1997 AIME Problems/Problem 14
Problem
Let and be distinct, randomly chosen roots of the equation . Let be the probability that , where and are relatively prime positive integers. Find .
Solution
Solution 1
By De Moivre's Theorem, we find that ()
Now, let be the root corresponding to , and let be the root corresponding to . The magnitude of is therefore:
We need . The cosine difference identity simplifies that to . Thus, .
Therefore, and cannot be more than away from each other. This means that for a given value of , there are values for that satisfy the inequality; of them , and of them . Since and must be distinct, can have possible values. Therefore, the probability is . The answer is then .
Solution 2
The solutions of the equation are the th roots of unity and are equal to for They are also located at the vertices of a regular -gon that is centered at the origin in the complex plane.
Without loss of generality, let Then
We want From what we just obtained, this is equivalent to This occurs when which is satisfied by (we don't include 0 because that corresponds to ). So out of the possible , work. Thus, So our answer is
Solution 3
We can solve a geometrical interpretation of this problem.
Without loss of generality, let . We are now looking for a point exactly one unit away from such that the point is at least units away from the origin. Note that the "boundary" condition is when the point will be exactly units away from the origin; these points will be the intersections of the circle centered at with radius and the circle centered at with radius . The equations of these circles are and . Solving for yields . Clearly, this means that the real part of is greater than . Solving, we note that possible s exist, meaning that . Therefore, the answer is .
See also
1997 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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