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I need to proof that given a regular compact surface $S$ in $\mathbb{R}^3$ there exists a line in $\mathbb{R}^3$ which intersects perpendicularly with $S$ twice. Could you help me?

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    $\begingroup$ I don't have a proof, but that looks related to Borsuk-Ulam's theorem... $\endgroup$ – emeu Jun 24 '15 at 12:31
  • $\begingroup$ a first try (I am not sure if this works): consider the map $\mathbb S^2 \rightarrow \mathbb R^2$ which sends $(a,b,c)$ to the projection of $argmax_S (a*x+b*y+c*z) - argmin_S(a*x+b*y+c*z)$ on the hyperplane defined by $a*x+b*y+c*z$. This is an odd function so if it is continuous, then it has a zero by Borsuk-Ulam which corresponds to the line we want. The problem is that it is not clear that it is continous (if the surface is the border of a convex body then I think that it is continous, but in the general case I do not know). $\endgroup$ – emeu Jun 24 '15 at 13:00
  • $\begingroup$ What do you mean with $argmax_S(a∗x+b∗y+c∗z)$? $\endgroup$ – Damaru Jun 24 '15 at 13:56
  • $\begingroup$ I meant the point $(x,y,z)\in S$ which maximizes the linear form (if uniquely defined, so this indeed requires extra assumptions on S). Thanks for the very nice problem and thanks Ted Shifrin for the very nice solution ! $\endgroup$ – emeu Jun 24 '15 at 20:55
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Nice question! If you fix $x\in S$ and find the point $y\in S$ farthest from $x$, you can check easily that the chord joining $x$ and $y$ must be normal to $S$ at $y$. By symmetry, then, if we take any pair of points $x,y\in S$ with maximal distance (which exists by compactness), the chord joining $x$ and $y$ must be normal to $S$ at both points.

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  • $\begingroup$ What a beautiful solution! Thank you very much! $\endgroup$ – Damaru Jun 24 '15 at 16:00
  • $\begingroup$ You're most welcome. $\endgroup$ – Ted Shifrin Jun 24 '15 at 16:02

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