Difference between revisions of "2022 AIME I Problems/Problem 4"
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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> | 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> | ||
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==Solution 1== | ==Solution 1== | ||
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== Solution 2 == | == Solution 2 == | ||
− | First we recognize that <math>w = \operatorname{cis}(30^{\circ})</math> and <math>z = \operatorname{cis}(120^{\circ})</math> because the cosine and sine sums of those angles give the values of <math>w</math> and <math>z</math>, respectively. By | + | First we recognize that <math>w = \operatorname{cis}(30^{\circ})</math> and <math>z = \operatorname{cis}(120^{\circ})</math> because the cosine and sine sums of those angles give the values of <math>w</math> and <math>z</math>, respectively. By De Moivre's theorem, <math>\operatorname{cis}(\theta)^n = \operatorname{cis}(n\theta)</math>. When you multiply by <math>i</math>, we can think of that as rotating the complex number <math>90^{\circ}</math> counterclockwise in the complex plane. Therefore, by the equation we know that <math>30r + 90</math> and <math>120s</math> land on the same angle. |
This means that | This means that | ||
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~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/MJ_M-xvwHLk?t=933 | ||
+ | |||
+ | ~ThePuzzlr | ||
+ | |||
+ | ==Video Solution by MRENTHUSIASM (English & Chinese)== | ||
+ | https://www.youtube.com/watch?v=1Z6GbkBFu4Q&ab_channel=MRENTHUSIASM | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | https://youtu.be/m1vg_DfHEX4 | ||
+ | |||
+ | ~AMC & AIME Training | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=I|num-b=3|num-a=5}} | {{AIME box|year=2022|n=I|num-b=3|num-a=5}} | ||
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+ | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:37, 23 February 2023
Contents
Problem
Let and where Find the number of ordered pairs of positive integers not exceeding that satisfy the equation
Solution 1
We rewrite and in polar form: The equation becomes for some integer
Since and we conclude that Note that the values for and the values for have one-to-one correspondence.
We apply casework to the values for
There are values for so there are values for It follows that so there are values for
There are ordered pairs in this case.
There are values for so there are values for It follows that so there are values for
There are ordered pairs in this case.
There are values for so there are values for It follows that so there are values for
There are ordered pairs in this case.
Together, the answer is
~MRENTHUSIASM
Solution 2
First we recognize that and because the cosine and sine sums of those angles give the values of and , respectively. By De Moivre's theorem, . When you multiply by , we can think of that as rotating the complex number counterclockwise in the complex plane. Therefore, by the equation we know that and land on the same angle.
This means that which we can simplify to Notice that this means that cycles by for every value of . This is because once hits , we get an angle of and the angle laps onto itself again. By a similar reasoning, laps itself every times, which is much easier to count. By listing the possible values out, we get the pairs : We have columns in total: values for the first column, for the second, for the third, and then for the fourth, for the fifth, for the sixth, etc. Therefore, this cycle repeats every columns and our total sum is .
~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)
Video Solution
https://youtu.be/MJ_M-xvwHLk?t=933
~ThePuzzlr
Video Solution by MRENTHUSIASM (English & Chinese)
https://www.youtube.com/watch?v=1Z6GbkBFu4Q&ab_channel=MRENTHUSIASM
~MRENTHUSIASM
Video Solution
~AMC & AIME Training
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.