How do I proove that a regular octagon has a centre?

By centre i mean a point equidistant from all the vertices?

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  • $\begingroup$ What's your definition of a regular octagon? $\endgroup$ – Jyrki Lahtonen Dec 22 '18 at 12:38
  • $\begingroup$ I think the definition is given in drawing: all sides and all angles are equal. $\endgroup$ – user Dec 22 '18 at 12:44
  • $\begingroup$ How do you know the octagon constructed in this way closes up? $\endgroup$ – Christian Blatter Dec 22 '18 at 18:48
  • $\begingroup$ @christianblatter thanks, that is a very good question, I did not think of this. $\endgroup$ – Rasmus Larsen Dec 23 '18 at 6:34

Take a line segment from one vertex to its opposite vertex. Bisect that segment. Clearly this point is equidistant from those two vertices.

Now rotate the octagon about that point, placing two new vertices onto the ends of the line segment. Rotations don't change distance; so those two vertices must be equidistant from the centre too.


  • $\begingroup$ I'm a bit uncomfortable with your argument. You seem to assume that the prescribed rotation exists. I rather think that you are to prove its existence! Of course, the OP should specify their definition of a regular octagon. $\endgroup$ – Jyrki Lahtonen Dec 22 '18 at 12:37
  • $\begingroup$ I was really internally assuming that $D_{16}$ can act on the thing, yes. $\endgroup$ – Patrick Stevens Dec 22 '18 at 12:41
  • $\begingroup$ @PatrickStevens Thanks. What does D_{16} refer to? Some kind classification of geometric functions? $\endgroup$ – Rasmus Larsen Jan 3 '19 at 9:33
  • 1
    $\begingroup$ @RasmusLarsen The dihedral group of order $16$. $\endgroup$ – Patrick Stevens Jan 3 '19 at 18:47

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