Difference between revisions of "2000 AMC 12 Problems/Problem 24"
Louisng114 (talk | contribs) (→Solution) |
m (→Solution) |
||
Line 20: | Line 20: | ||
<cmath>AF^2 + OF^2 = AO^2</cmath> | <cmath>AF^2 + OF^2 = AO^2</cmath> | ||
− | <cmath>(\frac {r_1}{2})^2 + r_2^2 = (r_1 - r_2)^2</cmath> | + | <cmath>\left(\frac {r_1}{2}\right)^2 + r_2^2 = (r_1 - r_2)^2</cmath> |
After simplification, <math>r_2 = \frac{3r_1}{8}</math>.) | After simplification, <math>r_2 = \frac{3r_1}{8}</math>.) |
Revision as of 17:11, 21 November 2017
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 .
(Alternatively, the Pythagorean Theorem can also be used to find in terms of . Notice that since AB is tangent to circle , is perpendicular to . Therefore,
After simplification, .)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.