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Difference between revisions of "1999 AIME Problems/Problem 9"
m (Minor Edits)
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<cmath>
 
<cmath>
 
\begin{eqnarray*} |(a + bi)z - z| & = & |(a + bi)z| \\
 
\begin{eqnarray*} |(a + bi)z - z| & = & |(a + bi)z| \\
−
|z(a - 1) + bzi| & = & |az + bzi| \\
+
|z(a - 1 + bi)| & = & |z(a + bi)| \\
−
|z||(a - 1) + bi| & = & |z||a + bi| \\
+
|z|\cdot|(a - 1) + bi| & = & |z|\cdot|a + bi| \\
 +
|(a - 1) + bi| & = & |a + bi| \\
 
(a - 1)^2 + b^2 & = & a^2 + b^2 \\
 
(a - 1)^2 + b^2 & = & a^2 + b^2 \\
 
& \Rightarrow & a = \frac 12 \end{eqnarray*}
 
& \Rightarrow & a = \frac 12 \end{eqnarray*}
 
</cmath>
 
</cmath>
−
Since <math>|a + bi| = 8,</math> <math>a^2 + b^2 = 64.</math> But <math>a = \frac 12,</math> thus <math>b^2 = \frac {255}4.</math>  So the answer is <math>259</math>.
+
Since <math>|a + bi| = 8,</math> <math>a^2 + b^2 = 64.</math> Because <math>a = \frac 12,</math> thus <math>b^2 = \frac {255}4.</math>  So the answer is <math>\boxed{259}</math>.
  
 
== Solution 4 ==
 
== Solution 4 ==

Revision as of 09:58, 8 June 2021

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 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

Plugging in $z=1$ yields $f(1) = a+bi$. This implies that $a+bi$ must fall on the line $Re(z)=a=\frac{1}{2}$, given the equidistant rule. By $|a+bi|=8$, we get $a^2 + b^2 = 64$, and plugging in $a=\frac{1}{2}$ yields $b^2=\frac{255}{4}$. The answer is thus $\boxed{259}$.

Solution 3

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 + bi)| & = & |z(a + bi)| \\ |z|\cdot|(a - 1) + bi| & = & |z|\cdot|a + bi| \\ |(a - 1) + bi| & = & |a + bi| \\ (a - 1)^2 + b^2 & = & a^2 + b^2 \\ & \Rightarrow & a = \frac 12 \end{eqnarray*} Since $|a + bi| = 8,$ $a^2 + b^2 = 64.$ Because $a = \frac 12,$ thus $b^2 = \frac {255}4.$ So the answer is $\boxed{259}$.

Solution 4

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 $259$.

Solution 5

Similarly to in Solution 3, we see that $|(a + bi)z - z| = |(a + bi)z|$. Letting the point $z = c + di$, we have $\sqrt{(ab+bc-d)^2+(ac-bd-c)^2} = \sqrt{(ac-bd)^2+(ad+bc)^2}$. Expanding both sides of this equation (after squaring, of course) and canceling terms, we get $(d^2+c^2)(-2a+1) = 0$. Of course, $(d^2+c^2)$ can't be zero because this property of the function holds for all complex $z$. Therefore, $a = \frac{1}{2}$ and we proceed as above to get $\boxed{259}$.

~ anellipticcurveoverq

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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