Difference between revisions of "2009 AIME I Problems/Problem 2"
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Find <math>n</math>. | Find <math>n</math>. | ||
− | == Solution == | + | ==Solution 1== |
− | |||
Let <math>z = a + 164i</math>. | Let <math>z = a + 164i</math>. | ||
Line 17: | Line 16: | ||
<cmath>a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.</cmath> | <cmath>a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.</cmath> | ||
− | + | By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation, | |
− | <cmath>a = -656</cmath> | + | |
− | + | we conclude that | |
+ | <cmath>a = -656.</cmath> | ||
+ | |||
+ | By equating the imaginary terms on each side of the equation, | ||
+ | |||
+ | we conclude that | ||
<cmath>164i = 4i \left (a + n \right ) = 4i \left (-656 + n \right ).</cmath> | <cmath>164i = 4i \left (a + n \right ) = 4i \left (-656 + n \right ).</cmath> | ||
Line 27: | Line 31: | ||
− | + | ==Solution 2== | |
<cmath>\frac {z}{z+n}=4i</cmath> | <cmath>\frac {z}{z+n}=4i</cmath> | ||
Line 39: | Line 43: | ||
<cmath>\frac {1+4i}{17}=\frac {z}{n}+1</cmath> | <cmath>\frac {1+4i}{17}=\frac {z}{n}+1</cmath> | ||
− | Since their | + | Since their imaginary part has to be equal, |
<cmath>\frac {4i}{17}=\frac {164i}{n}</cmath> | <cmath>\frac {4i}{17}=\frac {164i}{n}</cmath> | ||
Line 46: | Line 50: | ||
<cmath>n = \boxed{697}.</cmath> | <cmath>n = \boxed{697}.</cmath> | ||
+ | |||
+ | ==Solution 3 (Not Highly Recommended)== | ||
+ | Below is an image of the complex plane. Let <math>\operatorname{Im}(z)</math> denote the imaginary part of a complex number <math>z</math>. | ||
+ | <asy> | ||
+ | unitsize(1cm); | ||
+ | |||
+ | |||
+ | xaxis("Re",Arrows); | ||
+ | yaxis("Im",Arrows); | ||
+ | |||
+ | real f(real x) {return 164;} | ||
+ | pair F(real x) {return (x,f(x));} | ||
+ | |||
+ | draw(graph(f,-700,100),red,Arrows); | ||
+ | |||
+ | label("Im$(z)=164$",F(-330),N); | ||
+ | |||
+ | |||
+ | pair z = (-656,164); | ||
+ | dot(Label("$z$",align=N),z); | ||
+ | dot(Label("$z+n$",align=N),z+(697,0)); | ||
+ | |||
+ | draw(Label("$4x$"),z--(0,0)); | ||
+ | draw(Label("$x$"),(0,0)--z+(697,0)); | ||
+ | |||
+ | markscalefactor=2; | ||
+ | draw(rightanglemark(z,(0,0),z+(697,0))); | ||
+ | </asy> | ||
+ | <math>z</math> must lie on the line <math>\operatorname{Im}(z)=164</math>. <math>z+n</math> must also lie on the same line, since <math>n</math> is real and does not affect the imaginary part of <math>z</math>. | ||
+ | |||
+ | Consider <math>z</math> and <math>z+n</math> in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have <math>z_1z_2 = r_1\angle\theta_1 \cdot r_2\angle\theta_2 = r_1r_2\angle(\theta_1+\theta_2)</math> and <math>\frac{z_1}{z_2} = \frac{r_1\angle\theta_1}{r_2\angle\theta_2} = \frac{r_1}{r_2}\angle(\theta_1-\theta_2)</math>, where <math>r</math> is the magnitude and <math>\theta</math> is the phase, and <math>z_n=r_n\angle\theta_n</math>. | ||
+ | |||
+ | Since <math>4i</math> has magnitude <math>4</math> and phase <math>90^\circ</math> (since the positive imaginary axis points in a direction <math>90^\circ</math> counterclockwise from the positive real axis), <math>z</math> must have a magnitude <math>4</math> times that of <math>z+n</math>. We denote the length from the origin to <math>z+n</math> with the value <math>x</math> and the length from the origin to <math>z</math> with the value <math>4x</math>. Additionally, <math>z</math>, the origin, and <math>z+n</math> must form a right angle, with <math>z</math> counterclockwise from <math>z+n</math>. | ||
+ | |||
