# Number of closed paths formed by arcs of one fifth of a circle

**I was trying to solve the following issue:

Find the number of possible closed paths using one fifth of an arc (72 degrees), where at each time step we can move either clockwise or anti-clockwise. in that particular problem, it was assumed that the number of arcs should be 70.

In my solution, I assumed that all closed paths have no loops. I then proved that the number of arcs should necessarilly be even in order to get a closed path. Then, I simplified the problem to finding the number of possible EQUILATERAL POLYGONS, with interior angles: 72, 2*72, 3*72, and 4*72.

To do this I needed to find the the combinations of interior angles satisfying the following conditions:

1- Sum of cosines of the interior angles = (n/2), where n is the number sides of the polygon, which in turn is fixed and directly related to the number of arcs.

2- Sum of the interior angles = (n-2)*180 (equation 1)

I then assumed that x,y,u, and v are the number of angles of size 72, 2872, 3*72, and 4*72 respectively. These satisfy the following equations:

1- x+y+u+v = n 2- x+2y+3u+4v = (5/2)*(n-2)

Thus, I was left only with 2 unknowns: x and y!

After subsituting x and y in (equation 1) and doing sum math, and also noting that cos(72) = sqrt((3-sqrt(5))/2), the problem was down to the following:

Finding the solution of the equation:

A(1,x,y,n)cos(72) + B(1,x,y,n)[cos(72)]^2 +C(1,x,y,n)[cos(72)]^3 = D(x,y,n)

where A, B, and C are linear functions of 1,x,y, and n and D is a linear function of x,y, and n.

But I was unable to continue. Could anyone help me with this?