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

(revert edits by myself)
(and a rigorous solution by minsoens)
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== Problem ==
 
== Problem ==
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Let <math>S</math> be the set of points 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>.
  
 
__TOC__
 
__TOC__
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|}
 
|}
  
−
We can now graph the pairs of points which add up to <math>1</math>. Just using the first column of information gives us an interesting [[lattice]] pattern:
+
We can now graph the pairs of coordinates which add up to <math>1</math>. Just using the first column of information gives us an interesting [[lattice]] pattern:
  
 
[[Image:1997_AIME-13a.png]]
 
[[Image:1997_AIME-13a.png]]
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=== Solution 2 ===
 
=== 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 />
 +
<math>- 1 \le \big||x| - 2\big| - 1 \le 1</math><br />
 +
<math>0 \le \big||x| - 2\big| \le 2</math><br />
 +
<math>- 2 \le |x| - 2 \le 2</math><br />
 +
<math>- 4 \le x \le 4</math><br />
 +
Also <math>- 4 \le y \le 4</math>.
 +
 +
Define <math>f(a) = \Big|\big||a| - 2\big| - 1\Big|</math>.
 +
*If <math>0 \le a \le 1</math>:
 +
:<math>f(3 + a) = \Big|\big||3 + a| - 2\big| - 1\Big| = a</math>
 +
:<math>f(3 - a) = \Big|\big||3 - a| - 2\big| - 1\Big| = a</math>
 +
:<math>f(3 + a) = f(3 - a)</math>
 +
 +
*If <math>0 \le a \le 2</math>:
 +
:<math>f(2 + a) = \Big|\big||2 + a| - 2\big| - 1\Big| = a - 1</math>
 +
:<math>f(2 - a) = \Big|\big||2 - a| - 2\big| - 1\Big| = a - 1</math>
 +
:<math>f(2 + a) = f(2 - a)</math>
 +
 +
*If <math>0 \le a \le 4</math>:
 +
:<math>f(a) = \Big|\big||a| - 2\big| - 1\Big| = a - 3</math>
 +
:<math>f( - a) = \Big|\big|| - a| - 2\big| - 1\Big| = a - 3</math>
 +
:<math>f(a) = f( - a)</math>
 +
 +
*So the graph of <math>y(x)</math> at <math>3 \le x \le 4</math> is symmetric to <math>y(x)</math> at <math>2 \le x \le 3</math> (reflected over the line x=3)
 +
*And the graph of <math>y(x)</math> at <math>2 \le x \le 4</math> is symmetric to <math>y(x)</math> at <math>0 \le x \le 2</math> (reflected over the line x=2)
 +
*And the graph of <math>y(x)</math> at <math>0 \le x \le 4</math> is symmetric to <math>y(x)</math> at <math>- 4 \le x \le 0</math> (reflected over the line x=0)
 +
[this is also true for horizontal reflection, with <math>3 \le y \le 4</math>, etc]
 +
 +
So it is only necessary to find the length of the function at <math>3 \le x \le 4</math> and <math>3 \le y \le 4</math>:
 +
<math>\Big|\big||x| - 2\big| - 1\Big| + \Big|\big||y| - 2\big| - 1\Big| = 1</math><br />
 +
<math>x - 3 + y - 3 = 1</math><br />
 +
<math>y = - x + 7</math>
 +
(Length = <math>\sqrt {2}</math>)
 +
 +
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 <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{66}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 21:29, 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 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 $a + b = 64 + 2 = \boxed{66}$.

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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