2022 AIME I Problems/Problem 4: Difference between revisions

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

@@ Line 15: / Line 15: @@
 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 <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 \\
++r &= 4s+12k \\
++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

2022 AIME I (Problems • Answer Key • Resources)
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2022 AIME I Problems/Problem 4: Difference between revisions

Revision as of 16:46, 17 February 2022

Problem

Solution

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