1997 AIME Problems/Problem 13

Problem

Let $S$ be the set of points in the Cartesian plane that satisfy

$\Big|\big||x|-2\big|-1\Big|+\Big|\big||y|-2\big|-1\Big|=1.$

If a model of $S$ were built from wire of negligible thickness, then the total length of wire required would be $a\sqrt{b}$ , where $a$ and $b$ are positive integers and $b$ is not divisible by the square of any prime number. Find $a+b$ .

Solution

Solution 1

This solution is non-rigorous.

Let $f(x) = \Big|\big||x|-2\big|-1\Big|$ , $f(x) \ge 0$ . Then $f(x) + f(y) = 1 \Longrightarrow f(x), f(y) \le 1 \Longrightarrow x, y \le 4$ . We only have a $4\times 4$ area, so guessing points and graphing won't be too bad of an idea. Since $f(x) = f(-x)$ , there's a symmetry about all four quadrants, so just consider the first quadrant. We now gather some points:

$f(1) = 0$	$f(0.1) = 0.9$
$f(2) = 1$	$f(0.9) = 0.1$
$f(3) = 0$	$f(1.1) = 0.1$
$f(4) = 1$	$f(1.9) = 0.9$
$f(0.5) = 0.5$	$f(2.1) = 0.9$
$f(1.5) = 1.5$	$f(2.9) = 0.1$
$f(2.5) = 2.5$	$f(3.1) = 0.1$
$f(3.5) = 3.5$	$f(3.9) = 0.9$

We can now graph the pairs of points which add up to $1$ . Just using the first column of information gives us an interesting lattice pattern:

Plotting the remaining points and connecting lines, the graph looks like:

Calculating the lengths is now easy; each rectangle has sides of $\sqrt{2}, 3\sqrt{2}$ , so the answer is $4(\sqrt{2} + 3\sqrt{2}) = 16\sqrt{2}$ . For all four quadrants, this is $64\sqrt{2}$ , and $a+b=\boxed{066}$ .

1997 AIME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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1997 AIME Problems/Problem 13

Problem

Contents

Solution

Solution 1

Solution 2

See also