Difference between revisions of "2008 AMC 12B Problems/Problem 9"
Line 1: | Line 1: | ||
− | ==Problem | + | ==Problem== |
Points <math>A</math> and <math>B</math> are on a circle of radius <math>5</math> and <math>AB = 6</math>. Point <math>C</math> is the midpoint of the minor arc <math>AB</math>. What is the length of the line segment <math>AC</math>? | Points <math>A</math> and <math>B</math> are on a circle of radius <math>5</math> and <math>AB = 6</math>. Point <math>C</math> is the midpoint of the minor arc <math>AB</math>. What is the length of the line segment <math>AC</math>? | ||
Latest revision as of 12:48, 15 February 2021
Problem
Points and are on a circle of radius and . Point is the midpoint of the minor arc . What is the length of the line segment ?
Solutions
Solution 1
Let be the angle that subtends the arc . By the law of cosines, implies .
The half-angle formula says that . The law of cosines tells us , which is answer choice .
Solution 2
Define as the midpoint of line segment , and the center of the circle. Then , , and are collinear, and since is the midpoint of , and so . Since , , and so .
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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.