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

Revision as of 15:23, 19 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 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 $s+3k:$

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

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 the pairs (r,s):

$(1,1), (5, 2), \ldots, (97,25)$ $(1,4), (5, 5), \ldots, (97,28)$ $\vdots$ $(1,100), \ldots, (97, 100)$

We have 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 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)

@@ Line 3: / Line 3: @@
 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>
-==Solution==
+==Solution 1==
 We rewrite <math>w</math> and <math>z</math> in polar form:

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