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Let a circumference $K$ and $200$ points on $K$. Each point is at the end of a diameter corresponding to an integer number of degrees. Prove there are at least two points that are different ends of the same diameter.

There are $200$ points (pigeons $n=200$) and $180$ diameters (holes $m=180)$ :

$200/180 \approx 1.1 \therefore $ by the PHP there will be at least two points that are different ends of the same diameter

Is my proof correct? Is there another formal way to prove it?

B. David
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    Just to be clear. You are saying there are 180 potential diameters (a diameter goes from $k^{\circ}$ to $(k + 180^{\circ})$ fro $0 \le k < 180$). So there are 200 > 180 (not sure why you need 200/180 > 1 when 200 > 180 will suffice) holes. They can't all be on different diameters. Yes, you argument is good. – fleablood Sep 30 '17 at 22:35
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    You don't need an overly complicated or formal proof. All you need to say is that there are only $180$ diameters so the first $180$ points can be assigned to unique diameters. Then, the next $20$ points will have to be on diameters with a point already on them. You don't need to calculate $200/180$. – Jam Sep 30 '17 at 22:43

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