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

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~ anellipticcurveoverq
 
~ anellipticcurveoverq
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== Solution 6 ==
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This is a solution that minimizes the use of complex numbers, turning this into an introductory algebra analytic geometry problem.
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Consider any complex number <math>z=c+di</math>. Let <math>z</math> denote point <math>P</math> on the complex plane. Then <math>P=(c,d)</math> on the complex plane. The equation for the line <math>OP</math> is <math>y=\frac{d}{c}x</math>.
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Let the image of point <math>P</math> be <math>Q</math>, after the point undergoes the function. Since each image is equidistant from the preimage and the origin, <math>Q</math> must be on the perpendicular bisector of <math>OP</math>.Given <math>z=c+di</math>, <math>f(z)=(a+bi)(c+di)=(ac-bd)+(ad+bc)i</math>. Then <math>Q=(ac-bd,ad+bc)</math>. The midpoint of <math>OP</math> is <math>(0.5c, 0.5d)</math>. Since the slopes of two respectively nonvertical and nonhorizontal lines have a product of <math>-1</math>, using the point-slope form, the equation of the perpendicular line to <math>OP</math> is <math>y-0.5d=-\frac{c}{d}(x-0.5c)</math>. Rearranging, we have <math>y=-\frac{cx}{d}+\frac{c^2}{2d}+\frac{d}{2}</math>.
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Since we know that <math>Q=(ac-bd,ad+bc)</math>, thus we plug in <math>Q</math> into the line: <math>ad+bc=-\frac{ac^2-bcd}{d}+\frac{c^2}{2d}+\frac{d}{2}</math>.
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Let's start canceling. <math>2ad^2+2bcd=-2ac^2+2bcd+c^2+d^2</math>. Subtracting, <math>c^2+d^2-2ac^2=2ad^2</math>. Thus <math>c^2+d^2=2ac^2+2ad^2</math>. Since this is an identity for any <math>(c,d)</math>, thus <math>2a=1</math>. <math>a=\frac{1}{2}</math>. Since <math>|a+bi|=8</math>, thus <math>a^2+b^2=64</math> (or simply think of <math>a+bi</math> as the point <math>(a,b)</math>, and <math>|a+bi|</math> being the distance of <math>(a,b)</math> to the origin). Thus plug in <math>a=\frac{1}{2}, b^2=\frac{255}{4}</math>. Since <math>255</math> and <math>4</math> are relatively prime, the final result is <math>255+4=\boxed{259}</math>.
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~hastapasta
  
 
== See also ==
 
== See also ==

Latest revision as of 18:01, 24 May 2023

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

Solution 6

This is a solution that minimizes the use of complex numbers, turning this into an introductory algebra analytic geometry problem.

Consider any complex number $z=c+di$. Let $z$ denote point $P$ on the complex plane. Then $P=(c,d)$ on the complex plane. The equation for the line $OP$ is $y=\frac{d}{c}x$.

Let the image of point $P$ be $Q$, after the point undergoes the function. Since each image is equidistant from the preimage and the origin, $Q$ must be on the perpendicular bisector of $OP$.Given $z=c+di$, $f(z)=(a+bi)(c+di)=(ac-bd)+(ad+bc)i$. Then $Q=(ac-bd,ad+bc)$. The midpoint of $OP$ is $(0.5c, 0.5d)$. Since the slopes of two respectively nonvertical and nonhorizontal lines have a product of $-1$, using the point-slope form, the equation of the perpendicular line to $OP$ is $y-0.5d=-\frac{c}{d}(x-0.5c)$. Rearranging, we have $y=-\frac{cx}{d}+\frac{c^2}{2d}+\frac{d}{2}$.

Since we know that $Q=(ac-bd,ad+bc)$, thus we plug in $Q$ into the line: $ad+bc=-\frac{ac^2-bcd}{d}+\frac{c^2}{2d}+\frac{d}{2}$.

Let's start canceling. $2ad^2+2bcd=-2ac^2+2bcd+c^2+d^2$. Subtracting, $c^2+d^2-2ac^2=2ad^2$. Thus $c^2+d^2=2ac^2+2ad^2$. Since this is an identity for any $(c,d)$, thus $2a=1$. $a=\frac{1}{2}$. Since $|a+bi|=8$, thus $a^2+b^2=64$ (or simply think of $a+bi$ as the point $(a,b)$, and $|a+bi|$ being the distance of $(a,b)$ to the origin). Thus plug in $a=\frac{1}{2}, b^2=\frac{255}{4}$. Since $255$ and $4$ are relatively prime, the final result is $255+4=\boxed{259}$.

~hastapasta

See also

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