1987 AIME Problem #4 The Question: Find the area of the region enclosed by the graph of $|x-60|+|y|=|\frac x4|$.
Answer: 

 480

What I know: Because of all the absolute values I only need to find one side of the graph of this equation, the positive side. The equation can be rewritten as $y=|\frac x4|-|x-60|$, where $y≥0$. How do I proceed? 
Note: I have read the official solution for this problem, but I did not get how they found the vertices of the triangles.
 A: Your initial reasoning is correct, but could be formalized better.  We note that if $(x,y)$ is a solution to $$|x-60| + |y| = |x/4|,$$ then so also is $(x,-y)$, since $|-y| = |y|$ for all real $y$.  So the solution set is symmetric about the $x$-axis.  This allows us to assume without loss of generality that $y \ge 0$.  Now we write $$y = |x/4| - |x-60|,$$ hence under the aforementioned assumption, we conclude that $|x/4| \ge |x-60|$.  Under what circumstances does equality occur?  Clearly, this would imply either $x/4 = x-60$, or $x/4 = -(x-60)$.  In the first case, $x = 80$, and in the second, $x = 48$.  So all that remains is to investigate the sign of $|x/4| - |x-60|$ on the intervals $$x < 48, \quad 48 < x < 80, \quad x > 80.$$  We know that $|x/4| - |x-60|$ doesn't change sign within these intervals (for then we would have detected it when solving for equality).  It easily follows that $y \ge 0$ only when $48 \le x \le 80$.
From this, we note that on this interval, we have $$y = |x/4| - |x-60| = \begin{cases} x/4+x-60, & 48 \le x \le 60 \\ x/4-x+60, & 60 < x \le 80. \end{cases}.$$  This gives us the desired boundary, which is now easy to plot.  The vertices are therefore $$(48, 0), (60, \pm 15), (80, 0),$$ and the area is straightforward to compute.
A: You have six cases to consider:
1) $x\geq60, y\geq0\to y=-\frac{3}{4}x+60$.
2) $x\geq60, y\leq0\to y=\frac{3}{4}x-60$.
3) $0\leq x\leq60, y\geq0\to y=\frac{5}{4}x-60$.
4) $0\leq x\leq60, y\leq0\to y=-\frac{5}{4}x+60$.
5) $x\leq0, y\geq0\to y=\frac{3}{4}x+60$.
6) $x\leq0, y\leq0\to y=-\frac{3}{4}x-60$.
Now, the figure is the hexagon with vertices $(-80,0),(0,60),(60,15),(80,0),(60,-15),(0,-60)$
