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 "2021 Fall AMC 12A Problems/Problem 19"

(→‎Problem 19)
(→‎Solution 2)
Line 16: Line 16:
  
 
== Solution 2 ==
 
== Solution 2 ==
−
For choice A, we have
+
For choice <math>\textbf{(A)},</math> we have
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
−
\left| \sin x = \sin \left( x^2 \right) \right|
+
\left| \sin x - \sin \left( x^2 \right) \right|
−
& = \left| \sin 10^\circ = \sin \left( \left( 10^2 \right)^\circ \right) \right| \\
+
& = \left| \sin 10^\circ - \sin \left( \left( 10^2 \right)^\circ \right) \right| \\
 
& = \left| \sin 10^\circ - \sin 100^\circ \right| \\
 
& = \left| \sin 10^\circ - \sin 100^\circ \right| \\
 
& = \left| \sin 10^\circ - \sin 80^\circ \right| \\
 
& = \left| \sin 10^\circ - \sin 80^\circ \right| \\
Line 26: Line 26:
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
−
 
+
For choice <math>\textbf{(B)},</math> we have
−
For choice B, we have
 
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
−
\left| \sin x = \sin \left( x^2 \right) \right|
+
\left| \sin x - \sin \left( x^2 \right) \right|
−
& = \left| \sin 13^\circ = \sin \left( \left( 13^2 \right)^\circ \right) \right| \\
+
& = \left| \sin 13^\circ - \sin \left( \left( 13^2 \right)^\circ \right) \right| \\
 
& = \left| \sin 13^\circ - \sin 169^\circ \right| \\
 
& = \left| \sin 13^\circ - \sin 169^\circ \right| \\
 
& = \left| \sin 10^\circ - \sin 11^\circ \right| \\
 
& = \left| \sin 10^\circ - \sin 11^\circ \right| \\
Line 37: Line 36:
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
−
 
+
For choice <math>\textbf{(C)},</math> we have
−
For choice C, we have
 
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
−
\left| \sin x = \sin \left( x^2 \right) \right|
+
\left| \sin x - \sin \left( x^2 \right) \right|
−
& = \left| \sin 14^\circ = \sin \left( \left( 14^2 \right)^\circ \right) \right| \\
+
& = \left| \sin 14^\circ - \sin \left( \left( 14^2 \right)^\circ \right) \right| \\
 
& = \left| \sin 14^\circ - \sin 196^\circ \right| \\
 
& = \left| \sin 14^\circ - \sin 196^\circ \right| \\
 
& = \left| \sin 14^\circ + \sin 16^\circ \right| \\
 
& = \left| \sin 14^\circ + \sin 16^\circ \right| \\
Line 48: Line 46:
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
−
 
+
For choice <math>\textbf{(D)},</math> we have
−
For choice D, we have
 
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
−
\left| \sin x = \sin \left( x^2 \right) \right|
+
\left| \sin x - \sin \left( x^2 \right) \right|
−
& = \left| \sin 19^\circ = \sin \left( \left( 19^2 \right)^\circ \right) \right| \\
+
& = \left| \sin 19^\circ - \sin \left( \left( 19^2 \right)^\circ \right) \right| \\
 
& = \left| \sin 19^\circ - \sin 361^\circ \right| \\
 
& = \left| \sin 19^\circ - \sin 361^\circ \right| \\
 
& = \left| \sin 19^\circ - \sin 1^\circ \right| \\
 
& = \left| \sin 19^\circ - \sin 1^\circ \right| \\
Line 59: Line 56:
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
−
 
+
For choice <math>\textbf{(E)},</math> we have
−
For choice E, we have
 
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
−
\left| \sin x = \sin \left( x^2 \right) \right|
+
\left| \sin x - \sin \left( x^2 \right) \right|
−
& = \left| \sin 20^\circ = \sin \left( \left( 20^2 \right)^\circ \right) \right| \\
+
& = \left| \sin 20^\circ - \sin \left( \left( 20^2 \right)^\circ \right) \right| \\
 
& = \left| \sin 20^\circ - \sin 400^\circ \right| \\
 
& = \left| \sin 20^\circ - \sin 400^\circ \right| \\
 
& = \left| \sin 20^\circ - \sin 40^\circ \right| \\
 
& = \left| \sin 20^\circ - \sin 40^\circ \right| \\
Line 70: Line 66:
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
−
 
+
Therefore, the answer is <math>\boxed{\textbf{(B) }13},</math> as <math>\sin 11^\circ - \sin 10^\circ</math> is closest to <math>0.</math>
−
Therefore, the answer is <math>\boxed{\textbf{(B) }13}</math>.
 
  
 
~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)

Revision as of 01:07, 26 November 2021

Problem

Let $x$ be the least real number greater than $1$ such that $\sin(x)= \sin(x^2)$, where the arguments are in degrees. What is $x$ rounded up to the closest integer?

$\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20$

Solution 1

The smallest $x$ to make $\sin(x) = \sin(x^2)$ would require $x=x^2$, but since $x$ needs to be greater than $1$, these solutions are not valid.

The next smallest $x$ would require $x=180-x^2$, or $x^2+x=180$.

After a bit of guessing and checking, we find that $12^2+12=156$, and $13^2+13=182$, so the solution lies between $12{ }$ and $13$, making our answer $\boxed{\textbf{(B) } 13}.$

Note: One can also solve the quadratic and estimate the radical.

~kingofpineapplz

Solution 2

For choice $\textbf{(A)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 10^\circ - \sin \left( \left( 10^2 \right)^\circ \right) \right| \\ & = \left| \sin 10^\circ - \sin 100^\circ \right| \\ & = \left| \sin 10^\circ - \sin 80^\circ \right| \\ & = \sin 80^\circ - \sin 10^\circ . \end{align*} For choice $\textbf{(B)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 13^\circ - \sin \left( \left( 13^2 \right)^\circ \right) \right| \\ & = \left| \sin 13^\circ - \sin 169^\circ \right| \\ & = \left| \sin 10^\circ - \sin 11^\circ \right| \\ & = \sin 11^\circ - \sin 10^\circ . \end{align*} For choice $\textbf{(C)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 14^\circ - \sin \left( \left( 14^2 \right)^\circ \right) \right| \\ & = \left| \sin 14^\circ - \sin 196^\circ \right| \\ & = \left| \sin 14^\circ + \sin 16^\circ \right| \\ & = \sin 14^\circ + \sin 16^\circ . \end{align*} For choice $\textbf{(D)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 19^\circ - \sin \left( \left( 19^2 \right)^\circ \right) \right| \\ & = \left| \sin 19^\circ - \sin 361^\circ \right| \\ & = \left| \sin 19^\circ - \sin 1^\circ \right| \\ & = \sin 19^\circ - \sin 1^\circ . \end{align*} For choice $\textbf{(E)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 20^\circ - \sin \left( \left( 20^2 \right)^\circ \right) \right| \\ & = \left| \sin 20^\circ - \sin 400^\circ \right| \\ & = \left| \sin 20^\circ - \sin 40^\circ \right| \\ & = \sin 40^\circ - \sin 20^\circ . \end{align*} Therefore, the answer is $\boxed{\textbf{(B) }13},$ as $\sin 11^\circ - \sin 10^\circ$ is closest to $0.$

~Steven Chen (www.professorchenedu.com)

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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=2021_Fall_AMC_12A_Problems/Problem_19&oldid=166526"