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

m (Solution)
Line 10: Line 10:
 
*<math>x - 60 < 0</math>. Then <math>y = \frac{5}{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>.
+
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)) = \boxed{480}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 18:54, 14 February 2014

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 $2 \cdot \frac{1}{2}(80 - 48)(15 - (-15)) = \boxed{480}$.

See also

1987 AIME (ProblemsAnswer KeyResources)
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

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