Difference between revisions of "2020 AMC 12A Problems/Problem 9"
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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> | 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> | </ol> | ||
| − | The graphs of <math>y=\tan(2x)</math> and <math>y=\cos\left(\frac x2\right)</math> intersect once on each | + | 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> | <asy> | ||
/* Made by MRENTHUSIASM */ | /* Made by MRENTHUSIASM */ | ||
Revision as of 16:13, 17 September 2021
Problem
How many solutions does the equation 
have on the interval ![$[0,2\pi]?$](http://latex.artofproblemsolving.com/e/c/c/ecccefb4c460857891d99273237d9deb9ecf41c1.png)
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
![$[0,2\pi],$](//latex.artofproblemsolving.com/b/e/2/be287a7495ea16e7d9471055df3954dac7b0b0d8.png)
the graph has five branches: ![\[\biggl[0,\frac{\pi}{4}\biggr),\left(\frac{\pi}{4},\frac{3\pi}{4}\right),\left(\frac{3\pi}{4},\frac{5\pi}{4}\right),\left(\frac{5\pi}{4},\frac{7\pi}{4}\right),\left(\frac{7\pi}{4},2\pi\right].\]](//latex.artofproblemsolving.com/0/f/6/0f61c4e0ed515f1a18498e7b2922905c5f1e2a81.png)
Note that 
for the first branch, 
for the three middle branches, and ![$\tan(2x)\in(-\infty,0]$](//latex.artofproblemsolving.com/5/b/8/5b8d849c9e05ed072772ac711429ff9f30bb763d.png)
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 ![$\cos\left(\frac x2\right)\in[-1,1].$](//latex.artofproblemsolving.com/b/4/e/b4ed836dbb1efbe3523d09e1a13e10d81f07d20d.png)
Moreover, the graph is strictly decreasing.
asymptotes at 
and zeros at 
for some integer 
![$[0,2\pi],$](http://latex.artofproblemsolving.com/b/e/2/be287a7495ea16e7d9471055df3954dac7b0b0d8.png)
the graph has five branches: ![\[\biggl[0,\frac{\pi}{4}\biggr),\left(\frac{\pi}{4},\frac{3\pi}{4}\right),\left(\frac{3\pi}{4},\frac{5\pi}{4}\right),\left(\frac{5\pi}{4},\frac{7\pi}{4}\right),\left(\frac{7\pi}{4},2\pi\right].\]](http://latex.artofproblemsolving.com/0/f/6/0f61c4e0ed515f1a18498e7b2922905c5f1e2a81.png)
Note that 
for the first branch, 
for the three middle branches, and ![$\tan(2x)\in(-\infty,0]$](http://latex.artofproblemsolving.com/5/b/8/5b8d849c9e05ed072772ac711429ff9f30bb763d.png)
for the last branch. Moreover, all branches are strictly increasing.
and zeros at 
for some integer ![$\cos\left(\frac x2\right)\in[-1,1].$](http://latex.artofproblemsolving.com/b/4/e/b4ed836dbb1efbe3523d09e1a13e10d81f07d20d.png)
Moreover, the graph is strictly decreasing.
The graphs of and 
intersect once on each of the five branches of 
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]](http://latex.artofproblemsolving.com/c/1/4/c14719331d5c1b4c94bdc7825e40af9d20180888.png)
Therefore, the answer is 
~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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
