2022 AMC 12A Problems/Problem 17
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 |
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All AMC 12 Problems and Solutions |
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