## The problem of nodal points

Divide the sides of a triangle in n1 (5), n2 (7), n3 (9) equal parts respectively.
Determine the number N(n1, n2, n3) of points (like A and B above), where lines of all three families of joining-lines meet. Characterize the tripples with N=0. Which tripples have a large N?
How are N(k*n1, k*n2, k*n3) and N(n1, n2, n3) related? Generalize to convex polygons with odd number of sides.

By Ceva's theorem, assuming that the cevians divide the side on which they end in a ratio
AF/AB = a, BD/BC = b, CE/CA =c = >
abc = (1-a)(1-b)(1-c). (Steiner I. p.167)
Setting a = z1/n1, b = z2/n2, c = z3/n3, for positive integers and zi less than ni = >

z1*z2*z3 = (n1-z1)*(n2-z2)*(n3-z3). (***)

Thus the problem is equivalent to the number-theoretic problem:
Given the positive integers ( >2 say) n1, n2, n3, find all solutions of (***) in positive integers
z1, z2, z3 , with zi <ni. N(n1,n2,n3) is the total number of these solutions.

e.g. in the big figure (z1, z2, z3) = (2, 3, 6) and (n1-z1, n2-z2, n3-z3) = (3, 4, 3) are the only solutions. In general if (z1,z2,z3) is a solution, then (n1-z1,n2-z2,n3-z3) is also a solution. Thus the map (z1,z2,z3) |--- > (n1-z1, n2-z2, n3-z3) leaves the solution-space invariant. In particular, if all of (n1,n2,n3) are even then (n1/2, n2/2, n3/2) is always a solution. If there is some odd number among the numbers (n1, n2, n3), then the solutions must appear in pairs, hence N(n1,n2,n3) is an even number.

The case n1=2, n2=n3=x is easy (points on the median). Must z1*z2 = (x-z1)*(x-z2) = >
x = z1+z2. Thus N(2,x,x) = x-1. On the other side, for different n2, n3 there is no solution, since this would lead to the equation n1*n2 -z1*n2 - z2*n1 = 0. If n1, n2 are prime to each other then this is impossible. The non-prime case is also reducible to the prime.

The general equation (***) represents an algebraic (cubic) surface in space. We want the points of that surface that have (positive) integer coordinates and fall into a certain parallelepiped.