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1987 AIME Problems/Problem 4

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Problem

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

Solution

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 • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Category: Intermediate Algebra Problems

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