Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2009 AIME I Problems/Problem 2"

(→‎See also)
(→‎Solution 4 (fast))
 
(17 intermediate revisions by 11 users not shown)
Line 7: Line 7:
 
Find <math>n</math>.
 
Find <math>n</math>.
  
−
== Solution ==
+
==Solution 1==
−
'''1st Solution'''
 
  
 
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>
  
−
From this, we conclude that
+
By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation,
−
<cmath>a = -656</cmath>
+
 
−
and
+
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:
  
  
−
'''2nd Solution'''
+
==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 imaginery part has to be equal,
+
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 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 Number]]
+
[[Category:Complex numbers]]
 +
{{MAA Notice}}

Latest revision as of 01:00, 5 January 2025

Problem

There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that

\[\frac {z}{z + n} = 4i.\]

Find $n$.

Solution 1

Let $z = a + 164i$.

Then \[\frac {a + 164i}{a + 164i + n} = 4i\] and \[a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.\]

By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation,

we conclude that \[a = -656.\]

By equating the imaginary terms on each side of the equation,

we conclude that \[164i = 4i \left (a + n \right ) = 4i \left (-656 + n \right ).\]

We now have an equation for $n$: \[4i \left (-656 + n \right ) = 164i,\]

and this equation shows that $n = \boxed{697}.$


Solution 2

\[\frac {z}{z+n}=4i\]

\[1-\frac {n}{z+n}=4i\]

\[1-4i=\frac {n}{z+n}\]

\[\frac {1}{1-4i}=\frac {z+n}{n}\]

\[\frac {1+4i}{17}=\frac {z}{n}+1\]

Since their imaginary part has to be equal,

\[\frac {4i}{17}=\frac {164i}{n}\]

\[n=\frac {(164)(17)}{4}=697\]

\[n = \boxed{697}.\]

Solution 3 (Not Highly Recommended)

Below is an image of the complex plane. Let $\operatorname{Im}(z)$ denote the imaginary part of a complex number $z$. [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] $z$ must lie on the line $\operatorname{Im}(z)=164$. $z+n$ must also lie on the same line, since $n$ is real and does not affect the imaginary part of $z$.

Consider $z$ and $z+n$ 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 $z_1z_2 = r_1\angle\theta_1 \cdot r_2\angle\theta_2 = r_1r_2\angle(\theta_1+\theta_2)$ and $\frac{z_1}{z_2} = \frac{r_1\angle\theta_1}{r_2\angle\theta_2} = \frac{r_1}{r_2}\angle(\theta_1-\theta_2)$, where $r$ is the magnitude and $\theta$ is the phase, and $z_n=r_n\angle\theta_n$.

Since $4i$ has magnitude $4$ and phase $90^\circ$ (since the positive imaginary axis points in a direction $90^\circ$ counterclockwise from the positive real axis), $z$ must have a magnitude $4$ times that of $z+n$. We denote the length from the origin to $z+n$ with the value $x$ and the length from the origin to $z$ with the value $4x$. Additionally, $z$, the origin, and $z+n$ must form a right angle, with $z$ counterclockwise from $z+n$.

This means that $z$, the origin, and $z+n$ form a right triangle. The hypotenuse is the length from $z$ to $z+n$ and has length $n$, since $n$ is defined to be a positive integer. The area of the triangle can be expressed using the two legs, as $\frac{x \cdot 4x}{2}$, or using the hypotenuse and its corresponding altitude, as $\frac{164n}{2}$, so $\frac{x \cdot 4x}{2} = \frac{164n}{2} \implies x^2 = 41n$. By Pythagorean Theorem, $x^2+(4x)^2 = n^2 \implies 17x^2 = n^2$. Substituting out $x^2$ using the earlier equation, we get $17\cdot41n = n^2 \implies n = \boxed{697}$. ~emerald_block

Solution 4 (fast)

Taking the reciprocal of our equation gives us $1 + \frac{n}{z} = \frac{1}{4i}.$ Therefore, \[\frac{n}{z} = \frac{1-4i}{4i} = \frac{17}{-16+4i}.\] Since $z$ has an imaginary part of $164$, we must multiply both sides of our RHS fraction by $\frac{164}{4} = 41$ so that its denominator's imaginary part matches the LHS fraction's denominator's imaginary part. That looks like this: \[\frac{n}{z} = \frac{697}{-656 + 164i}.\] Therefore, we can conclude the real part of $z$ is $-656$ and $n = \boxed{697}.$ (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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_2&oldid=239008"