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

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

@@ Line 2: / Line 2: @@
 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>
-__TOC__
+== Solution 1 ==
+Suppose we pick an arbitrary point on the [[complex plane]], say <math>(1,1)</math>. According to the definition of <math>f(z)</math>,  <cmath>f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i,</cmath> this image must be equidistant to <math>(1,1)</math> and <math>(0,0)</math>. Thus the image must lie on the line with slope <math>-1</math> and which passes through <math>\left(\frac 12, \frac12\right)</math>, so its graph is <math>x + y = 1</math>. Substituting <math>x = (a-b)</math> and <math>y = (a+b)</math>, we get <math>2a = 1 \Rightarrow a = \frac 12</math>.
-== Solution ==
-=== Solution 1 ===
-Suppose we pick an arbitrary point on the [[complex plane]], say <math>(1,1)</math>. According to the definition of <math>f(z) = f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i</math>, this image must be equidistant to <math>(1,1)</math> and <math>(0,0)</math>. Thus the image must lie on the line with slope <math>-1</math> and which passes through <math>\left(\frac 12, \frac12\right)</math>, so its graph is <math>x + y = 1</math>. Substituting <math>x = (a-b)</math> and <math>y = (a+b)</math>, we get <math>2a = 1 \Rightarrow a = \frac 12</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>.
-===Solution 2===
+==Solution 2==
+Plugging in <math>z=1</math> yields <math>f(1) = a+bi</math>. This implies that <math>a+bi</math> must fall on the line <math>Re(z)=a=\frac{1}{2}</math>, given the equidistant rule. By <math>|a+bi|=8</math>, we get <math>a^2 + b^2 = 64</math>, and plugging in <math>a=\frac{1}{2}</math> yields <math>b^2=\frac{255}{4}</math>. The answer is thus <math>\boxed{259}</math>.
-=== Solution 3 ===
+== Solution 3 ==
 We are given that <math>(a + bi)z</math> is equidistant from the origin and <math>z.</math>  This translates to
 <cmath>
 \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 \\
 & \Rightarrow & a = \frac 12 \end{eqnarray*}
 </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 ==
 Let <math>P</math> and <math>Q</math> be the points in the complex plane represented by <math>z</math> and <math>(a+bi)z</math>, respectively.  <math>|a+bi| = 8</math> implies <math>OQ = 8OP</math>.  Also, we are given <math>OQ = PQ</math>, so <math>OPQ</math> is isosceles with base <math>OP</math>.  Notice that the base angle of this isosceles triangle is equal to the argument <math>\theta</math> of the complex number <math>a + bi</math>, because <math>(a+bi)z</math> forms an angle of <math>\theta</math> with <math>z</math>.  Drop the altitude/median from <math>Q</math> to base <math>OP</math>, and you end up with a right triangle that shows <math>\cos \theta = \frac{\frac{1}{2}OP}{8OQ} = \frac{\frac{1}{2}|z|}{8|z|} = \frac{1}{16}</math>.  Since <math>a</math> and <math>b</math> are positive, <math>z</math> lies in the first quadrant and <math>\theta < \pi/2</math>; hence by right triangle trigonometry <math>\sin \theta = \frac{\sqrt{255}}{16}</math>.  Finally, <math>b = |a+bi|\sin\theta = 8\frac{\sqrt{255}}{16} = \frac{\sqrt{255}}{2}</math>, and <math>b^2 = \frac{255}{4}</math>, so the answer is <math>259</math>.
+== Solution 5 ==
+Similarly to in Solution 3, we see that <math>|(a + bi)z - z| = |(a + bi)z|</math>. Letting the point <math>z = c + di</math>, we have <math>\sqrt{(ab+bc-d)^2+(ac-bd-c)^2} = \sqrt{(ac-bd)^2+(ad+bc)^2}</math>. Expanding both sides of this equation (after squaring, of course) and canceling terms, we get <math>(d^2+c^2)(-2a+1) = 0</math>. Of course, <math>(d^2+c^2)</math> can't be zero because this property of the function holds for all complex <math>z</math>. Therefore, <math>a = \frac{1}{2}</math> and we proceed as above to get <math>\boxed{259}</math>.
+~ 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 <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>.
+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>.
+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>.
+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>.
+~hastapasta
 == 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
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