1999 AIME Problems/Problem 9

Revision as of 16:42, 18 October 2007 by Azjps (talk | contribs) (solution)

Problem

A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ 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 $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$

Solution

Suppose we pick an arbitrary point on the line $x = y$ on the complex plane, say $(1,1)$. According to the definition of $f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i$, the image must be equidistant to $(1,1)$ and $(0,0)$. These points lie on the line with slope $-1$ and which passes through $\left\frac 12, \frac12\right)$ (Error compiling LaTeX. Unknown error_msg), so its graph is $x + y = 1$. Substituting $x = (a-b)$ and $y = (a+b)$, we get $2a = 1 \Rightarrow a = \frac 12$.

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 $\boxed{259}$.

See also

1999 AIME (ProblemsAnswer KeyResources)
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