Difference between revisions of "2021 AMC 12A Problems/Problem 19"
MRENTHUSIASM (talk | contribs) m |
Firebolt360 (talk | contribs) (→Solution 3 (Graphs and Analyses)) |
||
Line 157: | Line 157: | ||
~MRENTHUSIASM (credit given to TheAMCHub) | ~MRENTHUSIASM (credit given to TheAMCHub) | ||
+ | |||
+ | === Note (Solution 4) === | ||
+ | One can also find the values at <math>0, \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{5\pi}{6},</math> and <math>\pi</math> and then make a small graph by interpolating through. Small observations, like that one function is positive while the other is negative for <math>x>\frac{pi}{2}</math> can speed up the process. | ||
== Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry) == | == Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry) == |
Revision as of 11:56, 5 November 2022
Contents
[hide]Problem
How many solutions does the equation have in the closed interval ?
Solution 1 (Inverse Trigonometric Functions)
The ranges of and are both which is included in the range of so we can use it with no issues. This only happens at on the interval because one of and must be and the other Therefore, the answer is
~Tucker
Solution 2 (Cofunction Identity)
By the Cofunction Identity we rewrite the given equation: Recall that if then or for some integer Therefore, we have two cases:
- for some integer
We rearrange and simplify: By rough constraints, we know that from which The only possibility is so for some integer
We get for this case. Note that is an extraneous solution by squaring
- for some integer
Similar to Case 1, we conclude that so We get for this case.
Together, we obtain solutions:
~MRENTHUSIASM
Solution 3 (Graphs and Analyses)
This problem is equivalent to counting the intersections of the graphs of and in the closed interval We construct a table of values, as shown below: For note that:
- so
- so
For the graphs to intersect, we need This occurs when
By the Cofunction Identity we rewrite the given equation: Since and it follows that and
We can apply the arcsine function to both sides, then rearrange and simplify: From Case 1 in Solution 2, we conclude that and are the only points of intersection, as shown below: Therefore, the answer is
~MRENTHUSIASM (credit given to TheAMCHub)
Note (Solution 4)
One can also find the values at and and then make a small graph by interpolating through. Small observations, like that one function is positive while the other is negative for can speed up the process.
Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry)
~ pi_is_3.14
Video Solution (Quick and Easy)
~Education, the Study of Everything
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.