Difference between revisions of "2021 Fall AMC 12A Problems/Problem 19"
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== 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 | + | \left| \sin x - \sin \left( x^2 \right) \right| |
− | & = \left| \sin 10^\circ | + | & = \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 | + | \left| \sin x - \sin \left( x^2 \right) \right| |
− | & = \left| \sin 13^\circ | + | & = \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 | + | \left| \sin x - \sin \left( x^2 \right) \right| |
− | & = \left| \sin 14^\circ | + | & = \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 | + | \left| \sin x - \sin \left( x^2 \right) \right| |
− | & = \left| \sin 19^\circ | + | & = \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 | + | \left| \sin x - \sin \left( x^2 \right) \right| |
− | & = \left| \sin 20^\circ | + | & = \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 the closest to <math>0.</math> | ||
− | + | ~Steven Chen (www.professorchenedu.com) | |
− | + | ==Easy Solution with Intermediate Value Theorem== | |
+ | we know that x^2-x=180-2x+360k or x^2-x=360k from looking at the 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 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 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 B | ||
+ | |||
+ | emilyyunhanq@gmail.com | ||
+ | solution by Emily Q | ||
+ | |||
+ | ==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|num-b=18|num-a=20}} | {{AMC12 box|year=2021 Fall|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:27, 18 May 2024
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)
Easy Solution with Intermediate Value Theorem
we know that x^2-x=180-2x+360k or x^2-x=360k from looking at the 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 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 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 B
emilyyunhanq@gmail.com solution by Emily Q
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.