Difference between revisions of "1999 AIME Problems/Problem 9"
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+ | Plugging in <math>z=1</math> yields <math>f(1) = a+bi</math>. This implies that a+bi must fall on the line <math>Re(z)=a=\frac{1}{2}</math>, given the equidistant rule. By <math>|a+bi|=8, we get </math>a^2 + b^2 = 64<math>, and plugging in </math>a=\frac{1}{2}<math> yields </math>b^2=\frac{255}{4}<math>. The answer is thus </math>\boxed{259}$. | ||
=== Solution 3 === | === Solution 3 === |
Revision as of 16:28, 8 March 2014
Problem
A function is defined on the complex numbers by
where
and
are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that
and that
where
and
are relatively prime positive integers, find
Contents
Solution
Solution 1
Suppose we pick an arbitrary point on the complex plane, say . According to the definition of
, this image must be equidistant to
and
. Thus the image must lie on the line with slope
and which passes through
, so its graph is
. Substituting
and
, we get
.
By the Pythagorean Theorem, we have , and the answer is
.
Solution 2
Plugging in yields
. This implies that a+bi must fall on the line
, given the equidistant rule. By
a^2 + b^2 = 64
a=\frac{1}{2}
b^2=\frac{255}{4}
\boxed{259}$.
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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