(Assume that there is no friction and drag, and the ball follows the law of reflection.) Suppose we have an \(n\)-gon. There is a point-like ball at the midpoint of one of the sides of the \(n\)-gon (call that side Side A). It is launched to the midpoint of another side (call that Side B). Let \(k\) be the number of sides clockwise to Side A but counterclockwise to Side B (not including Side A and Side B). Define the function \(\delta_n(k)\) to be \(k\) if the ball bounces off every side of the \(n\)-gon before returning to the launch point, and otherwise \(0\). Find the closed form for the sum \[\sum^{n-2}_{k=0}{\delta_n(k)}\]

EDIT: Here is a hint: Euler's Totient Function

And if that doesn't make sense, here is a stepping stone: GCD. (Thanks Calvin L.)

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## Comments

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TopNewestA better hint would be Greatest Common Divisor, though it might give away too much.

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I assume you are talking about a regular n-gon?

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yes.

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Are you bobthesmartypants? :P

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Yeah that's what I thought.

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yep :P

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Consider a square. \(\delta_{4}(2) = 2\), because the ball will hit all sides before returning to its initial position. Also, \(\delta_{4}(0) = 0\), but that doesn't contribute to the sum. So \[\sum\limits_{k=0}^{2} \delta_{4}(k) = 2\] In general, \[\sum\limits_{k=0}^{n-2} \delta_{n}(k) \geq n - 2\] Another observation I made is that \[\sum\limits_{k=0}^{p-2} \delta_{p}(k) = \sum\limits_{k=0}^{p-2} k = \frac{(p-2)(p-1)}{2}\] for every prime number p. This is because the ball will never return to the initial position before hitting each side in such a p-gon, no matter which side you point the ball at.

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Sorry, but you answer is incorrect. Solving for the square does not solve it for every other n-gon.

HINT: Euler's totient function

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Yes, I already figured out I should use the Totient function. :) Also, I did not give an incorrect answer, because I did not yet give an answer… I just showed my observations. :)

I know that \[\sum\limits_{k=0}^{n-2} \delta_{n}(k) = - \phi(n) + \sum_{k=1}^{n-1} \begin{cases} k &\mbox{if } \gcd(n,k) = 1 \\ 0 &\mbox{otherwise} \end{cases}\] but that's all I got so far. ;-) Can it get better than this?

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Is the answer zero??

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no, sorry.

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