math>\triangle ACD</math> and <math>\triangle BCD</math> share the [[altitude]] (to <math>\overline{AB}</math>), their areas are the ratio of their bases, or <cmath>\frac{[ACD]}{[BCD]} = \frac{AD}{BD} = \frac{3}{4}</cmath> | Since <math>\triangle ACD</math> and <math>\triangle BCD</math> share the [[altitude]] (to <math>\overline{AB}</math>), their areas are the ratio of their bases, or <cmath>\frac{[ACD]}{[BCD]} = \frac{AD}{BD} = \frac{3}{4}</cmath> | ||
− | The semiperimeters are <math>s_A = \left( | + | The semiperimeters are <math>s_A = \left(3 + \frac{15}{7} + \frac{12\sqrt{2}}{7}\right)\left/\right.2 = \frac{18+6\sqrt{2}}{7}</math> and <math>s_B = \frac{24+ 6\sqrt{2}}{7}</math>. Thus, |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\frac{r_A}{r_B} &= \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A} = \frac{3}{4} \cdot \frac{(24+ 6\sqrt{2})/7}{(18+6\sqrt{2})/7} \\ | \frac{r_A}{r_B} &= \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A} = \frac{3}{4} \cdot \frac{(24+ 6\sqrt{2})/7}{(18+6\sqrt{2})/7} \\ |
Revision as of 16:26, 2 May 2015
Problem
Triangle has , , and . Point is on , and bisects the right angle. The inscribed circles of and have radii and , respectively. What is ?
Solution
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,
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.