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 "2022 AIME I Problems/Problem 4"
 
(39 intermediate revisions by 9 users not shown)
Line 1: Line 1:
−
Write <math>i=e^{i\frac{\pi}{2}}</math>, it turns to: <math>\frac{\pi}{6}(3+r)=\frac{4n\pi}{6}</math>, so <math>3+r=4s+12k</math>
+
==Problem==
  
−
it follows a pattern that <math>s=1,r=1,13....97</math> has 9 values; <math>s=2,r=5,17...89</math> 8 values and <math>s=3,r=9,21,...93</math> 8 values.
+
Let <math>w = \dfrac{\sqrt{3} + i}{2}</math> and <math>z = \dfrac{-1 + i\sqrt{3}}{2},</math> where <math>i = \sqrt{-1}.</math> Find the number of ordered pairs <math>(r,s)</math> of positive integers not exceeding <math>100</math> that satisfy the equation <math>i \cdot w^r = z^s.</math>
  
−
So the answer is <math>33*(9+8+8)+9=834</math>
+
==Solution 1==
  
−
~bluesoul
+
We rewrite <math>w</math> and <math>z</math> in polar form:
 +
<cmath>\begin{align*}
 +
w &= e^{i\cdot\frac{\pi}{6}}, \\
 +
z &= e^{i\cdot\frac{2\pi}{3}}.
 +
\end{align*}</cmath>
 +
The equation <math>i \cdot w^r = z^s</math> becomes
 +
<cmath>\begin{align*}
 +
e^{i\cdot\frac{\pi}{2}} \cdot \left(e^{i\cdot\frac{\pi}{6}}\right)^r &= \left(e^{i\cdot\frac{2\pi}{3}}\right)^s \\
 +
e^{i\left(\frac{\pi}{2}+\frac{\pi}{6}r\right)} &= e^{i\left(\frac{2\pi}{3}s\right)} \\
 +
\frac{\pi}{2}+\frac{\pi}{6}r &= \frac{2\pi}{3}s+2\pi k \\
 +
3+r &= 4s+12k \\
 +
3+r &= 4(s+3k).
 +
\end{align*}</cmath>
 +
for some integer <math>k.</math>
 +
 
 +
Since <math>4\leq 3+r\leq 103</math> and <math>4\mid 3+r,</math> we conclude that
 +
<cmath>\begin{align*}
 +
3+r &\in \{4,8,12,\ldots,100\}, \\
 +
s+3k &\in \{1,2,3,\ldots,25\}.
 +
\end{align*}</cmath>
 +
Note that the values for <math>s+3k</math> and the values for <math>r</math> have one-to-one correspondence.
 +
 
 +
We apply casework to the values for <math>s+3k:</math>
 +
<ol style="margin-left: 1.5em;">
 +
  <li><math>s+3k\equiv0\pmod{3}</math></li><p>
 +
There are <math>8</math> values for <math>s+3k,</math> so there are <math>8</math> values for <math>r.</math> It follows that <math>s\equiv0\pmod{3},</math> so there are <math>33</math> values for <math>s.</math> <p>
 +
There are <math>8\cdot33=264</math> ordered pairs <math>(r,s)</math> in this case.
 +
  <li><math>s+3k\equiv1\pmod{3}</math></li><p>
 +
There are <math>9</math> values for <math>s+3k,</math> so there are <math>9</math> values for <math>r.</math> It follows that <math>s\equiv1\pmod{3},</math> so there are <math>34</math> values for <math>s.</math> <p>
 +
There are <math>9\cdot34=306</math> ordered pairs <math>(r,s)</math> in this case.
 +
  <li><math>s+3k\equiv2\pmod{3}</math></li><p>
 +
There are <math>8</math> values for <math>s+3k,</math> so there are <math>8</math> values for <math>r.</math> It follows that <math>s\equiv2\pmod{3},</math> so there are <math>33</math> values for <math>s.</math> <p>
 +
There are <math>8\cdot33=264</math> ordered pairs <math>(r,s)</math> in this case.
 +
</ol>
 +
Together, the answer is <math>264+306+264=\boxed{834}.</math>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
== Solution 2 ==
 +
 
