Difference between revisions of "2014 AIME I Problems/Problem 7"
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And the answer is <math>1+99=\boxed{100}</math>. | And the answer is <math>1+99=\boxed{100}</math>. | ||
− | ==Video Solution== | + | ==Video Solution by Punxsutawney Phil== |
https://www.youtube.com/watch?v=yakhEuPy6Sg | https://www.youtube.com/watch?v=yakhEuPy6Sg | ||
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== See also == | == See also == |
Revision as of 21:25, 9 April 2021
Contents
Problem 7
Let and be complex numbers such that and . Let . The maximum possible value of can be written as , where and are relatively prime positive integers. Find . (Note that , for , denotes the measure of the angle that the ray from to makes with the positive real axis in the complex plane)
Solution
Let and . Then, .
Multiplying both the numerator and denominator of this fraction by gives us:
.
We know that is equal to the imaginary part of the above expression divided by the real part. Let . Then, we have that:
We need to find a maximum of this expression, so we take the derivative:
Thus, we see that the maximum occurs when . Therefore, , and . Thus, the maximum value of is , or , and our answer is .
Solution 2
Without the loss of generality one can let lie on the positive x axis and since is a measure of the angle if then and we can see that the question is equivalent to having a triangle with sides and and trying to maximize the angle
using the Law of Cosines we get: rearranging: solving for we get:
if we want to maximize we need to minimize , using AM-GM inequality we get that the minimum value for hence using the identity we get and our answer is .
Solution 3
Note that , and that . Thus is a complex number on the circle with radius and centered at on the complex plane. Let denote this circle.
Let and be the points that represent and respectively on the complex plane. Let be the origin. In order to maximize , we need to maximize . This angle is maximized when is tangent to . Using the Pythagorean Theorem, we get
Thus
And the answer is .
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=yakhEuPy6Sg
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.