Difference between revisions of "2021 Fall AMC 12A Problems/Problem 19"
(→See Also) |
m |
||
(9 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
− | ==Problem | + | ==Problem== |
− | Let <math>x</math> be the least real number greater than <math>1</math> such that | + | Let <math>x</math> be the least real number greater than <math>1</math> such that <math>\sin(x)= \sin(x^2)</math>, where the arguments are in degrees. What is <math>x</math> rounded up to the closest integer? |
<math>\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20</math> | <math>\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20</math> | ||
Line 14: | Line 14: | ||
~kingofpineapplz | ~kingofpineapplz | ||
+ | |||
+ | == Solution 2 == | ||
+ | For choice <math>\textbf{(A)},</math> 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 <math>\textbf{(B)},</math> 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 <math>\textbf{(C)},</math> 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 <math>\textbf{(D)},</math> 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 <math>\textbf{(E)},</math> 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> as <math>\sin 11^\circ - \sin 10^\circ</math> is the closest to <math>0.</math> | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | ==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 | ||
==See Also== | ==See Also== | ||
− | {{AMC12 box|year=2021 Fall|ab=A| | + | {{AMC12 box|year=2021 Fall|ab=A|num-b=18|num-a=20}} |
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:38, 7 April 2022
Contents
Problem
Let be the least real number greater than such that , where the arguments are in degrees. What is rounded up to the closest integer?
Solution 1
The smallest to make would require , but since needs to be greater than , these solutions are not valid.
The next smallest would require , or .
After a bit of guessing and checking, we find that , and , so the solution lies between and , making our answer
Note: One can also solve the quadratic and estimate the radical.
~kingofpineapplz
Solution 2
For choice we have For choice we have For choice we have For choice we have For choice we have Therefore, the answer is as is the closest to
~Steven Chen (www.professorchenedu.com)
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=H0pNJFbV4jE
Video Solution by TheBeautyofMath
Solved both Mentally and by writing things down
~IceMatrix
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.