2022 AIME I Problems/Problem 4: Difference between revisions

Revision as of 16:39, 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 $\frac{\pi}{2}+\frac{\pi}{6}r=\frac{2\pi}{3}s+2\pi k$ for some integer $k,$ or

~MRENTHUSIASM ~bluesoul

@@ Line 1: / Line 1: @@
-==solution 1==
+==Problem==
-Write <math>i=e^{i\frac{\pi}{2}}</math>, it turns to: <math>\frac{\pi}{6}(3+r)=\frac{4n\pi}{6}</math>, so <math>3+r=4s+12k</math>
+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>
-it follows a pattern that <math>s=1,r=1,13....97</math> has 9 values; <math>s=2,r=5,17...89</math> 8 values and <math>s=3,r=9,21,...93</math> 8 values.
+==Solution==
-So the answer is <math>33*(9+8+8)+9=834</math>
+We rewrite <math>w</math> and <math>z</math> in polar form:
+<cmath>\begin{align*}
+w &= e^{i\cdot\frac{\pi}{6}}, \\
+z &= e^{i\cdot\frac{2\pi}{3}}.
+\end{align*}</cmath>
+The equation <math>i \cdot w^r = z^s</math> becomes
+<cmath>\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*}</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
-~bluesoul
+~MRENTHUSIASM ~bluesoul
+==See Also==
+{{AIME box|year=2022|n=I|num-b=3|num-a=5}}
+{{MAA Notice}}

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

Page

Toolbox

Search

2022 AIME I Problems/Problem 4: Difference between revisions

Revision as of 16:39, 17 February 2022

Problem

Solution

See Also