Difference between revisions of "2000 AMC 12 Problems/Problem 24"
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<cmath>AF^2 = AE \cdot AD \Longrightarrow \left(\frac {r_1}2\right)^2 = (AD - 2r_2) \cdot AD.</cmath> | <cmath>AF^2 = AE \cdot AD \Longrightarrow \left(\frac {r_1}2\right)^2 = (AD - 2r_2) \cdot AD.</cmath> | ||
− | Since <math>AD = r_1</math>, then <math>\frac{r_1^2}{4} = r_1 (r_1 - 2r_2) \Longrightarrow r_2 = \frac{3r_1}{8}</math>. Since <math>ABC</math> is equilateral, <math>\angle BAC = 60^{\circ}</math>, and so <math>\stackrel{\frown}{BC} = 12 = \frac{60}{360} 2\pi r_1 \Longrightarrow r_1 = \frac{36}{\pi}</math>. Thus <math>r_2 = \frac{27}{2 | + | Since <math>AD = r_1</math>, then <math>\frac{r_1^2}{4} = r_1 (r_1 - 2r_2) \Longrightarrow r_2 = \frac{3r_1}{8}</math>. Since <math>ABC</math> is equilateral, <math>\angle BAC = 60^{\circ}</math>, and so <math>\stackrel{\frown}{BC} = 12 = \frac{60}{360} 2\pi r_1 \Longrightarrow r_1 = \frac{36}{\pi}</math>. Thus <math>r_2 = \frac{27}{2\pi}</math> and the circumference of the circle is <math>27\ \mathrm{(D)}</math>. |
== See also == | == See also == |
Revision as of 16:36, 21 June 2009
Problem
If circular arcs and have centers at and , respectively, then there exists a circle tangent to both and , and to . If the length of is , then the circumference of the circle is
Solution
Since are all radii, it follows that is an equilateral triangle.
Draw the circle with center and radius . Then let be the point of tangency of the two circles, and be the intersection of the smaller circle and . Let be the intersection of the smaller circle and . Also define the radii (note that is a diameter of the smaller circle, as is the point of tangency of both circles, the radii of a circle is perpendicular to the tangent, hence the two centers of the circle are collinear with each other and ).
By the Power of a Point Theorem,
Since , then . Since is equilateral, , and so . Thus and the circumference of the circle is .
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |