Difference between revisions of "1987 AIME Problems/Problem 4"

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== Problem ==
 
== Problem ==
Find the area of the region enclosed by the graph of <math>\displaystyle |x-60|+|y|=|x/4|.</math>
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Find the [[area]] of the region enclosed by the [[graph]] of <math>\displaystyle |x-60|+|y|=|x/4|.</math>
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== Solution ==
 
== Solution ==
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Since <math>|y|</math> is [[nonnegative]], <math>|\frac{x}{4}| \ge |x - 60|</math>. Solving this gives us two equations: <math>\frac{x}{4} \ge x - 60\ and \ -\frac{x}{4} \le x - 60</math>. Thus, <math>48 \le x \le 80</math>. The [[maximum]] and [[minimum]] y value is when <math>|x - 60| = 0</math>, which is when <math>x = 60</math> and <math>y = \pm 15</math>. The area of the region enclosed by the graph is that of the quadrilateral defined by the points <math>(48,0),\ (60,15),\ (80,0), \ (60,-15)</math>. Breaking it up into triangles and solving, we get <math>2 \cdot \frac{1}{2}(80 - 48)(15 - (-15)) = 480</math>.
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== See also ==
 
== See also ==
 
* [[1987 AIME Problems]]
 
* [[1987 AIME Problems]]
  
 
{{AIME box|year=1987|num-b=3|num-a=5}}
 
{{AIME box|year=1987|num-b=3|num-a=5}}

Revision as of 18:44, 15 February 2007

Problem

Find the area of the region enclosed by the graph of $\displaystyle |x-60|+|y|=|x/4|.$

Solution


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Since $|y|$ is nonnegative, $|\frac{x}{4}| \ge |x - 60|$. Solving this gives us two equations: $\frac{x}{4} \ge x - 60\ and \ -\frac{x}{4} \le x - 60$. Thus, $48 \le x \le 80$. The maximum and minimum y value is when $|x - 60| = 0$, which is when $x = 60$ and $y = \pm 15$. The area of the region enclosed by the graph is that of the quadrilateral defined by the points $(48,0),\ (60,15),\ (80,0), \ (60,-15)$. Breaking it up into triangles and solving, we get $2 \cdot \frac{1}{2}(80 - 48)(15 - (-15)) = 480$.

See also

1987 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions