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

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== Problem ==
 
== Problem ==
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Find the [[area]] of the region enclosed by the [[graph]] of <math>\displaystyle |x-60|+|y|=\left|\frac{x}{4}\right|.</math>
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Find the [[area]] of the region enclosed by the [[graph]] of <math>|x-60|+|y|=\left|\frac{x}{4}\right|.</math>
  
 
== Solution ==
 
== Solution ==
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{{image}}
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[[Image:1987_AIME-4.png]]
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Since <math>|y|</math> is [[nonnegative]], <math>\left|\frac{x}{4}\right| \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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 +
Since <math>|y|</math> is [[nonnegative]], <math>\left|\frac{x}{4}\right| \ge |x - 60|</math>. Solving this gives us two equations: <math>\frac{x}{4} \ge x - 60\ \mathrm{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>. Since the graph is [[symmetry|symmetric]] about the y-axis, we just need [[casework]] upon <math>x</math>. <math>\frac{x}{4} > 0</math>, so we break up the condition <math>|x-60|</math>:
 +
 
 +
*<math>x - 60 > 0</math>. Then <math>y = -\frac{3}{4}x+60</math>.
 +
*<math>x - 60 < 0</math>. Then <math>y = \frac{5}{4}x-60</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>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1987|num-b=3|num-a=5}}
 
{{AIME box|year=1987|num-b=3|num-a=5}}
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 +
[[Category:Intermediate Algebra Problems]]

Revision as of 20:16, 10 October 2007

Problem

Find the area of the region enclosed by the graph of $|x-60|+|y|=\left|\frac{x}{4}\right|.$

Solution

1987 AIME-4.png

Since $|y|$ is nonnegative, $\left|\frac{x}{4}\right| \ge |x - 60|$. Solving this gives us two equations: $\frac{x}{4} \ge x - 60\ \mathrm{and} \ -\frac{x}{4} \le x - 60$. Thus, $48 \le x \le 80$. The maximum and minimum y value is when $|x - 60| = 0$, which is when $x = 60$ and $y = \pm 15$. Since the graph is symmetric about the y-axis, we just need casework upon $x$. $\frac{x}{4} > 0$, so we break up the condition $|x-60|$:

  • $x - 60 > 0$. Then $y = -\frac{3}{4}x+60$.
  • $x - 60 < 0$. Then $y = \frac{5}{4}x-60$.

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 (Problems • Answer Key • Resources)
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
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