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Difference between revisions of "1999 AIME Problems/Problem 9"

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== Problem ==
 
== Problem ==
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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>
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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>
  
 
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__TOC__
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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===

Revision as of 11:46, 19 October 2007

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

Solution 1

Suppose we pick an arbitrary point on the complex plane, say $(1,1)$. According to the definition of $f(z) = f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i$, this image must be equidistant to $(1,1)$ and $(0,0)$. Thus the image must lie on the line with slope $-1$ and which passes through $\left(\frac 12, \frac12\right)$, 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 $\left(\frac{1}{2}\right)^2 + b^2 = 8^2 \Longrightarrow b^2 = \frac{255}{4}$, and the answer is $\boxed{259}$.

Solution 2

We are given that $(a + bi)z$ is equidistant from the origin and $z.$ This translates to \begin{eqnarray*} |(a + bi)z - z| & = & |(a + bi)z| \\ |z(a - 1) + bzi| & = & |az + bzi| \\ |z||(a - 1) + bi| & = & |z||a + bi| \\ (a - 1)^2 + b^2 & = & a^2 + b^2 \\ & \Rightarrow & a = \frac 12 \end{eqnarray*} Since $|a + bi| = 8,$ $a^2 + b^2 = 64.$ But $a = \frac 12,$ thus $b^2 = \frac {255}4.$ So the answer is $259$.

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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