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Difference between revisions of "2022 AIME I Problems/Problem 4"

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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>
 
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>
  
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==Solution==
+
==Solution 1==
  
 
We rewrite <math>w</math> and <math>z</math> in polar form:
 
We rewrite <math>w</math> and <math>z</math> in polar form:
Line 45: Line 45:
 
== Solution 2 ==
 
== Solution 2 ==
  
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First we recognize that <math>w = cis(30^{\circ})</math> and <math>z = cis(12^{\circ})</math> because the cosine and sine sums of those angles give the values of <math>w</math> and <math>z</math>, respectively. By Demoivre's theorem, <math>cis(\theta)^n = cis(n\theta)</math>. When you multiply by <math>i</math>, we can think of that as rotating the complex number 90 degrees 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.
+
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:
+
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>.
  
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<cmath>30r + 90 \equiv 120s \pmod{360}</cmath>
+
~KingRavi
  
−
Which we can simplify to
+
==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
  
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<cmath>r+3 \equiv 4s \pmod{12}</cmath>.
+
== Video Solution ==
  
−
<math>
+
https://youtu.be/m1vg_DfHEX4
−
\begin{tabular}[t]{cc|c|}
 
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\multicolumn{4}{c}{USAMTS Scores Round 1}\\\hline
 
−
Name&\#1&\#2\\\hline
 
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John Doe&5&5\\
 
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Jane Doe&5&5\\
 
−
Richard Feynman&5&5\\\hline
 
−
\end{tabular}
 
−
</math>
 
  
−
==Video Solution (Mathematical Dexterity)==
+
~AMC & AIME Training
−
https://www.youtube.com/watch?v=XiEaCq5jf5s
 
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=I|num-b=3|num-a=5}}
 
{{AIME box|year=2022|n=I|num-b=3|num-a=5}}
 +
 +
[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{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)
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Problem 3 		Followed by
Problem 5
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