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

(Added in detail explanation.)
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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>
 
\end{align*}</cmath>
Note that <math>\frac{\pi}{2}+\frac{\pi}{6}r=\frac{2\pi}{3}s+2\pi k</math> for some integer <math>k,</math> or
+
Note that  
 +
<cmath>\begin{align*}
 +
\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*}</cmath>
 +
for some integer <math>k.</math>
 +
 
 +
Since <math>4\leq 3+r\leq 103</math> and <math>4\mid 3+r,</math> we conclude that
  
 
~MRENTHUSIASM ~bluesoul
 
~MRENTHUSIASM ~bluesoul

Revision as of 16:46, 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)}. \end{align*} Note that \begin{align*} \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

~MRENTHUSIASM ~bluesoul

See Also

2022 AIME I (ProblemsAnswer KeyResources)
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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