2012 AIME II Problems/Problem 13
Problem 13
Equilateral has side length . There are four distinct triangles , , , and , each congruent to , with . Find .
Solution
Note that there are only two possible locations for points and , as they are both from point and from point , so they are the two points where a circle centered at with radius and a circle centered at with radius intersect. Let be the point on the opposite side of from , and the point on the same side of as .
Let be the measure of angle (which is also the measure of angle ); by the Law of Cosines,
There are two equilateral triangles with as a side; let be the third vertex that is farthest from , and be the third vertex that is nearest to .
Angle ; by the Law of Cosines, Angle ; by the Law of Cosines,
There are two equilateral triangles with as a side; let be the third vertex that is farthest from , and be the third vertex that is nearest to .
Angle ; by the Law of Cosines, Angle ; by the Law of Cosines,
The solution is: Substituting for gives the solution
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |