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**I was trying to solve the following issue:

Find the number of possible closed paths using one fifth of an arc ($72^o$), 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 necessarily 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 substituting $x$ and $y$ in (equation 1) and doing sum math, and also noting that $\cos(72) = \sqrt{\frac{3-\sqrt{5}}{2}}$, 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?

Thanks in advance! :) **

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This is Project Euler problem 208: – Ross Millikan May 16 '11 at 14:22
Do u have any clue about the solution? – rhseiki May 17 '11 at 6:21
No, I did a lot of them but never this one. – Ross Millikan May 17 '11 at 12:48

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