Difference between revisions of "2021 AIME I Problems/Problem 7"
(→Solution) |
|||
Line 2: | Line 2: | ||
Find the number of pairs <math>(m,n)</math> of positive integers with <math>1\le m<n\le 30</math> such that there exists a real number <math>x</math> satisfying<cmath>\sin(mx)+\sin(nx)=2.</cmath> | Find the number of pairs <math>(m,n)</math> of positive integers with <math>1\le m<n\le 30</math> such that there exists a real number <math>x</math> satisfying<cmath>\sin(mx)+\sin(nx)=2.</cmath> | ||
− | ==Solution== | + | ==Solution 1== |
The maximum value of <math>\sin \theta</math> is <math>1</math>, which is achieved at <math>\theta = \frac{\pi}{2}+2k\pi</math> for some integer <math>k</math>. This is left as an exercise to the reader. | The maximum value of <math>\sin \theta</math> is <math>1</math>, which is achieved at <math>\theta = \frac{\pi}{2}+2k\pi</math> for some integer <math>k</math>. This is left as an exercise to the reader. | ||
Line 19: | Line 19: | ||
This solution was brought to you by ~Leonard_my_dude~ | This solution was brought to you by ~Leonard_my_dude~ | ||
+ | |||
+ | ==Solution 2== | ||
+ | In order for <math>\sin(mx) + \sin(nx) = 2</math>, <math>\sin(mx) = \sin(nx) = 1</math>. This happens when | ||
+ | <math>mx \equiv nx \equiv \frac{\pi}{2} (mod 2\pi)</math> | ||
+ | I WILL FINISH THE SOLUTION SOON, PLEASE DO NOT EDIT THIS BEFORE THEN THANK YOU! | ||
+ | |||
+ | -KingRavi | ||
+ | |||
+ | |||
==See also== | ==See also== | ||
{{AIME box|year=2021|n=I|num-b=6|num-a=8}} | {{AIME box|year=2021|n=I|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:57, 13 March 2021
Contents
Problem
Find the number of pairs of positive integers with such that there exists a real number satisfying
Solution 1
The maximum value of is , which is achieved at for some integer . This is left as an exercise to the reader.
This implies that , and that and , for integers .
Taking their ratio, we have It remains to find all that satisfy this equation.
If , then . This corresponds to choosing two elements from the set . There are ways to do so.
If , by multiplying and by the same constant , we have that . Then either , or . But the first case was already counted, so we don't need to consider that case. The other case corresponds to choosing two numbers from the set . There are ways here.
Finally, if , note that must be an integer. This means that belong to the set , or . Taking casework on , we get the sets . Some sets have been omitted; this is because they were counted in the other cases already. This sums to .
In total, there are pairs of .
This solution was brought to you by ~Leonard_my_dude~
Solution 2
In order for , . This happens when I WILL FINISH THE SOLUTION SOON, PLEASE DO NOT EDIT THIS BEFORE THEN THANK YOU!
-KingRavi
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.