Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
 
(12 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 
==Problem==
 
==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
 
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 1==
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>.
 
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
 
- DavidHovey
Line 51: Line 50:
 
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}}</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
 
- 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 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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_17&oldid=217052"