Difference between revisions of "1999 AIME Problems/Problem 9"
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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> | 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> | ||
− | + | == 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>. | |
− | |||
− | |||
− | Suppose we pick an arbitrary point on the [[complex plane]], say <math>(1,1)</math>. According to the definition of <math>f(z) | ||
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>. | ||
− | === | + | ==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 == | |
We are given that <math>(a + bi)z</math> is equidistant from the origin and <math>z.</math> This translates to | We are given that <math>(a + bi)z</math> is equidistant from the origin and <math>z.</math> This translates to | ||
<cmath> | <cmath> | ||
\begin{eqnarray*} |(a + bi)z - z| & = & |(a + bi)z| \\ | \begin{eqnarray*} |(a + bi)z - z| & = & |(a + bi)z| \\ | ||
− | |z(a - 1) | + | |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> | + | 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 == | |
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>. | 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 == | == See also == |
Latest revision as of 19:01, 24 May 2023
Contents
Problem
A function is defined on the complex numbers by
where
and
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
and that
where
and
are relatively prime positive integers, find
Solution 1
Suppose we pick an arbitrary point on the complex plane, say . According to the definition of
,
this image must be equidistant to
and
. Thus the image must lie on the line with slope
and which passes through
, so its graph is
. Substituting
and
, we get
.
By the Pythagorean Theorem, we have , and the answer is
.
Solution 2
Plugging in yields
. This implies that
must fall on the line
, given the equidistant rule. By
, we get
, and plugging in
yields
. The answer is thus
.
Solution 3
We are given that is equidistant from the origin and
This translates to
Since
Because
thus
So the answer is
.
Solution 4
Let and
be the points in the complex plane represented by
and
, respectively.
implies
. Also, we are given
, so
is isosceles with base
. Notice that the base angle of this isosceles triangle is equal to the argument
of the complex number
, because
forms an angle of
with
. Drop the altitude/median from
to base
, and you end up with a right triangle that shows
. Since
and
are positive,
lies in the first quadrant and
; hence by right triangle trigonometry
. Finally,
, and
, so the answer is
.
Solution 5
Similarly to in Solution 3, we see that . Letting the point
, we have
. Expanding both sides of this equation (after squaring, of course) and canceling terms, we get
. Of course,
can't be zero because this property of the function holds for all complex
. Therefore,
and we proceed as above to get
.
~ 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 . Let
denote point
on the complex plane. Then
on the complex plane. The equation for the line
is
.
Let the image of point be
, after the point undergoes the function. Since each image is equidistant from the preimage and the origin,
must be on the perpendicular bisector of
.Given
,
. Then
. The midpoint of
is
. Since the slopes of two respectively nonvertical and nonhorizontal lines have a product of
, using the point-slope form, the equation of the perpendicular line to
is
. Rearranging, we have
.
Since we know that , thus we plug in
into the line:
.
Let's start canceling. . Subtracting,
. Thus
. Since this is an identity for any
, thus
.
. Since
, thus
(or simply think of
as the point
, and
being the distance of
to the origin). Thus plug in
. Since
and
are relatively prime, the final result is
.
~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 | ||
All AIME Problems and Solutions |
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