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

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up vote 18 down vote accepted

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.

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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 $O$.

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Thanks a lot for your answers which I was not quite able to figure out for myself. – Garabed Gulbenkian Apr 20 '10 at 20:50

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