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Difference between revisions of "2022 AIME I Problems/Problem 4"
Line 13: Line 13:
 
<cmath>\begin{align*}
 
<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\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)}.
+
e^{i\left(\frac{\pi}{2}+\frac{\pi}{6}r\right)} &= e^{i\left(\frac{2\pi}{3}s\right)} \\
−
\end{align*}</cmath>
 
−
Note that
 
−
<cmath>\begin{align*}
 
 
\frac{\pi}{2}+\frac{\pi}{6}r &= \frac{2\pi}{3}s+2\pi k \\
 
\frac{\pi}{2}+\frac{\pi}{6}r &= \frac{2\pi}{3}s+2\pi k \\
 
3+r &= 4s+12k \\
 
3+r &= 4s+12k \\
Line 24: Line 21:
  
 
Since <math>4\leq 3+r\leq 103</math> and <math>4\mid 3+r,</math> we conclude that
 
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>
 +
  <li><math>s+3k\equiv1\pmod{3}</math></li><p>
 +
  <li><math>s+3k\equiv2\pmod{3}</math></li><p>
 +
</ol>
 
~MRENTHUSIASM ~bluesoul
 
~MRENTHUSIASM ~bluesoul
  

Revision as of 18:02, 17 February 2022

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

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 $s.$

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

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

~MRENTHUSIASM ~bluesoul

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

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