Difference between revisions of "2022 AMC 12A Problems/Problem 17"
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We are given that <math>a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}</math> | We are given that <math>a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}</math> | ||
− | Using the sine double angle formula combine with the fact that <math>\sin{(3x)} = \sin{x}\cdot(4\cos^2{x} | + | Using the sine double angle formula combine with the fact that <math>\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)</math>, which can be derived using sine angle addition with <math>\sin{(2x + x)}</math>, we have <cmath>a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)</cmath> |
Since <math>\sin{x} \ne 0</math> as it is on the open interval <math>(0, \pi)</math>, we can divide out <math>\sin{x}</math> from both sides, leaving us with <cmath>a\cdot(1+2\cos{x})=4\cos^2{x}-1</cmath> | Since <math>\sin{x} \ne 0</math> as it is on the open interval <math>(0, \pi)</math>, we can divide out <math>\sin{x}</math> from both sides, leaving us with <cmath>a\cdot(1+2\cos{x})=4\cos^2{x}-1</cmath> | ||
Now, distributing <math>a</math> and rearranging, we achieve the equation <cmath>4\cos^2{x} - 2a\cos{x} - (1+a) = 0</cmath> which is a quadratic in <math>\cos{x}</math>. | Now, distributing <math>a</math> and rearranging, we achieve the equation <cmath>4\cos^2{x} - 2a\cos{x} - (1+a) = 0</cmath> which is a quadratic in <math>\cos{x}</math>. |
Revision as of 01:29, 14 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
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.