2000 AMC 12 Problems/Problem 24
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
Solutions
Solution 1
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, .
Solution 2 (Pythagorean Theorem)
First, note the triangle is equilateral. Next, notice that since the arc has length 12, it follows that we can find the radius of the sector centered at . . Next, connect the center of the circle to side , and call this length , and call the foot . Since is equilateral, it follows that , and (where O is the center of the circle) is . By the Pythagorean Theorem, you get . Finally, we see that the circumference 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.