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We are given a fixed point on a circle of radius $1$, and going from this point along the circumference in the positive direction on curved distances $0,1,2,\ldots$ from it we obtain points with abscissas $n = 0,1,2,\ldots$ respectively. How many points among them should we take to ensure that some two of them are less than the distance $1/5$ apart?

I have a few questions about this. Firstly, what does the question mean by abscissa? Are they talking about the abscissa of convergence? Secondly I don't really get the distance thing going $0,1,2,\ldots.$ Below is the official solution:

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Why don't they ever go more than $+1$ distance (except for going across the fixed point)? Finally, since we are worried about actual distance not curved, how can they be sure this is in fact the first such pair of points?

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  • $\begingroup$ I'D say the problem is simply ambiguously posed. For example if we need not take consecutive points, we can pick more than $20$ $\endgroup$ – Hagen von Eitzen Feb 8 '16 at 3:18
  • $\begingroup$ @HagenvonEitzen What does abscissa mean here? $\endgroup$ – user19405892 Feb 8 '16 at 3:22
  • $\begingroup$ It follows from the solution –abscissa is just curvilinear distance from initial point. They approximate $\pi$ by 3.14 $\endgroup$ – HEKTO Feb 8 '16 at 5:34
  • $\begingroup$ @HEKTO Are they even using the word abscissa correctly? $\endgroup$ – user19405892 Feb 8 '16 at 22:50
  • $\begingroup$ From Wikipedia (en.wikipedia.org/wiki/Abscissa): "In a somewhat obsolete variant usage, the abscissa of a point may also refer to any number that describes the point's location along some path, e.g. the parameter of a parametric equation." $\endgroup$ – HEKTO Feb 9 '16 at 1:50
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Starting with $p_0:=(1,0)$ we are marking points $p_k$ $(k\geq0)$ on the unit circle such that the signed angle between successive points is $1$. We stop when we get a mark $p_n$ such that the unsigned angle $\angle(p_n,p_k)$ between $p_n$ and some $p_k$ with $1\leq k<n$ is $<{1\over5}$. Since $$\angle(p_n,p_k)=\angle(p_{n-1},p_{k-1})\qquad(k>0)$$ we conclude that at the moment of stopping we necessarily have $\angle(p_n,p_0)<{1\over5}$. This is saying that $$|n-2r\pi|<{1\over5}\tag{1}$$ for a certain $r\geq1$. As $$2\pi\doteq6.28,\quad 4\pi\doteq12.57,\quad6\pi\doteq18.85$$ we see that $(1)$ is fulfilled for the first time when $r=3$ and $n=19$. So there are $20$ marks in all when we stop.

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  • $\begingroup$ Why does the question mention curved distances greater than $1$ if we are only going $1$ distance? Also what does the abscissa have to do with it? $\endgroup$ – user19405892 Feb 8 '16 at 16:15
  • $\begingroup$ My understanding of the problem is described in my answer. The problem as quoted is extremely badly formulated, maybe it has been translated by babelfish from a remote language. – If you are interpreting the problem differently, disregard my answer. At any rate there is no such thing as a "curved distance". $\endgroup$ – Christian Blatter Feb 8 '16 at 18:45
  • $\begingroup$ My question is how do you know the signed curved distance always goes up by $1$ if the question says $0,1,2,\ldots$ and what does abscissa mean here? I haven't heard that word out of context of Dirchlet series or algebra. What does it mean here? Also, you may be right about the poor translation. This question comes from the $1968$ IMO. $\endgroup$ – user19405892 Feb 8 '16 at 21:29

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