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Difference between revisions of "2020 AMC 12A Problems/Problem 9"

m (→‎Remark)
(Sol 1 is currently in fragments. Cleaned up the page by combining solutions in a generalized manner. Retained credits to everyone who contributed in this page. PM me if you think there is a better idea for major changes.)
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<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 1==
+
==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 branch of <math>y=\tan(2x),</math> as shown below:
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(800,200);
  
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Draw a graph of <math>\tan(2x)</math> and <math>\cos(\tfrac{x}{2})</math>
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real f(real x) { return tan(2*x); }
 +
real g(real x) { return cos(x/2); }
  
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<math>\tan(2x)</math> has a period of <math>\tfrac{\pi}{2},</math> asymptotes at <math>x = \tfrac{\pi}{4}+\tfrac{k\pi}{2},</math> and zeroes at <math>\tfrac{k\pi}{2}</math>. It is positive from <math>(0,\tfrac{\pi}{4}) \cup (\tfrac{\pi}{2},\tfrac{3\pi}{4}) \cup (\pi,\tfrac{5\pi}{4}) \cup (\tfrac{3\pi}{2},\tfrac{7\pi}{4})</math> and negative elsewhere.
+
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,"$y=\cos\left(\frac x2\right)$");
  
−
cos<math>(\tfrac{x}{2})</math> has a period of <math>4\pi</math> and zeroes at <math>\pi</math>. It is positive from <math>[0,\pi)</math> and negative elsewhere.
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real xMin = 0;
 +
real xMax = 9/4*pi;
 +
real yMin = -3;
 +
real yMax = 3;
  
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Drawing such a graph would get <math>\boxed{\textbf{E) }5}</math> ~lopkiloinm
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//Draws the horizontal gridlines
 +
void horizontalLines()
 +
{
 +
  for (real i = yMin+1; i < yMax; ++i)
 +
  {
 +
    draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
 +
  }
 +
}
  
−
Or you could see the points at which both graphs are positive or both are negative, again yielding 5 such areas. -hi13
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//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));
 +
  }
 +
}
  
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edited by - annabelle0913
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//Draws the horizontal ticks
 +
void horizontalTicks()
 +
{
 +
  for (real i = yMin+1; i < yMax; ++i)
 +
  {
 +
    draw((-1/8,i)--(1/8,i), black+linewidth(1));
 +
  }
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}
  
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==Solution 2==
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//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));
 +
  }
 +
}
  
−
To find the asymptotes of <math>\tan(2x)</math> we consider the behaviour of <math>\tan(x)</math> on <math>[0,4\pi]</math>. Then we see that there are five separate continuous parts of the graph splitting the plane into regions. Since <math>\cos(\frac{x}{2})</math> is continuous it must intersect each of the five pieces of <math>\tan</math> at least once. But since <math>\tan(2x)</math> is increasing and <math>\cos(\frac{x}{2})</math> is decreasing on the interval and continuous increasing functions and decreasing functions can intersect at most once, there are <math>\boxed{\textbf{E) }5}</math> intersections. -codecow
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horizontalLines();
 +
verticalLines();
 +
horizontalTicks();
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verticalTicks();
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draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
 +
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
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label("$x$",(xMax,0),(2,0));
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label("$y$",(0,yMax),(0,2));
  
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==Remark==
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pair A[], B[];
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The graphs of <math>f(x)=\tan(2x)</math> (in red) and <math>g(x)=\cos\left(\frac{x}{2}\right)</math> (in blue) are shown below.
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A[0] = (2pi,0);
 +
A[1] = (0,2);
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A[2] = (0,0);
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A[3] = (0,-2);
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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];
  
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[[File:2020 AMC 12A Problem 9.png|center]]
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label("$2\pi$",A[0],(0,-2.5));
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label("$2$",A[1],(-2.5,0));
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label("$0$",A[2],(-2.5,0));
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label("$-2$",A[3],(-2.5,0));
  
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Graph in Desmos: https://www.desmos.com/calculator/dyggbvywpt
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for (int i = 0; i < 5; ++i)
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{
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dot(B[i],black+linewidth(5));
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}
  
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~MRENTHUSIASM
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add(legend(),point(E),60E,UnFill);
 +
</asy>
 +
Therefore, the answer is <math>\boxed{\textbf{(E)}\ 5}.</math>
 +
 
 +
~MRENTHUSIASM ~lopkiloinm ~hi13 ~annabelle0913 ~codecow
  
 
==See Also==
 
==See Also==

Revision as of 17:09, 17 September 2021

Problem

How many solutions does the equation $\tan(2x)=\cos(\tfrac{x}{2})$ have on the interval $[0,2\pi]?$

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution

We count the intersections of the graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right):$

  1. The graph of $y=\tan(2x)$ has a period of $\frac{\pi}{2},$ asymptotes at $x=\frac{\pi}{4}+\frac{k\pi}{2},$ and zeros at $x=\frac{k\pi}{2}$ for some integer $k.$

    On the interval $[0,2\pi],$ the graph has five branches: \[\biggl[0,\frac{\pi}{4}\biggr),\left(\frac{\pi}{4},\frac{3\pi}{4}\right),\left(\frac{3\pi}{4},\frac{5\pi}{4}\right),\left(\frac{5\pi}{4},\frac{7\pi}{4}\right),\left(\frac{7\pi}{4},2\pi\right].\] Note that $\tan(2x)\in[0,\infty)$ for the first branch, $\tan(2x)\in(-\infty,\infty)$ for the three middle branches, and $\tan(2x)\in(-\infty,0]$ for the last branch. Moreover, all branches are strictly increasing.

  2. The graph of $y=\cos\left(\frac x2\right)$ has a period of $4\pi$ and zeros at $x=\pi+2k\pi$ for some integer $k.$

    On the interval $[0,2\pi],$ note that $\cos\left(\frac x2\right)\in[-1,1].$ Moreover, the graph is strictly decreasing.

The graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right)$ intersect once on each branch of $y=\tan(2x),$ 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 $\boxed{\textbf{(E)}\ 5}.$

~MRENTHUSIASM ~lopkiloinm ~hi13 ~annabelle0913 ~codecow

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

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