 +
First we recognize that <math>w = \operatorname{cis}(30^{\circ})</math> and <math>z = \operatorname{cis}(120^{\circ})</math> because the cosine and sine sums of those angles give the values of <math>w</math> and <math>z</math>, respectively. By De Moivre's theorem, <math>\operatorname{cis}(\theta)^n = \operatorname{cis}(n\theta)</math>. When you multiply by <math>i</math>, we can think of that as rotating the complex number <math>90^{\circ}</math> counterclockwise in the complex plane. Therefore, by the equation we know that <math>30r + 90</math> and <math>120s</math> land on the same angle.
 +
 
 +
This means that
 +
<cmath>30r + 90 \equiv 120s \pmod{360},</cmath>
 +
which we can simplify to
 +
<cmath>r+3 \equiv 4s \pmod{12}.</cmath>
 +
Notice that this means that <math>r</math> cycles by <math>12</math> for every value of <math>s</math>. This is because once <math>r</math> hits <math>12</math>, we get an angle of <math>360^{\circ}</math> and the angle laps onto itself again. By a similar reasoning, <math>s</math> laps itself every <math>3</math> times, which is much easier to count. By listing the possible values out, we get the pairs <math>(r,s)</math>:
 +
<cmath>\begin{array}{cccccccc}
 +
(1,1) & (5,2) & (9,3) & (13,1) & (17,2) & (21,3) & \ldots & (97,1) \\
 +
(1,4) & (5,5) & (9,6) & (13,4) & (17,5) & (21,6) & \ldots & (97,4) \\
 +
(1,7) & (5,8) & (9,9) & (13,7) & (17,8) & (21,9) & \ldots & (97,7) \\ [-1ex]
 +
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
 +
(1,100) & (5,98) & (9,99) & (13,100) & (17,98) & (21,99) & \ldots & (97,100)
 +
\end{array}</cmath>
 +
We have <math>25</math> columns in total: <math>34</math> values for the first column, <math>33</math> for the second, <math>33</math> for the third, and then <math>34</math> for the fourth, <math>33</math> for the fifth, <math>33</math> for the sixth, etc. Therefore, this cycle repeats every <math>3</math> columns and our total sum is <math>(34+33+33) \cdot 8 + 34 = 100 \cdot 8 + 34 = \boxed{834}</math>.
 +
 
 +
~KingRavi
 +
 
 +
==Video Solution (Mathematical Dexterity)==
 +
https://www.youtube.com/watch?v=XiEaCq5jf5s
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=qQ0TIhHuhnI
 +
 
 +
~Steven Chen (www.professorchenedu.com)
 +
 
 +
== Video Solution ==
 +
https://youtu.be/MJ_M-xvwHLk?t=933
 +
 
 +
~ThePuzzlr
 +
 
 +
==Video Solution by MRENTHUSIASM (English & Chinese)==
 +
https://www.youtube.com/watch?v=1Z6GbkBFu4Q&ab_channel=MRENTHUSIASM
 +
 
 +
~MRENTHUSIASM
 +
 
 +
== Video Solution ==
 +
 
 +
https://youtu.be/m1vg_DfHEX4
 +
 
 +
~AMC & AIME Training
 +
 
 +
==See Also==
 +
{{AIME box|year=2022|n=I|num-b=3|num-a=5}}
 +
 
 +
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 13:37, 23 February 2023

Problem

Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$

Solution 1

We rewrite $w$ and $z$ in polar form: \begin{align*} w &= e^{i\cdot\frac{\pi}{6}}, \\ z &= e^{i\cdot\frac{2\pi}{3}}. \end{align*} The equation $i \cdot w^r = z^s$ becomes \begin{align*} e^{i\cdot\frac{\pi}{2}} \cdot \left(e^{i\cdot\frac{\pi}{6}}\right)^r &= \left(e^{i\cdot\frac{2\pi}{3}}\right)^s \\ e^{i\left(\frac{\pi}{2}+\frac{\pi}{6}r\right)} &= e^{i\left(\frac{2\pi}{3}s\right)} \\ \frac{\pi}{2}+\frac{\pi}{6}r &= \frac{2\pi}{3}s+2\pi k \\ 3+r &= 4s+12k \\ 3+r &= 4(s+3k). \end{align*} for some integer $k.$

Since $4\leq 3+r\leq 103$ and $4\mid 3+r,$ we conclude that \begin{align*} 3+r &\in \{4,8,12,\ldots,100\}, \\ s+3k &\in \{1,2,3,\ldots,25\}. \end{align*} Note that the values for $s+3k$ and the values for $r$ have one-to-one correspondence.

