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For any compact set on a plane say C does there always exist a chord in C such that its end points are orthogonal to the boundary of C (assumed smooth)

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  • $\begingroup$ You should rule out the trivial counterexample of a finite set of points. $\endgroup$ Jul 27 '13 at 7:32
  • $\begingroup$ I talk about a continuous boundary. $\endgroup$
    – ARi
    Jul 27 '13 at 7:34
  • $\begingroup$ Also you should rule out things like $C$ being a circle without its interior (since then none of the desired chords "are in $C$), or rephrase "a chord in $C$" to maybe a chord between two points on the boundary of $C$. $\endgroup$
    – coffeemath
    Jul 27 '13 at 7:34
  • $\begingroup$ Does "compact set" on a plane satisfy these conditions. I dont know? $\endgroup$
    – ARi
    Jul 27 '13 at 7:43
  • $\begingroup$ A circle without its interior is a compact set. Maybe to rule such things out assume $C$ is a convex region with its boundary given by a smooth closed curve. That way all the chords joining two points of the boundary will lie inside $C$. Actually, since your whole question is only about points on the boundary, and orthogonality of chords joining such points, you could simply phrase the question in terms of smooth closed curves (ignoring inner regions if any). $\endgroup$
    – coffeemath
    Jul 27 '13 at 7:52
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Consider two points $p,q$ on the boundary at a maximal distance from each other. If the chord $pq$ is not orthogonal to the boundary, say at $p$, then one could move $p$ along the curve in one of the two directions, obtaining a new point $p'$ so that $d(p',q)>d(p,q)$.

In a comment, Neal has asked what guarantees the existence of points $p,q$ on the boundary at maximal distance from each other. First notation: Let $A$ be the set in question, whose boundary is say the closed smooth curve $\gamma$. Since $A$ is compact it has a diameter, and so there are points $p_1,q_1$ for which $d(p_1,q_1)=d$ where $d$ is the diameter of $A$. For this result see

diameter on a compact metric space

Now we claim that in fact $p_1$ lies on the boundary of $A$. For if not, it lies in the interior of $A$ and so lies in some disc $D$ contained in $A$. But then we could extend the segment $p_1p_2$ a bit into $D$, increasing its length, against $d(p_1,q_1)=d$. Similarly we can show that $q_1$ must lie on the boundary. Conclusion: with $p=p_1,q=q_1$ we have two points on the boundary at maximal distance from each other, and in addition achieving the diameter of the whole set $A$.

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  • $\begingroup$ @zyx A smooth boundary is by definition one with a parametrization with continuous nonzero velocity. Certainly a triangle is not considered to have a smooth boundary, even though one can parametrize it via a pair $(x(t),y(t))$ of functions with continuous second derivatives. Saying the boundary is smooth means it must be smooth as a 1-manifold. $\endgroup$
    – coffeemath
    Jul 27 '13 at 8:44
  • $\begingroup$ So the chord becomes orthogonal at P' but loses its orthogonality at q $\endgroup$
    – ARi
    Jul 27 '13 at 8:51
  • $\begingroup$ @Ari I think you missed the point of the argument. It showed that, assuming the chord was not orthogonal to the curve at $p$, then there is another point $p'$ with $d(p',q)>d(p,q)$. This is a contradiction to the definition of $p,q$ as two boundary points for which $d(p,q)$ is a maximum. Therefore chord $pq$ must be orthogonal to the curve at $p$, and an exactly similar argument shows it must also be orthogonal to the curve at $q$. $\endgroup$
    – coffeemath
    Jul 27 '13 at 8:58
  • $\begingroup$ I get it now ; so you prove that for C to be a closed curve there has to be at least one orthogonal chord in C, correct?. $\endgroup$
    – ARi
    Jul 27 '13 at 9:06
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    $\begingroup$ @Neal -- See the add-on to the answer (too long to place in comment). $\endgroup$
    – coffeemath
    Jul 27 '13 at 20:01
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The maximum distance argument is the essence of the matter (see answer by coffeemath) and I want to isolate here the minimal assumption needed for it to work, without imposing regularity conditions or topological restrictions on the boundary. The compact set could be the Hawaiian Earring space, or an earring with uncountably many circles, or the boundary could be a figure $8$, but the proof continues to work if...

  1. (infinitely near rays extend to infinitely near lines) for every boundary point $p$ of the boundary, if there is a sequence of points of the boundary $q_n$ converging to $p$ and asymptotic to a vector $v$ based at that point (so that the angle between $v$ and the vector from $p$ to $q_n$ goes to zero) then there are points of the boundary $r_n$ converging to $p$ and asymptotic to the vector opposite to $v$, AND
  2. (all points have infinitely near tangent-like lines) Every point on the boundary has such a sequence for at least one nonzero vector $v$.

This is true if the boundary is a union of differentiable curves with nonzero velocity vector (locally near any point, the curves are "branches", each one enclosing the point in an open smooth arc), which covers most cases. The proof works in any number of dimensions, or in a Riemannian manifold if you understand $v$ to live in the tangent space at $p$.

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  • $\begingroup$ How does this work for $p$ the vertex of say an equilateral triangle? Take $q_n \to p$ from one side, so the vector from $p$ to $q_n$ is constantly $v$ based at $p$ and going along the chosen side. But the vector opposite to $v$ points in the wrong direction for construction of the points $r_n$. (Of course in this triangle case, the meaning of orthogonal to the boundary has problems at the vertices anyway.) $\endgroup$
    – coffeemath
    Jul 27 '13 at 19:34
  • $\begingroup$ The conditions are violated at the vertex. The theorem is "if conditions hold at all points of the boundary, there is an orthogonal chord". $\endgroup$
    – zyx
    Jul 27 '13 at 19:50
  • $\begingroup$ @coffeemath, some clarifying material added. Maybe condition 2 was confusing: that it holds at every point is a constraint required for the proof, but the initial phrasing might have made it sound like it is claimed that condition 1 implies that 2 holds everywhere. $\endgroup$
    – zyx
    Jul 27 '13 at 23:42

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