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Sep 16, 2022 at 4:05 history edited person CC BY-SA 4.0
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Sep 15, 2022 at 22:54 comment added person I think my argument may be a bit faulty, I'll exchange that argument for a better one.
Sep 15, 2022 at 22:33 comment added Dan I think the question about $\lim\limits_{n\to\infty}\prod\limits_{k=1}^n l_k$ is an interesting question by itself, so I have posted it as another question: math.stackexchange.com/questions/4532364/…
Sep 15, 2022 at 22:10 comment added Narasimham to here? math.stackexchange.com/questions/4532364/…
Sep 15, 2022 at 21:40 comment added Dan How did you get $l_{k,\theta}-l_{k,\theta+\pi/n} \leq a(1+p)$ from the previous parts? And how did you get $\prod_{k=1}^n l_{k,\theta} = \prod_{k=1}^n l_{k,\theta+\pi/n} + o(|n^{n/2}|)$ ?
Sep 15, 2022 at 21:40 comment added Dan There was a mistake in my first comment: $(CD)(PD)$ should be $(CP)(PD)$.
Sep 15, 2022 at 5:58 history edited person CC BY-SA 4.0
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Sep 15, 2022 at 5:54 comment added person Thanks for the suggestion, I think I have it now. For some reason this theorem crossed my mind preciously but it did not occur to me to use it in this way. I think I should have a full proof now, although please look through to see if I've made some mistake.
Sep 15, 2022 at 5:52 history edited person CC BY-SA 4.0
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Sep 14, 2022 at 21:32 comment added Dan Consider diameter $AB$ through $P$, and an arbitrary chord $CD$ through $P$. If $n$ is even, then (using the intersecting chord theorem) the product of every pair of opposite line segments from $P$ is $(CD)(PD)=(AP)(PB)=(a-pa)(a+pa)=a^2(1-p^2)$. So $\left(\prod\limits_{k=1}^{n}l_k\right)^2=(a^2(1-p^2))^n=\left(\dfrac{n}{\pi(1-p^2)}(1-p^2)\right)^n=\left(\dfrac{n}{\pi}\right)^n$. Putting this into your expression for $L_n$, we get $\lim\limits_{n\to\infty}L_n=1$. But I don't know how to deal with the case with odd $n$.
Sep 13, 2022 at 23:51 history edited person CC BY-SA 4.0
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Sep 13, 2022 at 23:42 history edited person CC BY-SA 4.0
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Sep 13, 2022 at 23:30 history answered person CC BY-SA 4.0