Difference between revisions of "2021 AMC 12A Problems/Problem 19"
MRENTHUSIASM (talk | contribs) (Condensed Sol 2 a bit.) |
MRENTHUSIASM (talk | contribs) (Condensed Sol 3 considerably.) |
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==Solution 3 (Graphs and Analyses)== | ==Solution 3 (Graphs and Analyses)== | ||
− | + | This problem is equivalent to counting the intersections of the graphs of <math>y=\sin\left(\frac{\pi}{2}\cos x\right)</math> and <math>y=\cos\left(\frac{\pi}{2}\sin x\right)</math> in the closed interval <math>[0,\pi].</math> We construct a table of values, as shown below: | |
<cmath>\begin{array}{c|ccc} | <cmath>\begin{array}{c|ccc} | ||
& & & \\ [-2ex] | & & & \\ [-2ex] | ||
Line 50: | Line 50: | ||
\boldsymbol{\cos x} & 1 & 0 & -1 \\ [1.5ex] | \boldsymbol{\cos x} & 1 & 0 & -1 \\ [1.5ex] | ||
\boldsymbol{\frac{\pi}{2}\cos x} & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex] | \boldsymbol{\frac{\pi}{2}\cos x} & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex] | ||
− | \boldsymbol{ | + | \boldsymbol{\sin\left(\frac{\pi}{2}\cos x\right)} & 1 & 0 & -1 \\ [1.5ex] |
\hline | \hline | ||
& & & \\ [-1ex] | & & & \\ [-1ex] | ||
\boldsymbol{\sin x} & 0 & 1 & 0 \\ [1.5ex] | \boldsymbol{\sin x} & 0 & 1 & 0 \\ [1.5ex] | ||
\boldsymbol{\frac{\pi}{2}\sin x} & 0 & \frac{\pi}{2} & 0 \\ [1.5ex] | \boldsymbol{\frac{\pi}{2}\sin x} & 0 & \frac{\pi}{2} & 0 \\ [1.5ex] | ||
− | \boldsymbol{ | + | \boldsymbol{\cos\left(\frac{\pi}{2}\sin x\right)} & 1 & 0 & 1 \\ [1ex] |
\end{array}</cmath> | \end{array}</cmath> | ||
− | For | + | For <math>x\in[0,\pi],</math> note that: |
− | * | + | * <math>\frac{\pi}{2}\cos x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],</math> so <math>\sin\left(\frac{\pi}{2}\cos x\right)\in[-1,1].</math> |
− | * | + | * <math>\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],</math> so <math>\cos\left(\frac{\pi}{2}\sin x\right)\in[0,1].</math> |
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− | + | For the graphs to intersect, we need <math>\sin\left(\frac{\pi}{2}\cos x\right)\in[0,1].</math> This occurs when <math>\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right].</math> | |
− | + | By the Cofunction Identity <math>\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),</math> we have | |
+ | <cmath>\sin\left(\frac{\pi}{2}\cos x\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\sin x\right).</cmath> | ||
+ | Since <math>\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right],</math> we can apply the arcsine function to both sides and then simplify: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \frac{\pi}{2}\cos x | + | \frac{\pi}{2}\cos x &= \frac{\pi}{2}-\frac{\pi}{2}\sin x \\ |
− | \cos x + \ | + | \cos x &= 1 - \sin x \\ |
+ | \sin x + \cos x &= 1. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
From the last block of equations in Solution 2, we conclude that <math>(0,1)</math> and <math>\left(\frac{\pi}{2},0\right)</math> are the only points of intersection. So, the answer is <math>\boxed{\textbf{(C) }2}.</math> | From the last block of equations in Solution 2, we conclude that <math>(0,1)</math> and <math>\left(\frac{\pi}{2},0\right)</math> are the only points of intersection. So, the answer is <math>\boxed{\textbf{(C) }2}.</math> |
Revision as of 00:32, 17 September 2021
Contents
Problem
How many solutions does the equation have in the closed interval
?
Solution 1 (Inverse Trigonometric Functions)
The ranges of and
are both
which is included in the range of
so we can use it with no issues.
This only happens at
on the interval
because one of
and
must be
and the other
Therefore, the answer is
~Tucker
Solution 2 (Cofunction Identity)
By the Cofunction Identity we simplify the given equation:
for some integer
We rearrange and simplify:
By rough constraints, we know that
from which
The only possibility is
so we get
for some integer
The possible solutions in are
However,
is an extraneous solution by squaring
Therefore, the answer is
~MRENTHUSIASM
Solution 3 (Graphs and Analyses)
This problem is equivalent to counting the intersections of the graphs of and
in the closed interval
We construct a table of values, as shown below:
For
note that:
so
so
For the graphs to intersect, we need This occurs when
By the Cofunction Identity we have
Since
we can apply the arcsine function to both sides and then simplify:
From the last block of equations in Solution 2, we conclude that
and
are the only points of intersection. So, the answer is
~MRENTHUSIASM (credit given to TheAMCHub)
Remark
The graphs of (in red) and
(in blue) are shown below.
Graph in Desmos: https://www.desmos.com/calculator/brjh3gybcc
~MRENTHUSIASM
Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry)
~ pi_is_3.14
See also
2021 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.