1997 AIME Problems/Problem 13
Problem
Let be the set of points in the Cartesian plane that satisfy
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:
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 finally , 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 , and .
- 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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