Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
add comment

2 Answers

up vote 17 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.

share|improve this answer
add comment

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 $O$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.