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 .
Contents
Solution
Solution 1
Define . By De Moivre's Theorem the roots are given by
Now, let be the root corresponding to , and let be the root corresponding to . Then The cosine difference identity simplifies that to
We need , which simplifies 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 , where for Thus, they are located at uniform intervals on the unit circle in the complex plane.
The quantity is unchanged upon rotation around the origin, so, WLOG, we can assume after rotating the axis till lies on the real axis. Let . Since and , we have We want From what we just obtained, this is equivalent to 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 .
Solution 4
Since , the roots will have magnitude . Thus, the roots can be written as and for some angles and . We rewrite the requirement as , which can now be easily manipulated to .
WLOG, let . Thus, our inequality becomes , , and finally . Obviously, , and thus it follows that, on either side of a given point, points will work. The probability is , and thus our requested sum is ~SigmaPiE
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 | ||
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