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Consider a convex body $\Omega\subset \mathbb{R}^2$. Let $L(d)$ be the maximum over all curves $C$ of degree $d$ of the length of $C\cap\Omega$.

Is $L(d)\leq d P(E)/2$, where $P(E)$ is the perimeter of the longest ellipse inscribed in $\Omega$?

It is clear that $L(d)\geq dP(E)/2$ if $d$ is even. If $\Omega$ is a ball, then the matching upper bound follows from Buffon's noodle argument. In general, one has $L(d)\leq dP(\Omega)/2$, where $P(\Omega)$ is the perimeter of $\Omega$.

Is the answer for $\Omega=[-1,+1]^2$ really the same as $\Omega=B(0,1)$? Another attractive question is for equilateral triangle (see comments of Noam Elkies below).


In the answer below, jacob hints on an example that is likely to be a counterexample to the question above, namely, convex hull of the graph of $\{Tx^2 : x \in [-1,+1]\}$. Perhaps the right question is whether $L(d)$ is maximized by a quadratic curve.

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Do you mean a compact convex body? –  David Roberts Dec 17 '13 at 2:37
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Yes, in the above convex body means a compact convex set with non-empty interior. –  Boris Bukh Dec 17 '13 at 2:49
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I guess you mean for $\Omega$ to be the square $[-1,+1]^2$, not the line segment $[-1,+1]$. An equilateral triangle circumscribing the same circle seems like an even more tempting target (because an inscribed conic touches it at only three points, not four), at least if that circle is still the longest ellipse inside the triangle $-$ or are you allowing only centrally symmetric "bodies"? –  Noam D. Elkies Dec 17 '13 at 6:58
    
(I see that the inscribed circle is in fact not the longest ellipse $-$ an ellipse approaching a side of the triangle has length $4 \sqrt 3 > 2 \pi$ $-$ but the question still stands.) –  Noam D. Elkies Dec 17 '13 at 7:03
    
Well, I think that for the square longest ellipse inside is its incircle. Possibly, I may even prove it, though the proof is too long and technical for the comment. –  Fedor Petrov Dec 17 '13 at 17:11

1 Answer 1

I believe the answer is no. The example I have in mind is a rectangle ABCD where $|AB|=|CD|=1$ and $|BC|=|AD| = T$ where $T$ is some large parameter.

Now, I haven't calculated the optimal ellipse in here, but the 2 extreme ones seem to be the diagonal with multiplicity 2, and the stretched circle (i.e. the ellipse with major axis lengths 1 and $T$). Both of these have length $2T+O(T^{-1})$.

EDIT: This is false it turns out. The stretched circle has slightly higher length, so this might not work

Now consider a parabola whose vertex is at the midpoit of $AB$ and which passes through $C$ and $D$. This is just $y=4Tx^2$ for $\frac{-1}2\leq x\leq \frac12$ . Half of its length is $$\int_{t=0}^{\frac12} \sqrt{1+64T^2t^2}dt=(8T)^{-1}\int_{t=0}^{4T}\sqrt{1+t^2}dt$$ and thanks to Wolfram alpha, I'm not too lazy to compute this:

$$(16T)^{-1}\left(4T\sqrt{16T^2+1}+\sinh^{-1}(4T)\right)=T+\frac{\sinh^{-1}(4T)}{16T}+O(T^{-1})$$

While $\sinh^{-1}(4T)$ grows very slowly (logarithmically) it does got to infinty. So the Parabola seems to do better in this case.

Proof by "the other ellipses probably don't do better!"

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