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

Latest revision as of 20:27, 17 March 2024

Problem

Suppose $a$ 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 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

Solution 3

Use the sum to product formula to obtain $2a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{3x}$ . Use the double angle formula on the RHS to obtain $a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{\frac{3x}{2}}\cos{\frac{3x}{2}}$ . From here, it is obvious that $x=\frac{2\pi}{3}$ is always a solution, and thus we divide by $\sin{\frac{3x}{2}}$ to get $a\cdot\cos{\frac{x}{2}}=\cos{\frac{3x}{2}}$ We wish to find all $a$ such that there is at least one more solution to this equation distinct from $x=\frac{2\pi}{3}$ . Letting $y=\cos{\frac{x}{2}}$ , and noting that $\cos{\frac{3x}{2}}=4y^3-3y$ , we can rearrange our equation to $4y^3=y(3+a)$ The smallest value $x$ where $y=0$ is $\pi$ , which is not in our domain so we divide by $y$ to obtain $4y^2=a+3$ . By the trivial inequality, $a+3\ge{0}$ . Furthermore, $y\neq{0}$ , so $a+3>0$ . Also, if $a=-2$ , then the solution to this equation would be shared with $x=\frac{2\pi}{3}$ , so there would only be one distinct solution. Finally, because $y\le{1}$ due to the restrictions of a sine wave, and that $y\neq{1}$ due to the restrictions on $x$ , we have $-3<a<1$ with $a\neq{-2}$ . Thus, $p=-3,q=-2, r=1$ , so our final answer is $-3+(-2)+1=\boxed{\textbf{(A) } {-}4}$ .

~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

@@ Line 1: / Line 1: @@
 ==Problem==
-Supppose <math>a</math> is a real number such that the equation <cmath>a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}</cmath>
+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
 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>?
-<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>
-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>
 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>.
@@ Line 28: / Line 27: @@
 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>
+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
+==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 ==
 {{AMC12 box|year=2022|ab=A|num-b=16|num-a=18}}
+[[Category:Intermediate Trigonometry Problems]]
 {{MAA Notice}}

2022 AMC 12A (Problems • Answer Key • Resources)
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