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Difference between revisions of "2022 AMC 12A Problems/Problem 17"
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==Solution==
 
==Solution==
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Solution 1:
  
 
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>
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- DavidHovey
 
- DavidHovey
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Solution 2:
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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
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 +
<cmath>a = \frac{4\cos^2{x}-1}{(1+2\cos{x})} = 2\cos x - 1</cmath>
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 +
and then get
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<cmath>
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\cos x = \frac{a+1}{2}.
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</cmath>
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Now, solving for our two solutions,  <math>\cos{x} = -\frac{1}{2}</math> and <math>\cos{x} = \frac{a+1}{2}</math>.
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 +
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>
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 +
- Dan
  
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 03:42, 13 November 2022

Problem

Supppose $a$ is a real number such that the equation \[a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}\] has more than one solution in the interval $(0, \pi)$. The set of all such $a$ that can be written in the form \[(p,q) \cup (q,r),\] where $p, q,$ and $r$ are real numbers with $p < q< r$. What is $p+q+r$?

$\textbf{(A) } -4 \qquad \textbf{(B) } -1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 4$


Solution

Solution 1:

We are given that $a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}$

Using the sine double angle formula combine with the fact that $\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}+1)$, which can be derived using sine angle addition with $\sin{(2x + x)}$, we have \[a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)\] Since $\sin{x} \ne 0$ as it is on the open interval $(0, \pi)$, we can divide out $\sin{x}$ from both sides, leaving us with \[a\cdot(1+2\cos{x})=4\cos^2{x}-1\] Now, distributing $a$ and rearranging, we achieve the equation \[4\cos^2{x} - 2a\cos{x} - (1+a) = 0\] which is a quadratic in $\cos{x}$.

Applying the quadratic formula to solve for $\cos{x}$, we get \[\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}\] and expanding the terms under the radical, we get \[\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}\] Factoring, since $4a^2+16a+16 = (2a+4)^2$, we can simplify our expression even further to \[\cos{x} =\frac{a\pm(a+2)}{4}\]

Now, solving for our two solutions, $\cos{x} = -\frac{1}{2}$ and $\cos{x} = \frac{a+1}{2}$.

Since $\cos{x} = -\frac{1}{2}$ yields a solution that is valid for all $a$, that being $x = \frac{2\pi}{3}$, we must now solve for the case where $\frac{a+1}{2}$ yields a valid value.

As $x\in (0, \pi)$, $\cos{x}\in (-1, 1)$, and therefore $\frac{a+1}{2}\in (-1, 1)$, and $a\in(-3,1)$.

There is one more case we must consider inside this interval though, the case where $\frac{a+1}{2} = -\frac{1}{2}$, as this would lead to a double root for $\cos{x}$, yielding only one valid solution for $x$. Solving for this case, $a \ne -2$.

Therefore, combining this fact with our solution interval, $a\in(-3, -2) \cup (-2, 1)$, so the answer is $-3-2+1 = \boxed{\textbf{(A) -4}}$

- DavidHovey


Solution 2:

We can optimize from the step from \[a\cdot(1+2\cos{x})=4\cos^2{x}-1\] in solution 1 by writing

\[a = \frac{4\cos^2{x}-1}{(1+2\cos{x})} = 2\cos x - 1\]

and then get \[\cos x = \frac{a+1}{2}.\]

Now, solving for our two solutions, $\cos{x} = -\frac{1}{2}$ and $\cos{x} = \frac{a+1}{2}$.

Since $\cos{x} = -\frac{1}{2}$ yields a solution that is valid for all $a$, that being $x = \frac{2\pi}{3}$, we must now solve for the case where $\frac{a+1}{2}$ yields a valid value.

As $x\in (0, \pi)$, $\cos{x}\in (-1, 1)$, and therefore $\frac{a+1}{2}\in (-1, 1)$, and $a\in(-3,1)$.

There is one more case we must consider inside this interval though, the case where $\frac{a+1}{2} = -\frac{1}{2}$, as this would lead to a double root for $\cos{x}$, yielding only one valid solution for $x$. Solving for this case, $a \ne -2$.

Therefore, combining this fact with our solution interval, $a\in(-3, -2) \cup (-2, 1)$, so the answer is $-3-2+1 = \boxed{\textbf{(A) -4}}$

- 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

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