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 * * * * / * * * * / / * * * * / * * * * 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. > > > >