Difference between revisions of "2022 AIME I Problems/Problem 4"
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== Solution 2 == | == Solution 2 == | ||
− | First we recognize that <math>w = cis(30^{\circ})</math> and <math>z = cis( | + | 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 |
+ | <cmath>30r + 90 \equiv 120s \pmod{360},</cmath> | ||
+ | which we can simplify to | ||
+ | <cmath>r+3 \equiv 4s \pmod{12}.</cmath> | ||
+ | Notice that this means that <math>r</math> cycles by <math>12</math> for every value of <math>s</math>. This is because once <math>r</math> hits <math>12</math>, we get an angle of <math>360^{\circ}</math> and the angle laps onto itself again. By a similar reasoning, <math>s</math> laps itself every <math>3</math> times, which is much easier to count. By listing the possible values out, we get the pairs <math>(r,s)</math>: | ||
+ | <cmath>\begin{array}{cccccccc} | ||
+ | (1,1) & (5,2) & (9,3) & (13,1) & (17,2) & (21,3) & \ldots & (97,1) \\ | ||
+ | (1,4) & (5,5) & (9,6) & (13,4) & (17,5) & (21,6) & \ldots & (97,4) \\ | ||
+ | (1,7) & (5,8) & (9,9) & (13,7) & (17,8) & (21,9) & \ldots & (97,7) \\ [-1ex] | ||
+ | \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ | ||
+ | (1,100) & (5,98) & (9,99) & (13,100) & (17,98) & (21,99) & \ldots & (97,100) | ||
+ | \end{array}</cmath> | ||
+ | We have <math>25</math> columns in total: <math>34</math> values for the first column, <math>33</math> for the second, <math>33</math> for the third, and then <math>34</math> for the fourth, <math>33</math> for the fifth, <math>33</math> for the sixth, etc. Therefore, this cycle repeats every <math>3</math> columns and our total sum is <math>(34+33+33) \cdot 8 + 34 = 100 \cdot 8 + 34 = \boxed{834}</math>. | ||
− | + | ~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 | + | == Video Solution == |
− | |||
− | + | https://youtu.be/m1vg_DfHEX4 | |
− | https:// | ||
− | ~ | + | ~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}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:37, 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.