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Difference between revisions of "2022 AMC 12A Problems/Problem 17"

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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>
 
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>
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Now, distributing <math>a</math> and rearranging, we achieve the equation <cmath>4\cos^2{x} - 2a\cos{x} - (1+a)</cmath> which is a quadratic in <math>\cos{x}</math>.
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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>.
  
 
Applying the quadratic formula to solve for <math>\cos{x}</math>, we get <cmath>\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}</cmath> and expanding the terms under the radical, we get <cmath>\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}</cmath>
 
Applying the quadratic formula to solve for <math>\cos{x}</math>, we get <cmath>\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}</cmath> and expanding the terms under the radical, we get <cmath>\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}</cmath>

Revision as of 02:28, 12 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

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

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