2021 Fall AMC 12A Problems/Problem 11
Contents
Problem
Consider two concentric circles of radius and The larger circle has a chord, half of which lies inside the smaller circle. What is the length of the chord in the larger circle?
Solution 1 (Power of a Point)
Draw the diameter perpendicular to the chord. Call the intersection between that diameter and the chord . In the circle of radius , let the shorter piece of the diameter cut by the chord would be of length , making the longer piece In that same circle, let the be the length of the portion of the chord in the smaller circle that is cut by the diameter we drew. Thus, in the circle of radius , the shorter piece of the diameter cut by the chord would be of length , making the longer piece and length of the piece of the chord cut by the diameter would be (as given in the problem). By Power of a Point, we can construct the system of equations Expanding both equations, we get and in which the and terms magically cancel when we subtract the first equation from the second equation. Thus, now we have .
-fidgetboss_4000
Solution 2 (Pythagorean Theorem)
Label the intersection of the chord with the smaller and larger circle as A and B respectively.
Construct the radius perpendicular to the chord and label its intersection with the chord as M. Because a radius that is perpendicular to a chord also bisects the chord and because half of the chord in the larger circle lies in the smaller circle, .
Construct segments AO and BO. These are radii with lengths 17 and 19 respectively.
Then, use the Pythagorean Theorem on right triangles and to get the following system of equations:
So, , and the length of the chord in the larger circle .
Solution 3
Denote by the center of both circles. Denote by this chord. Let this chord intersect the smaller circle at and , where is between and . Denote by the midpoint of .
In , following from the Pythagorean theorem, we have
In , following from the Pythagorean theorem, we have
Equating these two equations, we get
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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.