1997 AIME Problems/Problem 13
Problem
Let be the set of points in the Cartesian plane that satisfy
![$\Big|\big||x|-2\big|-1\Big|+\Big|\big||y|-2\big|-1\Big|=1.$](http://latex.artofproblemsolving.com/8/6/1/861bc503fcf3e136f3357993d4d30ded9c04a837.png)
If a model of were built from wire of negligible thickness, then the total length of wire required would be
, where
and
are positive integers and
is not divisible by the square of any prime number. Find
.
Solution
Solution 1
- This solution is non-rigorous.
Let ,
. Then
. We only have a
area, so guessing points and graphing won't be too bad of an idea. Since
, there's a symmetry about all four quadrants, so just consider the first quadrant. We now gather some points:
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We can now graph the pairs of coordinates which add up to . 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 , so the answer is
. For all four quadrants, this is
, and
.
Solution 2
Since and
Also .
Define .
- If
:
- If
:
- If
:
- So the graph of
at
is symmetric to
at
(reflected over the line x=3)
- And the graph of
at
is symmetric to
at
(reflected over the line x=2)
- And the graph of
at
is symmetric to
at
(reflected over the line x=0)
[this is also true for horizontal reflection, with , etc]
So it is only necessary to find the length of the function at and
:
(Length =
)
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 doublings =
=
, the total length =
, and
.
Solution 3 (FASTEST)
We make use of several consecutive substitutions.
Let and similarly with
.
Therefore, our graph is
. This is a diamond with perimeter
. Now, we make use of the following fact for a function of two variables
and
: Suppose we have
. Then
is equal to the graph of
reflected across the y axis and x axis, and the reflection across the y axis across the x axis, therefore the perimeter of of
is 4 times the perimeter of
. Now, we continue making substitutions at each absolute value sign (
and so forth), noting that the constants don't matter and each absolute value sign increases the perimeter 4 times as much. Therefore, the answer is $4^2 \times 4\sqrt{2} = \box{64/sqrt{2}}$ (Error compiling LaTeX. Unknown error_msg)
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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