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

Revision as of 11:16, 19 August 2025

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 the closest to $0.$

~Steven Chen (www.professorchenedu.com)

Solution 3

Knowing that $\sin x = \sin (180^\circ - x)$ , the original equation can be written as $\sin (180^\circ - x) = \sin (x^2)$ . We will obtain $180^\circ - x = x^2$ by taking $\arcsin$ on both sides.

Solving $180^\circ - x = x^2$ : $x^2 + x -180^\circ = 0$ $x = \frac{-1 \pm \sqrt{1 + 4 \cdot 180^\circ}}{2}$

Since $x>1$ , $x = \frac{-1 + \sqrt{1 + 4 \cdot 180^\circ}}{2}$ .

To find the integer that is the closest to $x$ : $\begin{align*} x & = \frac{-1 + \sqrt{1 + 4 \cdot 180^\circ}}{2}\\ & \approx \frac{\sqrt{4 \cdot 180^\circ}}{2}\\ & \approx \boxed{\textbf{(B) } 13^\circ}\\ \end{align*}$

~Andy_li0805

Easy Solution with Intermediate Value Theorem

We know that $x^2-x=180-2x+360k$ or $x^2-x=360k$ from looking at the $\text{sin}$ values on the unit circle. With these equations, we can just insert the values that are given in the $5$ answers.

Let's try $\text{A}$ first.

We know that one of the equations should be greater on the left, and another equation we make should be greater on the right for the solution to be in between the values inserted into the equations. The two numbers it is in between is $9$ and $10$ since it is rounded up.

$100-10$ vs $180-20,$

$81-9$ vs $180-18.$

This does not satisfy what we would like to find.

Now let's try $\text{B}$ .

Between $12$ and $13$ ;

$144-12$ vs $180-24,$

$169-13$ vs $180-26.$

Now this satisfies our hunt for the solution. The answer is $\boxed{\text{(B)}}$

emilyyunhanq@gmail.com

Solution by Emily Q

(edited by A7456321)

Video Solution by Mathematical Dexterity

https://www.youtube.com/watch?v=H0pNJFbV4jE

Video Solution by TheBeautyofMath

Solved both Mentally and by writing things down

https://youtu.be/o2MAmtgBbKc

~IceMatrix

@@ Line 71: / Line 71: @@
 == Solution 3 ==
+Knowing that <math>\sin x = \sin (180^\circ - x)</math>, the original equation can be written as <math>\sin (180^\circ - x) = \sin (x^2)</math>. We will obtain <math>180^\circ - x = x^2</math> by taking <math>\arcsin</math> on both sides.
+Solving <math>180^\circ - x = x^2</math>:
+<cmath>x^2 + x -180^\circ = 0</cmath>
+<cmath>x = \frac{-1 \pm \sqrt{1 + 4 \cdot 180^\circ}}{2} </cmath>
+Since <math>x>1</math>, <math>x = \frac{-1 + \sqrt{1 + 4 \cdot 180^\circ}}{2}</math>.
+To find the integer that is the closest to <math>x</math>:
+<cmath>
+\begin{align*}
+x
+& = \frac{-1 + \sqrt{1 + 4 \cdot 180^\circ}}{2}\\
+& \approx \frac{\sqrt{4 \cdot 180^\circ}}{2}\\
+& \approx \boxed{\textbf{(B) } 13^\circ}\\
+\end{align*}
+</cmath>
+~[[Andy_li0805]]
 ==Easy Solution with Intermediate Value Theorem==

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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