most recent 30 from http://mathoverflow.net 2013-05-23T13:18:29Z http://mathoverflow.net/feeds/question/21961 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21961/can-the-circle-be-characterized-by-the-following-property Garabed Gulbenkian 2010-04-20T14:21:29Z 2010-04-20T17:18:20Z

<p>In the Euclidean plane, is the circle the only simple closed curve that has an axis of symmetry in every direction?</p>

http://mathoverflow.net/questions/21961/can-the-circle-be-characterized-by-the-following-property/21962#21962 Robin Chapman 2010-04-20T14:31:13Z 2010-04-20T14:31:13Z

<p>If a figure had an axis of symmetry in three non-parallel but non-concurrent axes, then composing these suitably would give a translative symmetry, which is impossible if the figure is bounded. So all the axes of symmetry of your putative curve are concurrent through a point $O$ which we shall call a centre. Then all rotations about the centre $O$ are symmetries. The only simple closed curves with this property are circles centred at $O$.</p>

http://mathoverflow.net/questions/21961/can-the-circle-be-characterized-by-the-following-property/21989#21989 Roland Bacher 2010-04-20T17:16:02Z 2010-04-20T17:16:02Z

<p>A slightly different argument is as follows. Choose two symmetries $\sigma,\tau$ with axes intersecting at a point $P$ and forming an angle of $2\pi \lambda$ with $\lambda$ irrational. The composition $\rho=\sigma\circ \tau$ is then a rotation of infinite order generating a dense subgroup of the group of all rotations centered at $P$. Any closed subset left invariant under $\rho$ is thus a union of concentric circles centered at $P$. A simple closed curve invariant under $\rho$ is thus such a circle.</p>