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== | ||
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>. | ||
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There is one more case we must consider inside this interval though, the case where <math>\frac{a+1}{2} = -\frac{1}{2}</math>, as this would lead to a double root for <math>\cos{x}</math>, yielding only one valid solution for <math>x</math>. Solving for this case, <math> a \ne -2</math>. | There is one more case we must consider inside this interval though, the case where <math>\frac{a+1}{2} = -\frac{1}{2}</math>, as this would lead to a double root for <math>\cos{x}</math>, yielding only one valid solution for <math>x</math>. Solving for this case, <math> a \ne -2</math>. | ||
− | Therefore, combining this fact with our solution interval, <math>a\in(-3, -2) \cup (-2, 1)</math>, so the answer is <math>-3-2+1 = \boxed{\textbf{(A) -4 | + | Therefore, combining this fact with our solution interval, <math>a\in(-3, -2) \cup (-2, 1)</math>, so the answer is <math>-3-2+1 = \boxed{\textbf{(A) } {-}4}</math>. |
- DavidHovey | - DavidHovey | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | We can optimize from the step from <cmath>a\cdot(1+2\cos{x})=4\cos^2{x}-1</cmath> in solution 1 by writing | ||
+ | |||
+ | <cmath>a = \frac{4\cos^2{x}-1}{1+2\cos{x}} = 2\cos x - 1</cmath> | ||
+ | |||
+ | and then get | ||
+ | <cmath> | ||
+ | \cos x = \frac{a+1}{2}. | ||
+ | </cmath> | ||
+ | |||
+ | Now, solving for our two solutions, <math>\cos{x} = -\frac{1}{2}</math> and <math>\cos{x} = \frac{a+1}{2}</math>. | ||
+ | |||
+ | Since <math>\cos{x} = -\frac{1}{2}</math> yields a solution that is valid for all <math>a</math>, that being <math>x = \frac{2\pi}{3}</math>, we must now solve for the case where <math>\frac{a+1}{2}</math> yields a valid value. | ||
+ | |||
+ | As <math>x\in (0, \pi)</math>, <math>\cos{x}\in (-1, 1)</math>, and therefore <math>\frac{a+1}{2}\in (-1, 1)</math>, and <math>a\in(-3,1)</math>. | ||
+ | |||
+ | There is one more case we must consider inside this interval though, the case where <math>\frac{a+1}{2} = -\frac{1}{2}</math>, as this would lead to a double root for <math>\cos{x}</math>, yielding only one valid solution for <math>x</math>. Solving for this case, <math> a \ne -2</math>. | ||
+ | |||
+ | Therefore, combining this fact with our solution interval, <math>a\in(-3, -2) \cup (-2, 1)</math>, so the answer is <math>-3-2+1 = \boxed{\textbf{(A) } {-}4}</math>. | ||
+ | |||
+ | - Dan | ||
+ | |||
+ | ==Solution 3== | ||
+ | Use the sum to product formula to obtain <math>2a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{3x}</math>. Use the double angle formula on the RHS to obtain <math>a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{\frac{3x}{2}}\cos{\frac{3x}{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{3x}{2}}</math> to get <cmath>a\cdot\cos{\frac{x}{2}}=\cos{\frac{3x}{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{x}{2}}</math>, and noting that <math>\cos{\frac{3x}{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\le{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>. | ||
+ | |||
+ | ~sigma | ||
+ | |||
+ | ==Video Solution 1 (Quick and Simple)== | ||
+ | https://youtu.be/Tl5hBEkHzbA | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution 2 by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=0gCMvUmZtpI | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2022|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2022|ab=A|num-b=16|num-a=18}} | ||
+ | |||
+ | [[Category:Intermediate Trigonometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:27, 17 March 2024
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
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=0gCMvUmZtpI
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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