+ | This means that <math>z</math>, the origin, and <math>z+n</math> form a right triangle. The hypotenuse is the length from <math>z</math> to <math>z+n</math> and has length <math>n</math>, since <math>n</math> is defined to be a positive integer. The area of the triangle can be expressed using the two legs, as <math>\frac{x \cdot 4x}{2}</math>, or using the hypotenuse and its corresponding altitude, as <math>\frac{164n}{2}</math>, so <math>\frac{x \cdot 4x}{2} = \frac{164n}{2} \implies x^2 = 41n</math>. By Pythagorean Theorem, <math>x^2+(4x)^2 = n^2 \implies 17x^2 = n^2</math>. Substituting out <math>x^2</math> using the earlier equation, we get <math>17\cdot41n = n^2 \implies n = \boxed{697}</math>. ~[[User:emerald_block|emerald_block]] | ||
+ | |||
+ | ==Solution 4 (fast)== | ||
+ | Taking the reciprocal of our equation gives us <math>1 + \frac{n}{z} = \frac{1}{4i}.</math> Therefore, <cmath>\frac{n}{z} = \frac{1-4i}{4i} = \frac{17}{-16+4i}.</cmath> Since <math>z</math> has an imaginary part of <math>164</math>, we must multiply both sides of our RHS fraction by <math>\frac{164}{4} = 41</math> so that its denominator's imaginary part matches the LHS fraction's denominator's imaginary part. That looks like this: | ||
+ | <cmath>\frac{n}{z} = \frac{697}{-656 + 164i}.</cmath> Therefore, we can conclude the the real part of <math>z</math> is <math>-656</math> and <math>n = \boxed{697}.</math> (it wasn't necessary to find the real part) | ||
+ | |||
+ | ~Maximilian113 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/NL79UexadzE | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=P00iOJdQiL4 | ||
+ | |||
+ | ~Shreyas S | ||
== See also == | == See also == | ||
{{AIME box|year=2009|n=I|num-b=1|num-a=3}} | {{AIME box|year=2009|n=I|num-b=1|num-a=3}} | ||
− | [[Category:Complex | + | [[Category:Complex numbers]] |
+ | {{MAA Notice}} |
Latest revision as of 12:53, 23 December 2023
Contents
Problem
There is a complex number with imaginary part
and a positive integer
such that
Find .
Solution 1
Let .
Then
and
By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation,
we conclude that
By equating the imaginary terms on each side of the equation,
we conclude that
We now have an equation for :
and this equation shows that
Solution 2
Since their imaginary part has to be equal,
Solution 3 (Not Highly Recommended)
Below is an image of the complex plane. Let denote the imaginary part of a complex number
.
must lie on the line
.
must also lie on the same line, since
is real and does not affect the imaginary part of
.
Consider and
in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have
and
, where
is the magnitude and
is the phase, and
.
Since has magnitude
and phase
(since the positive imaginary axis points in a direction
counterclockwise from the positive real axis),
must have a magnitude
times that of
. We denote the length from the origin to
with the value
and the length from the origin to
with the value
. Additionally,
, the origin, and
must form a right angle, with
counterclockwise from
.
This means that , the origin, and
form a right triangle. The hypotenuse is the length from
to
and has length
, since
is defined to be a positive integer. The area of the triangle can be expressed using the two legs, as
, or using the hypotenuse and its corresponding altitude, as
, so
. By Pythagorean Theorem,
. Substituting out
using the earlier equation, we get
. ~emerald_block
Solution 4 (fast)
Taking the reciprocal of our equation gives us Therefore,
Since
has an imaginary part of
, we must multiply both sides of our RHS fraction by
so that its denominator's imaginary part matches the LHS fraction's denominator's imaginary part. That looks like this:
Therefore, we can conclude the the real part of
is
and
(it wasn't necessary to find the real part)
~Maximilian113
Video Solution
~IceMatrix
Video Solution
https://www.youtube.com/watch?v=P00iOJdQiL4
~Shreyas S
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.