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Consider a regular n-gon with side length $A$.

Let $p$ be a point in the polygon. Let the distances from $p$ to the corners of the n-gon be $x_1,x_2,...,x_n$

Are there solutions with $A,x_1,x_2,...x_n$ all positive integers and $gcd(A,x_1,x_2,...,x_n) = 1$.

For the triangle ( $n=3$) this question has been answered already here

Do there exist an infinite number of 'rational' points in the equilateral triangle $ABC$?

https://mathoverflow.net/questions/180191/rational-distance-from-vertices-of-an-equilateral-triangle

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I assume for sufficiently large $n$ there are no solutions ?

In particular im intrested in $n=4,5,6$.

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  • $\begingroup$ Another useful source of information: mathoverflow.net/questions/180191/… $\endgroup$
    – lulu
    Nov 11, 2015 at 12:33
  • $\begingroup$ Is $A$ rational? Should $ a,b,c $ be individually be rational ? Or should they all be positive integers? The Fermat point is fixed for the minimum sum. $\endgroup$
    – Narasimham
    Nov 11, 2015 at 12:47
  • $\begingroup$ Im considering a generalization to avoid the duplicate. But I have to read and think first. $\endgroup$
    – mick
    Nov 11, 2015 at 18:49
  • $\begingroup$ I edited too generalize. Therefore voted to reopen. Plz vote reopen too :) $\endgroup$
    – mick
    Nov 11, 2015 at 21:57

1 Answer 1

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Use Viviani's theorem that says $ x + y + z $ is invariant.

Compute $x=y=z$ for $A=1$.

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    $\begingroup$ Vivani theorem says that theant sum of distances to the sides is constant. $x,y,z$ are distances to the corners. $\endgroup$
    – Wojowu
    Nov 11, 2015 at 12:31
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    $\begingroup$ In Viviani's Theorem, the invariant is the sum of the distances to the sides, not the corners. No? $\endgroup$
    – lulu
    Nov 11, 2015 at 12:31
  • $\begingroup$ Hmm. So it is the distance to the middle of the sides ? Can this still be useful ? $\endgroup$
    – mick
    Nov 11, 2015 at 18:44
  • $\begingroup$ @mick Not the distance to the middle of the sides. It's a length of the shortest line segment connecting the point and the line segment. $\endgroup$
    – Wojowu
    Nov 11, 2015 at 18:54
  • $\begingroup$ Thanks Wojowu. I need to think. $\endgroup$
    – mick
    Nov 11, 2015 at 19:16

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