A nice problem I made

(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.) Note by Daniel Liu
6 years, 4 months ago

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Are you bobthesmartypants? :P

- 6 years, 4 months ago

Yeah that's what I thought.

- 6 years, 4 months ago

yep :P

- 6 years, 4 months ago

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

- 6 years, 4 months ago

yes.

- 6 years, 4 months ago

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

Staff - 6 years, 4 months ago

Is the answer zero??

- 6 years, 4 months ago

no, sorry.

- 6 years, 4 months ago

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.

- 6 years, 4 months ago

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

HINT: Euler's totient function

- 6 years, 4 months ago

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?

- 6 years, 4 months ago

yep, there is actually a closed form for the second sum on the RHS. Hint on finding the closed sum: What is Gauss famous for as a schoolchild?

- 6 years, 4 months ago

The second term: $\sum_{k=1}^{n-1} \begin{cases} k &\mbox{if } \gcd(n,k) = 1 \\ 0 &\mbox{otherwise} \end{cases}$ is the sum of all values $k$ smaller than $n$ that are coprime to $n$. These values $k$ seem symmetric about $\frac{n}{2}$. For example, if $n=6$ then $k=1$ or $k=5$. And (intuitively) this works for any $n$. So basically, the average for all values $k$ is $\frac{n}{2}$. So the second term in my previous solution is equal to the average $\left(\frac{n}{2}\right)$ multiplied by the number of terms $\left(\phi(n)\right)$: \begin{aligned} \sum\limits^{n-2}_{k=0} \delta_{n}(k) &= \phi(n) * \frac{n}{2} - \phi(n) \\ &= \phi(n) \left(\frac{n}{2} - 1\right) \end{aligned} I'm guessing this is the answer? :) That was a very entertaining problem, thanks.

- 6 years, 4 months ago