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

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== Problem ==
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==Problem==
 
Let <math>S</math> be the [[set]] of [[point]]s in the [[Cartesian plane]] that satisfy <center><math>\Big|\big||x|-2\big|-1\Big|+\Big|\big||y|-2\big|-1\Big|=1.</math></center> If a model of <math>S</math> were built from wire of negligible thickness, then the total length of wire required would be <math>a\sqrt{b}</math>, where <math>a</math> and <math>b</math> are positive integers and <math>b</math> is not divisible by the square of any prime number. Find <math>a+b</math>.
 
Let <math>S</math> be the [[set]] of [[point]]s in the [[Cartesian plane]] that satisfy <center><math>\Big|\big||x|-2\big|-1\Big|+\Big|\big||y|-2\big|-1\Big|=1.</math></center> If a model of <math>S</math> were built from wire of negligible thickness, then the total length of wire required would be <math>a\sqrt{b}</math>, where <math>a</math> and <math>b</math> are positive integers and <math>b</math> is not divisible by the square of any prime number. Find <math>a+b</math>.
  
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__TOC__
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==Solution==
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== Solution ==
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===Solution 1===
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=== Solution 1 ===
 
 
:''This solution is non-rigorous.''
 
:''This solution is non-rigorous.''
 
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:
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Calculating the lengths is now easy; each rectangle has sides of <math>\sqrt{2}, 3\sqrt{2}</math>, so the answer is <math>4(\sqrt{2} + 3\sqrt{2}) = 16\sqrt{2}</math>. For all four quadrants, this is <math>64\sqrt{2}</math>, and <math>a+b=\boxed{066}</math>.
 
Calculating the lengths is now easy; each rectangle has sides of <math>\sqrt{2}, 3\sqrt{2}</math>, so the answer is <math>4(\sqrt{2} + 3\sqrt{2}) = 16\sqrt{2}</math>. For all four quadrants, this is <math>64\sqrt{2}</math>, and <math>a+b=\boxed{066}</math>.
  
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=== Solution 2 ===
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===Solution 2===
 
Since <math>0 \le \Big|\big||x| - 2\big| - 1\Big| \le 1</math> and <math>0 \le \Big|\big||y| - 2\big| - 1\Big| \le 1</math><br />
 
Since <math>0 \le \Big|\big||x| - 2\big| - 1\Big| \le 1</math> and <math>0 \le \Big|\big||y| - 2\big| - 1\Big| \le 1</math><br />
 
<math>- 1 \le \big||x| - 2\big| - 1 \le 1</math><br />
 
<math>- 1 \le \big||x| - 2\big| - 1 \le 1</math><br />
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So a total of <math>6</math> doublings = <math>2^6</math> = <math>64</math>, the total length = <math>64 \cdot \sqrt {2} = a\sqrt {b}</math>, and <math>a + b = 64 + 2 = \boxed{066}</math>.
 
So a total of <math>6</math> doublings = <math>2^6</math> = <math>64</math>, the total length = <math>64 \cdot \sqrt {2} = a\sqrt {b}</math>, and <math>a + b = 64 + 2 = \boxed{066}</math>.
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=== Solution 3 (FASTEST) ===
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===Solution 3 (FASTEST)===
 
We make use of several consecutive substitutions.  
 
We make use of several consecutive substitutions.  
 
Let <math>||x| - 2|= x_1</math> and similarly with <math>y</math>.
 
Let <math>||x| - 2|= x_1</math> and similarly with <math>y</math>.
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== See also ==
 
== See also ==
 
{{AIME box|year=1997|num-b=12|num-a=14}}
 
{{AIME box|year=1997|num-b=12|num-a=14}}
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:21, 10 March 2020

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:

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

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 $a + b = 64 + 2 = \boxed{066}$.

Solution 3 (FASTEST)

We make use of several consecutive substitutions. Let $||x| - 2|= x_1$ and similarly with $y$. Therefore, our graph is $|x_1 - 1| + |y_1 - 1| = 1$. This is a diamond with perimeter $4\sqrt{2}$. Now, we make use of the following fact for a function of two variables $x$ and $y$: Suppose we have $f(x, y) = c$. Then $f(|x|, |y|)$ is equal to the graph of $f(x, y)$ reflected across the y axis and x axis, and the reflection across the y axis across the x axis, therefore the perimeter of of $f(|x|, |y|)$ is 4 times the perimeter of $f(x, y)$. Now, we continue making substitutions at each absolute value sign ($|x| - 2 = x_2$ and finally $x = x_3$, similarly for y as well. ), noting that the constants don't matter as they just translate the graph and each absolute value sign increases the perimeter 4 times as much. Therefore, the length is $4^2 \times 4\sqrt{2} = 64\sqrt{2} = a\sqrt{b}$, and $a+b = \boxed{066}$.

- whatRthose

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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