Difference between revisions of "1999 AIME Problems/Problem 9"
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== Problem == | == Problem == | ||
| − | A function <math> | + | A function <math>f</math> is defined on the [[complex number]]s by <math>f(z)=(a+bi)z,</math> where <math>a_{}</math> and <math>b_{}</math> 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 <math>|a+bi|=8</math> and that <math>b^2=m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are relatively prime positive integers. Find <math>m+n.</math> |
== Solution == | == Solution == | ||
| + | Suppose we pick an arbitrary point on the line <math>x = y</math> on the [[complex plane]], say <math>(1,1)</math>. According to the definition of <math>f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i</math>, the image must be equidistant to <math>(1,1)</math> and <math>(0,0)</math>. These points 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>. | ||
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| + | By the [[Pythagorean Theorem]], we have <math>\left(\frac{1}{2}\right_^2 + b^2 = 8^2 \Longrightarrow b^2 = \frac{255}{4}</math>, and the answer is <math>\boxed{259}</math>. | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1999|num-b=8|num-a=10}} | |
| − | + | ||
| − | + | [[Category:Intermediate Complex Numbers Problems]] | |
| + | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 16:42, 18 October 2007
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
Suppose we pick an arbitrary point on the line 
on the complex plane, say 
. According to the definition of 
, the image must be equidistant to and 
. These points lie on the line with slope 
and which passes through $\left\frac 12, \frac12\right)$ (Error compiling LaTeX. Unknown error_msg), so its graph is 
. Substituting 
and 
, we get 
.
By the Pythagorean Theorem, we have $\left(\frac{1}{2}\right_^2 + b^2 = 8^2 \Longrightarrow b^2 = \frac{255}{4}$ (Error compiling LaTeX. Unknown error_msg), and 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 | ||
