Difference between revisions of "1999 AIME Problems/Problem 9"
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Suppose we pick an arbitrary point on the [[complex plane]], say <math>(1,1)</math>. According to the definition of <math>f(z) = f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i</math>, this image must be equidistant to <math>(1,1)</math> and <math>(0,0)</math>. Thus the image must lie on the line with slope <math>-1</math> and which passes through <math>\left(\frac 12, \frac12\right)</math>, so its graph is <math>x + y = 1</math>. Substituting <math>x = (a-b)</math> and <math>y = (a+b)</math>, we get <math>2a = 1 \Rightarrow a = \frac 12</math>. | Suppose we pick an arbitrary point on the [[complex plane]], say <math>(1,1)</math>. According to the definition of <math>f(z) = f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i</math>, this image must be equidistant to <math>(1,1)</math> and <math>(0,0)</math>. Thus the image must lie on the line with slope <math>-1</math> and which passes through <math>\left(\frac 12, \frac12\right)</math>, so its graph is <math>x + y = 1</math>. Substituting <math>x = (a-b)</math> and <math>y = (a+b)</math>, we get <math>2a = 1 \Rightarrow a = \frac 12</math>. |
Revision as of 21:58, 26 April 2016
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 But 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 .
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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