1987 AIME, Problem #4
Find the area of the region enclosed by the graph of $|x − 60| + |y| = |x/4|$.
My attempt:
We have \begin{equation} |y|=|x/4|-|x-60| = \begin{cases}-\frac{3x}{4}+60, &\text{if $x\ge60$}. \\ \frac{5x}{4}-60, &\text{if $60>x\ge0$}. \\ \frac{3x}{4}-60, &\text{if $0>x$}. \end{cases} \end{equation} Clearly these are really only 4 cases because of the symmetry of the signs of the first and third cases, and our equations eventually turn out to be \begin{equation} y = -3x + 60 \\ y = \frac{3}{4}x-60 \\ y=\frac{5}{4}x-60 \\ y = -\frac{5}{4}x+60 \end{equation} If we plot these lines, we have in the area enclosed by all of these lines, a concave shape which can be described as a triangle cut out from a larger triangle. However, if we just plot the equation of the original problem statement, it is only a quadrilateral that is a smaller enclosure which is plotted. What did I do wrong?