Difference between revisions of "2020 AMC 12A Problems/Problem 9"
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==Problem== | ==Problem== | ||
− | How many solutions does the equation | + | How many solutions does the equation <math>\tan(2x)=\cos(\tfrac{x}{2})</math> have on the interval <math>[0,2\pi]?</math> |
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math> | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math> | ||
+ | |||
+ | ==Solution== | ||
+ | We count the intersections of the graphs of <math>y=\tan(2x)</math> and <math>y=\cos\left(\frac x2\right):</math> | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>The graph of <math>y=\tan(2x)</math> has a period of <math>\frac{\pi}{2},</math> asymptotes at <math>x=\frac{\pi}{4}+\frac{k\pi}{2},</math> and zeros at <math>x=\frac{k\pi}{2}</math> for some integer <math>k.</math> <p> | ||
+ | On the interval <math>[0,2\pi],</math> the graph has five branches: <cmath>\biggl[0,\frac{\pi}{4}\biggr),\left(\frac{\pi}{4},\frac{3\pi}{4}\right),\left(\frac{3\pi}{4},\frac{5\pi}{4}\right),\left(\frac{5\pi}{4},\frac{7\pi}{4}\right),\left(\frac{7\pi}{4},2\pi\right].</cmath> | ||
+ | Note that <math>\tan(2x)\in[0,\infty)</math> for the first branch, <math>\tan(2x)\in(-\infty,\infty)</math> for the three middle branches, and <math>\tan(2x)\in(-\infty,0]</math> for the last branch. Moreover, all branches are strictly increasing. | ||
+ | </li><p> | ||
+ | <li>The graph of <math>y=\cos\left(\frac x2\right)</math> has a period of <math>4\pi</math> and zeros at <math>x=\pi+2k\pi</math> for some integer <math>k.</math> <p> | ||
+ | On the interval <math>[0,2\pi],</math> note that <math>\cos\left(\frac x2\right)\in[-1,1].</math> Moreover, the graph is strictly decreasing.</li><p> | ||
+ | </ol> | ||
+ | The graphs of <math>y=\tan(2x)</math> and <math>y=\cos\left(\frac x2\right)</math> intersect once on each of the five branches of <math>y=\tan(2x),</math> as shown below: | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(800,200); | ||
+ | |||
+ | real f(real x) { return tan(2*x); } | ||
+ | real g(real x) { return cos(x/2); } | ||
+ | |||
+ | draw(graph(f,0,atan(3)/2),red,"$y=\tan(2x)$"); | ||
+ | draw(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),red); | ||
+ | draw(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),red); | ||
+ | draw(graph(f,-atan(3)/2+3*pi/2,atan(3)/2+3*pi/2),red); | ||
+ | draw(graph(f,-atan(3)/2+4*pi/2,2*pi),red); | ||
+ | draw(graph(g,0,2pi),blue,"$y=\cos\left(\frac x2\right)$"); | ||
+ | |||
+ | real xMin = 0; | ||
+ | real xMax = 9/4*pi; | ||
+ | real yMin = -3; | ||
+ | real yMax = 3; | ||
+ | |||
+ | //Draws the horizontal gridlines | ||
+ | void horizontalLines() | ||
+ | { | ||
+ | for (real i = yMin+1; i < yMax; ++i) | ||
+ | { | ||
+ | draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); | ||
+ | } | ||
+ | } | ||
+ | |||
+ | //Draws the vertical gridlines | ||
+ | void verticalLines() | ||
+ | { | ||
+ | for (real i = xMin+pi/2; i < xMax; i+=pi/2) | ||
+ | { | ||
+ | draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); | ||
+ | } | ||
+ | } | ||
+ | |||
+ | //Draws the horizontal ticks | ||
+ | void horizontalTicks() | ||
+ | { | ||
+ | for (real i = yMin+1; i < yMax; ++i) | ||
+ | { | ||
+ | draw((-1/8,i)--(1/8,i), black+linewidth(1)); | ||
+ | } | ||
+ | } | ||
+ | |||
+ | //Draws the vertical ticks | ||
+ | void verticalTicks() | ||
+ | { | ||
+ | for (real i = xMin+pi/2; i < xMax; i+=pi/2) | ||
+ | { | ||
+ | draw((i,-1/8)--(i,1/8), black+linewidth(1)); | ||
+ | } | ||
+ | } | ||
+ | |||
+ | horizontalLines(); | ||
+ | verticalLines(); | ||
+ | horizontalTicks(); | ||
+ | verticalTicks(); | ||
+ | draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); | ||
+ | draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); | ||
+ | label("$x$",(xMax,0),(2,0)); | ||
+ | label("$y$",(0,yMax),(0,2)); | ||
+ | |||
+ | pair A[], B[]; | ||
+ | A[0] = (2pi,0); | ||
+ | A[1] = (0,2); | ||
+ | A[2] = (0,0); | ||
+ | A[3] = (0,-2); | ||
+ | B[0] = intersectionpoints(graph(f,0,atan(3)/2),graph(g,0,2pi))[0]; | ||
+ | B[1] = intersectionpoints(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),graph(g,0,2pi))[0]; | ||
+ | B[2] = intersectionpoints(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),graph(g,0,2pi))[0]; | ||
+ | B[3] = intersectionpoints(graph(f,-atan(3)/2+3*pi/2,atan(3)/2+3*pi/2),graph(g,0,2pi))[0]; | ||
+ | B[4] = intersectionpoints(graph(f,-atan(3)/2+4*pi/2,atan(3)/2+4*pi/2),graph(g,0,2pi))[0]; | ||
+ | |||
+ | label("$2\pi$",A[0],(0,-2.5)); | ||
+ | label("$2$",A[1],(-2.5,0)); | ||
+ | label("$0$",A[2],(-2.5,0)); | ||
+ | label("$-2$",A[3],(-2.5,0)); | ||
+ | |||
+ | for (int i = 0; i < 5; ++i) | ||
+ | { | ||
+ | dot(B[i],black+linewidth(5)); | ||
+ | } | ||
+ | |||
+ | add(legend(),point(E),60E,UnFill); | ||
+ | </asy> | ||
+ | Therefore, the answer is <math>\boxed{\textbf{(E)}\ 5}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM ~lopkiloinm ~hi13 ~annabelle0913 ~codecow | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/ZdVut0V1O4g with tips on graphing trig functions fast ~MathProblemSolvingSkills.com | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC12 box|year=2020|ab=A|num-b=8|num-a=10}} | ||
+ | |||
+ | [[Category:Introductory Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 04:07, 8 July 2022
Contents
Problem
How many solutions does the equation have on the interval
Solution
We count the intersections of the graphs of and
- The graph of
has a period of
asymptotes at
and zeros at
for some integer
On the interval
the graph has five branches:
Note that
for the first branch,
for the three middle branches, and
for the last branch. Moreover, all branches are strictly increasing.
- The graph of
has a period of
and zeros at
for some integer
On the interval
note that
Moreover, the graph is strictly decreasing.
The graphs of and
intersect once on each of the five branches of
as shown below:
Therefore, the answer is
~MRENTHUSIASM ~lopkiloinm ~hi13 ~annabelle0913 ~codecow
Video Solution
https://youtu.be/ZdVut0V1O4g with tips on graphing trig functions fast ~MathProblemSolvingSkills.com
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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.