Difference between revisions of "2005 AMC 12B Problems/Problem 7"
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== Problem == | == Problem == | ||
| + | What is the area enclosed by the graph of <math>|3x|+|4y|=12</math>? | ||
| − | == Solution == | + | <math> |
| + | \mathrm{(A)}\ 6 \qquad | ||
| + | \mathrm{(B)}\ 12 \qquad | ||
| + | \mathrm{(C)}\ 16 \qquad | ||
| + | \mathrm{(D)}\ 24 \qquad | ||
| + | \mathrm{(E)}\ 25 | ||
| + | </math> | ||
| + | |||
| + | == Solution 1== | ||
| + | If we get rid of the absolute values, we are left with the following 4 equations (using the logic that if <math>|a|=b</math>, then <math>a</math> is either <math>b</math> or <math>-b</math>): | ||
| + | |||
| + | <cmath>\begin{align*} 3x+4y=12 \\ -3x+4y=12 \\ 3x-4y=12 \\ -3x-4y=12 \end{align*}</cmath> | ||
| + | |||
| + | We can then put these equations in slope-intercept form in order to graph them. | ||
| + | |||
| + | <cmath>\begin{align*} 3x+4y=12 \,\implies\, y=-\dfrac{3}{4}x+3\\ -3x+4y=12\,\implies\, y=\dfrac{3}{4}x+3\\ 3x-4y=12\,\implies\, y=\dfrac{3}{4}x-3\\ -3x-4y=12\,\implies\, y=-\dfrac{3}{4}x-3\end{align*}</cmath> | ||
| + | |||
| + | Now you can graph the lines to find the shape of the graph: | ||
| + | |||
| + | <asy> | ||
| + | Label f; | ||
| + | f.p=fontsize(6); | ||
| + | xaxis(-8,8,Ticks(f, 4.0)); | ||
| + | yaxis(-6,6,Ticks(f, 3.0)); | ||
| + | fill((0,-3)--(4,0)--(0,3)--(-4,0)--cycle,grey); | ||
| + | draw((-4,-6)--(8,3), Arrows(4)); | ||
| + | draw((4,-6)--(-8,3), Arrows(4)); | ||
| + | draw((-4,6)--(8,-3), Arrows(4)); | ||
| + | draw((4,6)--(-8,-3), Arrows(4));</asy> | ||
| + | |||
| + | We can easily see that it is a rhombus with diagonals of <math>6</math> and <math>8</math>. The area is <math>\dfrac{1}{2}\times 6\times8</math>, or <math>\boxed{\mathrm{(D)}\ 24}</math> | ||
| + | |||
| + | == Solution 2== | ||
| + | You can also assign <math>x</math> and <math>y</math> to be <math>0</math>. Then you can easily see that the diagonals are <math>6</math> and <math>8</math>. Multiply and divide by <math>2</math> to get D. <math>24</math> | ||
| + | |||
| + | ==Solution 3== | ||
| + | The graph is symmetric with respect to both coordinate axes, and in the first quadrant it coincides with the graph of the line <math>3x + 4y = 12.</math> Therefore the region is a rhombus, and the area is | ||
| + | |||
| + | <cmath>\text{Area} = 4\left(\frac{1}{2}(4\cdot 3)\right) = 24 \rightarrow \boxed{D}</cmath> | ||
| + | |||
| + | <asy> | ||
| + | draw((-5,0)--(5,0),Arrow); | ||
| + | draw((0,-4)--(0,4),Arrow); | ||
| + | label("$x$",(5,0),S); | ||
| + | label("$y$",(0,4),E); | ||
| + | label("4",(4,0),S); | ||
| + | label("-4",(-4,0),S); | ||
| + | label("3",(0,3),NW); | ||
| + | label("-3",(0,-3),SW); | ||
| + | draw((4,0)--(0,3)--(-4,0)--(0,-3)--cycle,linewidth(0.7)); | ||
| + | </asy> | ||
| + | |||
| + | ~Alcumus | ||
== See also == | == See also == | ||
| − | + | {{AMC12 box|year=2005|ab=B|num-b=6|num-a=8}} | |
| + | {{MAA Notice}} | ||
Latest revision as of 19:15, 27 December 2020
Problem
What is the area enclosed by the graph of 
?
Solution 1
If we get rid of the absolute values, we are left with the following 4 equations (using the logic that if 
, then 
is either 
or 
):
We can then put these equations in slope-intercept form in order to graph them.
Now you can graph the lines to find the shape of the graph:
![[asy] Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 4.0)); yaxis(-6,6,Ticks(f, 3.0)); fill((0,-3)--(4,0)--(0,3)--(-4,0)--cycle,grey); draw((-4,-6)--(8,3), Arrows(4)); draw((4,-6)--(-8,3), Arrows(4)); draw((-4,6)--(8,-3), Arrows(4)); draw((4,6)--(-8,-3), Arrows(4));[/asy]](http://latex.artofproblemsolving.com/a/1/9/a19def037694543a1771b3897d68d519dd066fcf.png)
We can easily see that it is a rhombus with diagonals of 
and 
. The area is 
, or 
Solution 2
You can also assign 
and 
to be 
. Then you can easily see that the diagonals are and . Multiply and divide by 
to get D. 
Solution 3
The graph is symmetric with respect to both coordinate axes, and in the first quadrant it coincides with the graph of the line 
Therefore the region is a rhombus, and the area is
![\[\text{Area} = 4\left(\frac{1}{2}(4\cdot 3)\right) = 24 \rightarrow \boxed{D}\]](http://latex.artofproblemsolving.com/e/f/0/ef0949e38da672b8251f851ec7d51ec63056bfa9.png)
![[asy] draw((-5,0)--(5,0),Arrow); draw((0,-4)--(0,4),Arrow); label("$x$",(5,0),S); label("$y$",(0,4),E); label("4",(4,0),S); label("-4",(-4,0),S); label("3",(0,3),NW); label("-3",(0,-3),SW); draw((4,0)--(0,3)--(-4,0)--(0,-3)--cycle,linewidth(0.7)); [/asy]](http://latex.artofproblemsolving.com/f/6/f/f6fb4f11d88c24aa05f736f273aa28bf02060d80.png)
~Alcumus
See also
| 2005 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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. 
