# 1987 AIME Problems/Problem 4

## 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 or using shoelace, we get $2 \cdot \frac{1}{2}(80 - 48)(15) = \boxed{480}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 