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ANSWER TO
PLACEMATS
by Bill Graham
Restate the problem in the language of mathematics so that you can work
mathematical magic on it. The rectangular placemat's dimensions are
N × M circles. The border is 2M + 2N - 4 circles.
The entire area is N×M. After subtracting the border, the
interior is NM - 2M - 2N + 4 circles.
Equate border and interior to get the following equation.
2M + 2N - 4 = NM - 2M - 2N + 4
Rearrange.
NM = 4N + 4M - 8
This Diophantine equation is non-linear.
To skip down to the answer, press here.
Find other shapes discussed briefly here.
Often, you can solve Diophantine equations by using odd-even logic.
The solution here demonstrates this strategy.
Before starting, note that 0×0 will always work mathematically because
the inside and outside will both be zero.
However, these placemats would be useless.
On to the solution.
First, note that the area, NM, must be divisible by four.
Thus, either both N and M are even or one of them, say N, is divisible by four.
In the latter case, M must be odd if the solution is to differ from the
former case. Take these cases one at a time.
Both N and M even
N = 2n, M = 2m
4nm = 8n + 8m - 8
nm = 2n + 2m - 2
The right side of this equation is even so that nm is also even.
Either n or m is even. Take n even so that N is divisible by four.
You'd get the same results with m even; it makes no difference.
Redo substitutions with N = 4n.
N = 4n, M = 2m
8nm = 16n + 8m - 8
nm = 2n + m - 1
If m is even, then nm must be odd -- impossible with m even.
So, m must be odd making (2n + m -1) and nm even.
But, m is odd so that n must be even.
n = 2n', m = 2m' + 1
4n'm' + 2m' = 4n' + 2m' + 1 - 1
4n'm' = 4n'
You must conclude that (unless n' = 0), m' = 1.
m' = 1
m = 2m' + 1 = 3
M = 2m = 6
Substitute into the original equation.
NM = 4N + 4M - 8
6N = 4N + 24 - 8
2N = 16
N = 8
The required rectangle is unique (for the above assumptions) and is
6 × 8 circles.
It's border is 24 circles, which matches the interior size.
Press here to skip to next answer.
N divisible by 4, M odd
N = 4n
M = 2m + 1
4n(2m+1) = 16n + 8m + 4
8nm = 12n + 8m - 4
2nm = 3n + 2m - 1
Note that n must be odd because 3n-1 (= 2nm - 2m) must be even.
n = 2n' + 1
4n'm + 2m = 6n' + 3 + 2m - 1
4n'm = 6n' + 2
2n'm = 3n' + 1
Solve for n'.
n' = 1/(2m-3)
Here, n' can only be an integer because you can't have fractional circular
pieces. Also, m' must be an integer. No matter what integer value you select
for m, you'll never get a value of n' greater than one.
Thus, n' is one, and m is two.
n'=1, m=2
M = 2m+1 = 5
n = 2n' + 1 = 3
N = 4n = 12
The rectangle is 5 × 12 with a border of 30 circles, which
again matches the interior.
Exactly two rectangular placemats
6 × 8
5 × 12
Other shapes
Many other shapes are possible: triangles, trapezoids, rhomboids, hexagons, and
so on. Most of these would make rather unsatisfactory placemats.
You might use a stretched hexagon, where the top and bottom edges are longer than
the end edges. Take the length of the top side as M and the length of the other
edges as N.
The number of circles in the border is 2M + 4N - 6.
The total area is 2NM - (N+M-1) + N(N-1).
The interior area is the total area less the border, which is
INTERIOR = 2NM - (N+M-1) + N(N-1) - 2M - 4N + 6
INTERIOR = 2NM + NN - 6N - 3M + 7
Set the interior and border equal.
2M + 4N - 6 = 2NM + NN - 6N - 3M + 7
2NM + NN = 5M + 10N - 13
What the heck! This is really tough.
By inspection, I found one solution: M=8, N=3. The outer border has 22 circles as
does the inner part.
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Contents copyright 2004 by Bill Graham and ParaComp, Inc.
All rights reserved.
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Vol 2, No 3 Answers
