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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 19"

(→‎Problem 19)
Line 14: Line 14:
  
 
~kingofpineapplz
 
~kingofpineapplz
 +
 +
== Solution 2 ==
 +
For choice A, we have
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
For choice B, we have
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
For choice C, we have
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
For choice D, we have
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
For choice E, we have
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(B) }13}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=18|num-a=20}}
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:51, 25 November 2021

Problem 19

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 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 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 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 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 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 $\boxed{\textbf{(B) }13}$.

~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

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