2021 Fall AMC 12A Problems/Problem 19

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 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)

2021 Fall AMC 12A (Problems • Answer Key • Resources)
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2021 Fall AMC 12A Problems/Problem 19

Contents

Problem

Solution 1

Solution 2

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