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Difference between revisions of "1999 AIME Problems/Problem 9"
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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>.
 
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>.
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===Solution 2===
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Revision as of 15:24, 8 March 2014

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

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 $259$.

Solution 3

Let $P$ and $Q$ be the points in the complex plane represented by $z$ and $(a+bi)z$, respectively. $|a+bi| = 8$ implies $OQ = 8OP$. Also, we are given $OQ = PQ$, so $OPQ$ is isosceles with base $OP$. Notice that the base angle of this isosceles triangle is equal to the argument $\theta$ of the complex number $a + bi$, because $(a+bi)z$ forms an angle of $\theta$ with $z$. Drop the altitude/median from $Q$ to base $OP$, and you end up with a right triangle that shows $\cos \theta = \frac{\frac{1}{2}OP}{8OQ} = \frac{\frac{1}{2}|z|}{8|z|} = \frac{1}{16}$. Since $a$ and $b$ are positive, $z$ lies in the first quadrant and $\theta < \pi/2$; hence by right triangle trigonometry $\sin \theta = \frac{\sqrt{255}}{16}$. Finally, $b = |a+bi|\sin\theta = 8\frac{\sqrt{255}}{16} = \frac{\sqrt{255}}{2}$, and $b^2 = \frac{255}{4}$, so the answer is $259$.

See also

1999 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AIME Problems and Solutions

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