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

Revision as of 18:36, 4 July 2013

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

Solution 2

Since $0 \le \Big|\big||x| - 2\big| - 1\Big| \le 1$ and $0 \le \Big|\big||y| - 2\big| - 1\Big| \le 1$
$- 1 \le \big||x| - 2\big| - 1 \le 1$
$0 \le \big||x| - 2\big| \le 2$
$- 2 \le |x| - 2 \le 2$
$- 4 \le x \le 4$
Also $- 4 \le y \le 4$ .

Define $f(a) = \Big|\big||a| - 2\big| - 1\Big|$ .

If $0 \le a \le 1$ :

$f(3 + a) = \Big|\big||3 + a| - 2\big| - 1\Big| = a$

$f(3 - a) = \Big|\big||3 - a| - 2\big| - 1\Big| = a$

$f(3 + a) = f(3 - a)$

If $0 \le a \le 2$ :

$f(2 + a) = \Big|\big||2 + a| - 2\big| - 1\Big| = a - 1$

$f(2 - a) = \Big|\big||2 - a| - 2\big| - 1\Big| = a - 1$

$f(2 + a) = f(2 - a)$

If $0 \le a \le 4$ :

$f(a) = \Big|\big||a| - 2\big| - 1\Big| = a - 3$

$f( - a) = \Big|\big|| - a| - 2\big| - 1\Big| = a - 3$

$f(a) = f( - a)$

So the graph of $y(x)$ at $3 \le x \le 4$ is symmetric to $y(x)$ at $2 \le x \le 3$ (reflected over the line x=3)
And the graph of $y(x)$ at $2 \le x \le 4$ is symmetric to $y(x)$ at $0 \le x \le 2$ (reflected over the line x=2)
And the graph of $y(x)$ at $0 \le x \le 4$ is symmetric to $y(x)$ at $- 4 \le x \le 0$ (reflected over the line x=0)

[this is also true for horizontal reflection, with $3 \le y \le 4$ , etc]

So it is only necessary to find the length of the function at $3 \le x \le 4$ and $3 \le y \le 4$ : $\Big|\big||x| - 2\big| - 1\Big| + \Big|\big||y| - 2\big| - 1\Big| = 1$
$x - 3 + y - 3 = 1$
$y = - x + 7$ (Length = $\sqrt {2}$ )

This graph is reflected over the line y=3, the quantity of which is reflected over y=2,

the quantity of which is reflected over y=0,

the quantity of which is reflected over x=3,

the quantity of which is reflected over x=2,

the quantity of which is reflected over x=0..

So a total of $6$ doublings = $2^6$ = $64$ , the total length = $64 \cdot \sqrt {2} = a\sqrt {b}$ , and $a + b = 64 + 2 = \boxed{66}$ .

Revision as of 21:29, 23 November 2007 (view source) Azjps (talk \| contribs) (and a rigorous solution by minsoens) ← Older edit		Revision as of 18:36, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
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	[[Category:Intermediate Algebra Problems]]		[[Category:Intermediate Algebra Problems]]
		+	{{MAA Notice}}

1997 AIME (Problems • Answer Key • Resources)
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