| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
How many triangles are there where each side has length a positive integer, and the perimeter is n? ([3,4,5] & [3,5,4] & [4,5,3] are all the same triangle.) Regards, Andrew.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1211.1 | ALLVAX::JROTH | It's a bush recording... | Thu Mar 22 1990 10:18 | 19 | |
The exact number may be a difficult question in additive number theory...
The (a,b,c) lengths will be integer lattice points lying on the
plane a+b+c = n and satisfying the further inequalities
a,b,c > 0 positive lengths lie in equilateral triangle in plane
c < a+b triangle inequality
a <= b
a <= c
b <= c make the triangle unique up to symmetry
It shouldn't be hard to have an asymptotic estimate of the number of
triangles, since the limiting density of lattice points in the
plane a+b+c = n should be easy to estimate, (I don't know the number
offhand) and the bounds could be used to compute the area of the subset
we're interested in.
- Jim
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| 1211.2 | ALLVAX::JROTH | It's a bush recording... | Thu Mar 22 1990 11:10 | 14 | |
I tried a simple minded program and it seems that
log(k)/log(n) ~= 1.4 or so; this doesn't make sense, since the density
of lattice points in a plane x+y+z = n is proprortional to n^2
(the constant is about .5771). If the points were equidistributed
then only 1/6 of them would be our triangles on account of the
inequalities and symmetry.
The density of lattice points in the plane must not be uniform.
I may try plotting the pattern and see if it looks interesting.
fwiw, for n = 20000 I saw 8333333 unique triangles.
- Jim
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| 1211.3 | 1/6 done? | IOSG::CARLIN | Dick Carlin IOSG | Thu Mar 22 1990 14:55 | 10 |
Is it really that difficult? I calculate that for
n = 12*m k = 3*m^2
n = 12*m+6 k = 3*m*(m+1)+1
and the other 5/6 should be similar. On the other hand I could be wrong
as I'm rushing to do this before my ALL-IN-1 link completes.
dick
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| 1211.4 | ALLVAX::JROTH | It's a bush recording... | Thu Mar 22 1990 15:26 | 20 | |
Yes, my face should be red for making it overly complex...
The cases could be tabulated as:
n k
---------- ------------------
12*m 3*m^2
12*m + 1 3*m^2 + 2*m
12*m + 2 3*m^2 + m
12*m + 3 3*m^2 + 3*m + 1
12*m + 4 3*m+2 + 2*m
12*m + 5 3*m+2 + 4*m + 1
12*m + 6 3*m+2 + 3*m + 1
12*m + 7 3*m+2 + 5*m + 2
12*m + 8 3*m+2 + 4*m + 1
12*m + 9 3*m+2 + 6*m + 3
12*m + 10 3*m^2 + 5*m + 2
12*m + 11 3*m^2 + 7*m + 4
- Jim
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| 1211.5 | yes, that's it | HERON::BUCHANAN | combinatorial bomb squad | Fri Mar 23 1990 11:33 | 96 |
| 1211.6 | ALLVAX::JROTH | It's a bush recording... | Sun Apr 15 1990 14:58 | 9 | |
Here is an even simpler expression for the number
n even: floor((n^2+21)/48)
n odd: floor((n^2+6*n+21)/48)
Actually, the problem would have been simpler if you hadn't required that
congruent triangles be considered equivalent :-)
- Jim
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