From: "Ignacio Larrosa Canestro"
Subject: RE: Checkerboard.
Date: Sun, 22 Oct 2000 18:48:25 +0200
Newsgroups: sci.math
For a 4*4 grid, the calculations are easy:
Comb(16, 9)=560 triads of points
There are 10 straight lines with four points on then: 4 verticals, 4
horizontals an 2 main diagonals. There ara 4 triads (i.e., Comb(4, 3)) of
points in each one of them --> 40
Aditionally, the four diagonals (2 of then indicated below) close the main
diagonals also contain 3 points.
Then, there are 44 triads of points collinears, and the number of triangles
is 560 - 44=516
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In general, let B(m,n) the number of triangles in a grid of m*n points.
Let C(j,k) the number of triangles in a grid of j*k points that cannot be
put on a grid of j'*k' points if j' For m=n see
>
> http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?An
> um=045996
>
> There n is the number of points, not the squares. So, for n=4 (3 for you)
> the number is 516.
>
> a(n) = ((n - 1)^2*n^2*(n + 1)^2)/6 - 2*Sum(Sum((n - k + 1)*(n - l + 1)*gcd(k
> - 1, l - 1), k, 2, n), l, 2, n)
>
>
> Ignacio Larrosa Canestro
> > How many triangles(of any size)are there in an m by n checkerboard
> whith
> > (m+1)*(n+1) vertices?For m = n = 3,the numbers of vertices is
> (3+1)*(3+1)=16
> > and I have two diferents solutions a)520 triangles b)516 triangles.What´s
> > the right and general solution?.Why?
> >
> >
> > S.Lomas.
> >
> >