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

Revision as of 22:25, 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 $s+3k:$

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

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.

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

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.

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

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

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

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 $360$ degrees and the angle laps onto itself again. By a similar reasoning, $s$ laps itself every 3 times - this is much easier to count. By listing the possible values out, we get:

Video Solution (Mathematical Dexterity)

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

@@ Line 55: / Line 55: @@
 <cmath>r+3 \equiv 4s \pmod{12}</cmath>.
-<math>
+Notice that this means that <math>r</math> cycles by 12 for every value of <math>s</math>. This is because once <math>r</math> hits 12, we get an angle of <math>360</math> degrees and the angle laps onto itself again. By a similar reasoning, <math>s</math> laps itself every 3 times - this is much easier to count. By listing the possible values out, we get:
-\begin{tabular}[t]{cc|c|}
-\multicolumn{4}{c}{USAMTS Scores Round 1}\\\hline
-Name&\#1&\#2\\\hline
-John Doe&5&5\\
-Jane Doe&5&5\\
-Richard Feynman&5&5\\\hline
-\end{tabular}
-</math>
 ==Video Solution (Mathematical Dexterity)==

2022 AIME I (Problems • Answer Key • Resources)
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