1999 AIME Problems/Problem 9
Contents
Problem
A function is defined on the complex numbers by where and are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that and that where and are relatively prime positive integers, find
Solution 1
Suppose we pick an arbitrary point on the complex plane, say . According to the definition of , this image must be equidistant to and . Thus the image must lie on the line with slope and which passes through , so its graph is . Substituting and , we get .
By the Pythagorean Theorem, we have , and the answer is .
Solution 2
Plugging in yields . This implies that must fall on the line , given the equidistant rule. By , we get , and plugging in yields . The answer is thus .
Solution 3
We are given that is equidistant from the origin and This translates to Since Because thus So the answer is .
Solution 4
Let and be the points in the complex plane represented by and , respectively. implies . Also, we are given , so is isosceles with base . Notice that the base angle of this isosceles triangle is equal to the argument of the complex number , because forms an angle of with . Drop the altitude/median from to base , and you end up with a right triangle that shows . Since and are positive, lies in the first quadrant and ; hence by right triangle trigonometry . Finally, , and , so the answer is .
Solution 5
Similarly to in Solution 3, we see that . Letting the point , we have . Expanding both sides of this equation (after squaring, of course) and canceling terms, we get . Of course, can't be zero because this property of the function holds for all complex . Therefore, and we proceed as above to get .
~ anellipticcurveoverq
Solution 6
This is a solution that minimizes the use of complex numbers, turning this into an introductory algebra analytic geometry problem.
Consider any complex number . Let denote point on the complex plane. Then on the complex plane. The equation for the line is .
Let the image of point be , after the point undergoes the function. Since each image is equidistant from the preimage and the origin, must be on the perpendicular bisector of .Given , . Then . The midpoint of is . Since the slopes of two respectively nonvertical and nonhorizontal lines have a product of , using the point-slope form, the equation of the perpendicular line to is . Rearranging, we have .
Since we know that , thus we plug in into the line: .
Let's start canceling. . Subtracting, . Thus . Since this is an identity for any , thus . . Since , thus (or simply think of as the point , and being the distance of to the origin). Thus plug in . Since and are relatively prime, the final result is .
~hastapasta
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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