Difference between revisions of "2022 AMC 12A Problems/Problem 17"
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==Problem== | ==Problem== | ||
− | + | Suppose <math>a</math> is a real number such that the equation <cmath>a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}</cmath> | |
has more than one solution in the interval <math>(0, \pi)</math>. The set of all such <math>a</math> that can be written | has more than one solution in the interval <math>(0, \pi)</math>. The set of all such <math>a</math> that can be written | ||
in the form <cmath>(p,q) \cup (q,r),</cmath> | in the form <cmath>(p,q) \cup (q,r),</cmath> | ||
where <math>p, q,</math> and <math>r</math> are real numbers with <math>p < q< r</math>. What is <math>p+q+r</math>? | where <math>p, q,</math> and <math>r</math> are real numbers with <math>p < q< r</math>. What is <math>p+q+r</math>? | ||
− | <math>\textbf{(A) } -4 \qquad \textbf{(B) } -1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4</math> | + | <math>\textbf{(A) } {-}4 \qquad \textbf{(B) } {-}1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4</math> |
− | |||
==Solution 1== | ==Solution 1== |
Revision as of 18:00, 26 December 2022
Contents
Problem
Suppose is a real number such that the equation has more than one solution in the interval . The set of all such that can be written in the form where and are real numbers with . What is ?
Solution 1
We are given that
Using the sine double angle formula combine with the fact that , which can be derived using sine angle addition with , we have Since as it is on the open interval , we can divide out from both sides, leaving us with Now, distributing and rearranging, we achieve the equation which is a quadratic in .
Applying the quadratic formula to solve for , we get and expanding the terms under the radical, we get Factoring, since , we can simplify our expression even further to
Now, solving for our two solutions, and .
Since yields a solution that is valid for all , that being , we must now solve for the case where yields a valid value.
As , , and therefore , and .
There is one more case we must consider inside this interval though, the case where , as this would lead to a double root for , yielding only one valid solution for . Solving for this case, .
Therefore, combining this fact with our solution interval, , so the answer is
- DavidHovey
Solution 2
We can optimize from the step from in solution 1 by writing
and then get
Now, solving for our two solutions, and .
Since yields a solution that is valid for all , that being , we must now solve for the case where yields a valid value.
As , , and therefore , and .
There is one more case we must consider inside this interval though, the case where , as this would lead to a double root for , yielding only one valid solution for . Solving for this case, .
Therefore, combining this fact with our solution interval, , so the answer is
- Dan
Solution 3
Use the sum to product formula to obtain . Use the double angle formula on the RHS to obtain . From here, it is obvious that is always a solution, and thus we divide by to get We wish to find all such that there is at least one more solution to this equation distinct from . Letting , and noting that , we can rearrange our equation to The smallest value where is , which is not in our domain so we divide by to obtain . By the trivial inequality, . Furthermore, , so . Also, if , then the solution to this equation would be shared with , so there would only be one distinct solution. Finally, because due to the restrictions of a sine wave, and that due to the restrictions on , we have with . Thus, , so our final answer is .
~sigma
Video Solution 1 (Quick and Simple)
~Education, the Study of Everything
See Also
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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.