Difference between revisions of "2022 AMC 12A Problems/Problem 17"
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+ | Use the sum to product formula to obtain <math>a\cdot\sin{\frac{3\pi*x}{2}}\cos{\frac{1\pi*x}{2}}=\sin{3\pi*x}</math>. Use the double angle formula on the RHS to obtain <math>a\cdot\sin{\frac{3\pi*x}{2}}\cos{\frac{1\pi*x}{2}}=\sin{\frac{3\pi*x}{2}}\cos{\frac{3\pi*x}{2}}</math>. From here, it is obvious that <math>x=\frac{2\pi}{3}</math> is always a solution, and thus we divide by <math>\sin{\frac{3\pi*x}{2}}</math> to get <cmath>a\cdot\cos{\frac{1\pi*x}{2}}=\cos{\frac{3\pi*x}{2}}</cmath> We wish to find all <math>a</math> such that there is at least one more solution to this equation distinct from <math>x=\frac{2\pi}{3}</math>. Letting <math>y=\cos{\frac{1\pi*x}{2}}</math>, and noting that <math>\cos{\frac{3\pi*x}{2}}=4y^3-3y</math>, we can rearrange our equation to <math>4y^3=y(3+a)</math> The smallest value <math>x</math> where <math>y=0</math> is <math>\pi</math>, which is not in our domain so we divide by <math>y</math> to obtain <math>4y^2=a+3</math>. By the trivial inequality, <math>a+3\ge{0}</math>. Furthermore, <math>y\neq{0}</math>, so <math>a+3>0</math>. Also, if <math>a=-2</math>, then the solution to this equation would be shared with <math>x=\frac{2\pi}{3}</math>, so there would only be one distinct solution. Finally, because <math>y<1</math> due to the restrictions of a sine wave, and that <math>y\neq{1}</math> due to the restrictions on <math>x</math>, we have <math>-3<a<1</math> with <math>a\neq{-2}</math>. Thus, <math>p=-3,q=-2, r=1</math>, so our final answer is <math>-3+(-2)+1=\boxed{\textbf{(A) -4}}</math>. | ||
==Video Solution 1 (Quick and Simple)== | ==Video Solution 1 (Quick and Simple)== |
Revision as of 22:57, 15 November 2022
Contents
Problem
Supppose 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 .
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 |
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