We apply casework to the values for $s+3k:$

  1. $s+3k\equiv0\pmod{3}$

    2. There are $8$ values for $s+3k,$ so there are $8$ values for $r.$ It follows that $s\equiv0\pmod{3},$ so there are $33$ values for $s.$

    There are $8\cdot33=264$ ordered pairs $(r,s)$ in this case.

  2. $s+3k\equiv1\pmod{3}$

    3. There are $9$ values for $s+3k,$ so there are $9$ values for $r.$ It follows that $s\equiv1\pmod{3},$ so there are $34$ values for $s.$

    There are $9\cdot34=306$ ordered pairs $(r,s)$ in this case.

  3. $s+3k\equiv2\pmod{3}$

    4. There are $8$ values for $s+3k,$ so there are $8$ values for $r.$ It follows that $s\equiv2\pmod{3},$ so there are $33$ values for $s.$

    There are $8\cdot33=264$ ordered pairs $(r,s)$ in this case.

Together, the answer is $264+306+264=\boxed{834}.$

~MRENTHUSIASM

Solution 2

First we recognize that $w = \operatorname{cis}(30^{\circ})$ and $z = \operatorname{cis}(120^{\circ})$ because the cosine and sine sums of those angles give the values of $w$ and $z$, respectively. By De Moivre's theorem, $\operatorname{cis}(\theta)^n = \operatorname{cis}(n\theta)$. When you multiply by $i$, we can think of that as rotating the complex number $90^{\circ}$ counterclockwise in the complex plane. Therefore, by the equation we know that $30r + 90$ and $120s$ land on the same angle.

This means that \[30r + 90 \equiv 120s \pmod{360},\] which we can simplify to \[r+3 \equiv 4s \pmod{12}.\] Notice that this means that $r$ cycles by $12$ for every value of $s$. This is because once $r$ hits $12$, we get an angle of $360^{\circ}$ and the angle laps onto itself again. By a similar reasoning, $s$ laps itself every $3$ times, which is much easier to count. By listing the possible values out, we get the pairs $(r,s)$: \[\begin{array}{cccccccc} (1,1) & (5,2) & (9,3) & (13,1) & (17,2) & (21,3) & \ldots & (97,1) \\ (1,4) & (5,5) & (9,6) & (13,4) & (17,5) & (21,6) & \ldots & (97,4) \\ (1,7) & (5,8) & (9,9) & (13,7) & (17,8) & (21,9) & \ldots & (97,7) \\ [-1ex] \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ (1,100) & (5,98) & (9,99) & (13,100) & (17,98) & (21,99) & \ldots & (97,100) \end{array}\] We have $25$ columns in total: $34$ values for the first column, $33$ for the second, $33$ for the third, and then $34$ for the fourth, $33$ for the fifth, $33$ for the sixth, etc. Therefore, this cycle repeats every $3$ columns and our total sum is $(34+33+33) \cdot 8 + 34 = 100 \cdot 8 + 34 = \boxed{834}$.

~KingRavi

Video Solution (Mathematical Dexterity)

https://www.youtube.com/watch?v=XiEaCq5jf5s

Video Solution

https://www.youtube.com/watch?v=qQ0TIhHuhnI

~Steven Chen (www.professorchenedu.com)

Video Solution

https://youtu.be/MJ_M-xvwHLk?t=933

~ThePuzzlr

Video Solution by MRENTHUSIASM (English & Chinese)

https://www.youtube.com/watch?v=1Z6GbkBFu4Q&ab_channel=MRENTHUSIASM

~MRENTHUSIASM

Video Solution

https://youtu.be/m1vg_DfHEX4

~AMC & AIME Training

See Also

2022 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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=2022_AIME_I_Problems/Problem_4&oldid=190175"