The sides of an equilateral triangle ABC are divided into n equal parts (n≥2). For each point on a side, we draw the lines parallel to other sides of the triangle ABC, e.g. for n=3 we have the following diagram:
[asy]
unitsize(150);
defaultpen(linewidth(0.7));
int n = 3; /* # of vertical lines, including AB */
pair A = (0,0), B = dir(-30), C = dir(30);
draw(A--B--C--cycle,linewidth(2)); dot(A,UnFill(0)); dot(B,UnFill(0)); dot(C,UnFill(0));
label("A",A,W); label("C",C,NE); label("B",B,SE);
for(int i = 1; i < n; ++i) {
draw((i*A+(n-i)*B)/n--(i*A+(n-i)*C)/n);
draw((i*B+(n-i)*A)/n--(i*B+(n-i)*C)/n);
draw((i*C+(n-i)*A)/n--(i*C+(n-i)*B)/n);
}
[/asy]
For each n≥2, find the number of existing parallelograms. geometryparallelogramcombinatorics proposedcombinatorics