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2022 AIME I Problems/Problem 4

Revision as of 22:17, 17 February 2022 by Kingravi (talk | contribs) (→‎Solution 2)

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 $s.$

    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 $w = cis(30^{\circ})$ and $z = cis(12^{\circ})$ because the cosine and sine sums of those angles give the values of $w$ and $z$, respectively. By Demoivre's theorem, $cis(\theta)^n = cis(n\theta)$. When you multiply by $i$, we can think of that as rotating the complex number 90 degrees 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

\[r+3 \equiv 4s \pmod{12}\].

$\begin{tabular}[t]{|l|l|cccc|c|}\hline \multicolumn{7}{|c|}{USAMTS Final Scores by Round}\\\hline Medal&Name&\#1&\#2&\#3&\#4&Total\\\hline\hline &Richard Feynman&25&25&25&25&100\\\cline{3-7} Gold&Albert Einstein&25&25&25&25&100\\\cline{3-7} &Marie Curie&25&24&24&25&98\\\hline Silver&John Nash&20&20&25&24&89\\\hline &Jane Doe&23&\multicolumn{2}{c}{None}&25&48\\\cline{3-7} None&John Doe&\multicolumn{2}{c}{None}&25&20&45\\\cline{3-7} &Lazy Person&5&\multicolumn{3}{c|}{None}&5\\\hline \end{tabular}$

Video Solution (Mathematical Dexterity)

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

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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