2013 AIME II Problems/Problem 5
Problem 5
In equilateral let points and trisect . Then can be expressed in the form , where and are relatively prime positive integers, and is an integer that is not divisible by the square of any prime. Find .
Solution
Without loss of generality, assume the triangle sides have length 3. Then the trisected side is partitioned into segments of length 1, making your computation easier.
Let be the midpoint of . Then is a 30-60-90 triangle with , and . Since the triangle is right, then we can find the length of by pythagorean theorem, . Therefore, since is a right triangle, we can easily find and . So we can use the double angle formula for sine, . Therefore, .
Solution 2
We find that, as before, , and also the area of is 1/3 the area of . Thus, using the area formula, , and . Therefore,
Solution 3
We notice that . We can find , to be , so our answer is .
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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