2013 AMC 12B Problems/Problem 24
Problem
Let be a triangle where is the midpoint of , and is the angle bisector of with on . Let be the intersection of the median and the bisector . In addition is equilateral with . What is ?
Solution
Let and . From the conditions, let's deduct some convenient conditions that seems sufficient to solve the problem.
is the midpoint of side .
This implies that . Given that angle is degrees and angle is degrees, we can use the area formula to get
So, .....(1)
is angle bisector.
In the triangle , one has , therefore .....(2)
Furthermore, triangle is similar to triangle , so , therefore ....(3)
By (2) and (3) and the subtraction law of ratios, we get
Therefore , or . So .
Finally, using the law of cosine for triangle , we get
See also
2013 AMC 12B (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.