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 nn-gon. There is a point-like ball at the midpoint of one of the sides of the nn-gon (call that side Side A). It is launched to the midpoint of another side (call that Side B). Let kk be the number of sides clockwise to Side A but counterclockwise to Side B (not including Side A and Side B). Define the function δn(k)\delta_n(k) to be kk if the ball bounces off every side of the nn-gon before returning to the launch point, and otherwise 00. Find the closed form for the sum k=0n2δn(k)\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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4 votes

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Are you bobthesmartypants? :P

Michael Tang - 6 years, 4 months ago

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Yeah that's what I thought.

Ryan Soedjak - 6 years, 4 months ago

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yep :P

Daniel Liu - 6 years, 4 months ago

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

Tim Vermeulen - 6 years, 4 months ago

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yes.

Daniel Liu - 6 years, 4 months ago

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

Calvin Lin Staff - 6 years, 4 months ago

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Is the answer zero??

Shubham Srivastava - 6 years, 4 months ago

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no, sorry.

Daniel Liu - 6 years, 4 months ago

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Consider a square. δ4(2)=2\delta_{4}(2) = 2, because the ball will hit all sides before returning to its initial position. Also, δ4(0)=0\delta_{4}(0) = 0, but that doesn't contribute to the sum. So k=02δ4(k)=2\sum\limits_{k=0}^{2} \delta_{4}(k) = 2 In general, k=0n2δn(k)n2\sum\limits_{k=0}^{n-2} \delta_{n}(k) \geq n - 2 Another observation I made is that k=0p2δp(k)=k=0p2k=(p2)(p1)2\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.

Tim Vermeulen - 6 years, 4 months ago

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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

Daniel Liu - 6 years, 4 months ago

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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 k=0n2δn(k)=ϕ(n)+k=1n1{kif gcd(n,k)=10otherwise\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?

Tim Vermeulen - 6 years, 4 months ago

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@Tim Vermeulen 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?

Daniel Liu - 6 years, 4 months ago

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@Daniel Liu The second term: k=1n1{kif gcd(n,k)=10otherwise\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 kk smaller than nn that are coprime to nn. These values kk seem symmetric about n2\frac{n}{2}. For example, if n=6n=6 then k=1k=1 or k=5k=5. And (intuitively) this works for any nn. So basically, the average for all values kk is n2\frac{n}{2}. So the second term in my previous solution is equal to the average (n2)\left(\frac{n}{2}\right) multiplied by the number of terms (ϕ(n))\left(\phi(n)\right): k=0n2δn(k)=ϕ(n)n2ϕ(n)=ϕ(n)(n21) \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.

Tim Vermeulen - 6 years, 4 months ago

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@Tim Vermeulen that might be a bad hint, so think about this: pairing values up

Daniel Liu - 6 years, 4 months ago

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