On 1/6/10, Dan Asimov <dasimov@earthlink.net> wrote:

Clarification:

I wrote:

<< In the plane:

Let C be a C^oo simple closed curve.

Let a "double-normal" be a line segment whose endpoints lie on C and which is normal to C at each of them.

C must have a double-normal. (Proof: Consider the longest segment from C to C).

Question: Let a "simple" double-normal be one that intersects C only at its endpoints. Must C have at least one simple double-normal? Prove or find a counterexample.

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Oops --- my ingenious device --- taking the shortest join instead of the longest --- depends on the curvature being everywhere negative. Back to the drawing board!

Veit's counterexample (of a smoothed {7/3} star polygon) is a brilliant example and maybe even the simplest one possible, if the question had not required that the closed curve be simple, i.e., non-self-intersecting.

I can't trace this reference, but note that the radiussing of the outermost vertices has to be kept quite close to them, otherwise the construction fails.

But it does, so the original puzzle is still open for solutions.

If there is an argument establishing this conjecture, and which breaks down only when there are (at least) 14 intersections, then it's going to be pretty hairy! WFL