Difference between revisions of "2018 AMC 12A Problems/Problem 9"
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<cmath>\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \frac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi </cmath> | <cmath>\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \frac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi </cmath> | ||
| − | == Solution == | + | == Solution 1 == |
On the interval <math>[0, \pi]</math> sine is nonnegative; thus <math>\sin(x + y) = \sin x \cos y + \sin y \cos x \le \sin x + \sin y</math> for all <math>x, y \in [0, \pi]</math>. The answer is <math>\boxed{\textbf{(E) } 0\le y\le \pi}</math>. (CantonMathGuy) | On the interval <math>[0, \pi]</math> sine is nonnegative; thus <math>\sin(x + y) = \sin x \cos y + \sin y \cos x \le \sin x + \sin y</math> for all <math>x, y \in [0, \pi]</math>. The answer is <math>\boxed{\textbf{(E) } 0\le y\le \pi}</math>. (CantonMathGuy) | ||
| + | |||
| + | ==Solution 2== | ||
| + | Expanding as before, <cmath>\cos y \sin x + \cos x \sin y \le \sin x + \sin y</cmath> Let <math>\sin x =a \ge 0</math>, <math>\sin y = b \ge 0</math>. We have that <cmath>(\cos y)a+(\cos x)b \le a+b</cmath> Comparing coefficients of <math>a</math> and <math>b</math> gives a clear solution: both <math>\cos y</math> and <math>\cos x</math> are less than or equal to one, so the coefficients of <math>a</math> and <math>b</math> on the left are less than on the right. Since <math>a, b \ge 0</math>, that means that this equality is always satisfied over this interval, or <math>\boxed{\textbf{(E) } 0\le y\le \pi}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2018|ab=A|num-b=8|num-a=10}} | {{AMC12 box|year=2018|ab=A|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:55, 9 February 2018
Contents
Problem
Which of the following describes the largest subset of values of 
within the closed interval ![$[0,\pi]$](http://latex.artofproblemsolving.com/e/c/e/ece27ba559bd93cdf2416aa0f88fc8d7d85e5791.png)
for which
![\[\sin(x+y)\leq \sin(x)+\sin(y)\]](http://latex.artofproblemsolving.com/b/9/b/b9b2c0b1c3b4bd2cd0535f1bcd6552a064d603ca.png)
for every 
between 
and 
, inclusive?
![\[\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \frac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi\]](http://latex.artofproblemsolving.com/0/1/8/01880704ce76279becf4da3cfbee26e326d89573.png)
Solution 1
On the interval ![$[0, \pi]$](http://latex.artofproblemsolving.com/0/a/1/0a151f5b93ce5735f72cdd79c46051f52b9d820d.png)
sine is nonnegative; thus 
for all ![$x, y \in [0, \pi]$](http://latex.artofproblemsolving.com/4/4/7/447a91e1512333fc639da02e03bf52a259f38101.png)
. The answer is 
. (CantonMathGuy)
Solution 2
Expanding as before, ![\[\cos y \sin x + \cos x \sin y \le \sin x + \sin y\]](http://latex.artofproblemsolving.com/6/2/a/62a916ee8c05e199ecfa5fb02e05d7868542a433.png)
Let 
, 
. We have that ![\[(\cos y)a+(\cos x)b \le a+b\]](http://latex.artofproblemsolving.com/8/1/9/81993ddeff536aa854fdcff82c41fd83230cb359.png)
Comparing coefficients of 
and 
gives a clear solution: both 
and 
are less than or equal to one, so the coefficients of and on the left are less than on the right. Since 
, that means that this equality is always satisfied over this interval, or .
See Also
| 2018 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. 
