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Suppose J is a Jordan curve in the Euclidean plane E and that there are two perpendicular straight lines in E, each of which is an axis of symmetry of J. Does the intersection point of these two lines always necessarily lie in the interior of J?

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Here is a more general fact: if $J$ is central-symmetric w.r.t. a point $O$ (in our case, this is the intersection of the axes), then $O$ is inside.

Indeed, let $D$ be the domain bounded by $J$. We know it's a topological disc. And it's central symmetric w.r.t. $O$. Now if $O$ is outside, then the symmetry is a map from $D$ to itself without fixed points, contrary to Brouwer fixed point theorem.

If $O$ is on the boundary (i.e. on $J$), the the central symmetry is an involution of $J$ with exactly one fixed point. But there are no such involutions of the circle.

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Yes. Suppose without loss of generality that the two lines are the x- and y-axes. Some point of J lies in some quadrant, so corresponding points a, b, c, d of J exist in every quadrant by symmetry. Suppose a, b are the points in the northeast and southeast quadrants, respectively. Then the portion of J passing between a and b cannot intersect the y-axis (or else, by symmetry, its reflection across the y-axis intersects itself), so it is homotopic to the straight line between a and b in in a half plane $x > \epsilon$ for some $\epsilon > 0$, hence homotopic to the straight line between a and b in $\mathbb{R}^2 - \{ (0, 0) \}$. Similar statements are true of every other pair of adjacent points, so J is homotopic to a rectangle in $\mathbb{R}^2 - \{ (0, 0) \}$ containing the origin.

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Thanks for your answers which have helped me better understand the reply to my previous question about a property that characterizes the circle among all plane Jordan curves. –  Garabed Gulbenkian Apr 24 '10 at 20:56
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