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Difference between revisions of "1997 AIME Problems/Problem 13"

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Let <math>f(x) = \Big|\big||x|-2\big|-1\Big|</math>, <math>f(x) \ge 0</math>. Then <math>f(x) + f(y) = 1 \Longrightarrow f(x), f(y) \le 1 \Longrightarrow x, y \le 4</math>. We only have a <math>4\times 4</math> area, so guessing points and graphing won't be too bad of an idea. Since <math>f(x) = f(-x)</math>, there's a symmetry about all four [[quadrant]]s, so just consider the first quadrant. We now gather some points:
 
Let <math>f(x) = \Big|\big||x|-2\big|-1\Big|</math>, <math>f(x) \ge 0</math>. Then <math>f(x) + f(y) = 1 \Longrightarrow f(x), f(y) \le 1 \Longrightarrow x, y \le 4</math>. We only have a <math>4\times 4</math> area, so guessing points and graphing won't be too bad of an idea. Since <math>f(x) = f(-x)</math>, there's a symmetry about all four [[quadrant]]s, so just consider the first quadrant. We now gather some points:
  
−
{| class="wikitable" style="background:none;padding:10px;"
+
{| class="wikitable" style="background:none;"
 
|-  
 
|-  
−
| <math>f(1) = 0</math> || <math>f(0.1) = 0.9</math>
+
| style="width:15%;" | <math>f(1) = 0</math> || <math>f(0.1) = 0.9</math>
 
|-
 
|-
 
| <math>f(2) = 1</math> || <math>f(0.9) = 0.1</math>
 
| <math>f(2) = 1</math> || <math>f(0.9) = 0.1</math>

Revision as of 21:18, 23 November 2007

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:

1997 AIME-13a.png

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

1997 AIME-13b.png

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 $a+b=\boxed{066}$.

Solution 2

See also

1997 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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