1999 AIME Problems/Problem 9
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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 
Contents |
Solution
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
We are given that 
is equidistant from the origin and 
This translates to
Since 
But 
thus 
So the answer is 
.
Solution 3
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 • Resources) | ||
| Preceded by Problem 8 | Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
