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Given a circle $C$, and an infinite set $S$ of mutually disjoint ellipses which are inside and tangent to $C$, prove that there must exist a disk $D$ which lies inside $C$ but outside every ellipse.

It seems like there should be an elegant proof.

Note that if degenerate ellipses, i.e. line segments, are allowed, then the conclusion does not follow -- radial segments can be used that get arbitrarily close to the circle's center in every sector.

(This question was inspired by this one about filling the plane with parabolas.)

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Matt
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  • Excellent question, but what does it have to do with fractals? – Moishe Kohan Jul 09 '14 at 03:52
  • @studiosus: You can view the question like this: Is there a figure like the Apollonian gasket, but (1) using ellipses in the interior rather than circles, so that (2) every ellipse is tangent to the outer circle? The ellipses need not touch -- in fact, if any two (say X and Y) touch, you can fairly easily prove the result, since you can consider an ellipse Z reaching into a small disk very close to where X and Y touch, and there will be a region "under the cusp of Z" which cannot see the outer circle and therefore cannot be in any ellipse. That is why the question assumes disjointness. – Matt Jul 09 '14 at 05:49
  • Yes, that much I know. But unlike the AG, I do not see a way to iterate in your setting since there is no analogue of "inversion" in an ellipse. One has a birational transformation of the projective plane fixing ellipse, but not an honest map. However, maybe birational transforms are enough to generate a fractal. – Moishe Kohan Jul 09 '14 at 06:14
  • @studiosus: You could also think of the question like this: Consider a shape like a snowflake curve, but adding tangent ellipses to the original line rather than triangles. The thing to prove is that this cannot be done without leaving gaps between the ellipses. Of course there is no particular shape under discussion -- not an Apollonian gasket, nor a modified snowflake curve, nor any other, fractal or otherwise, because the problem is to prove that no such disk-filling arrangement of ellipses exists. The tag's purpose is to allow people who are interested in such concepts to see the question. – Matt Jul 09 '14 at 11:19

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This is hard to make rigorous, but if assume by contradiction that all the space inside $C$ is taken, then there are couples and triples of ellipses arbitrarily close. Assume that we have $3$ ellipses very close each other, like in figure: ellipses arrangement then they define a region $D$ with a concave boundary with the property that any entering ellipse cannot have a minor axis greater than the separation between the ellipses. Hence I would bet that the measure of the subset of $D$ taken by entering ellipses cannot exceed the sum of the areas of the three depicted trapezoids, and such a sum is stricly less than $\mu(D)$, hence in $D$ there must be a neighbourhood of a point that is not taken by any ellipse, as wanted.

However, this argument just gives an idea and a bet, and it is still far from being a proof.

Jack D'Aurizio
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  • Nice picture! If three ellipses are close enough so that (as in your diagram) there is no line separating one from the other two, then there is a small region D (much smaller than the one shown) near each "non-orthogonal" trapezoid side which cannot "see" any of C (i.e. any segment from a point in D to C must pass through one of the three ellipses). Therefore, no part of D can be inside any other ellipse E, since if it were, it would be able to "see" the point of tangency of E and C. So... there cannot be such a configuration of 3 ellipses. (Was there a reason to expect close triples?) – Matt Aug 18 '14 at 18:18
  • Oops, my previous comment ignored the possibility of ellipses coming in through the "far" gap! But the idea remains: If the three ellipses are close enough (as determined by considering the configuration of mutual internal tangent lines), then there will be a region D which cannot see C, and thus satisfies the problem statement. But I don't see why we should expect there to be any such triples of close ellipses. – Matt Aug 18 '14 at 19:07
  • Assume that all the space is took. Then you can find two ellipses arbitrarily close, say $\Gamma_1$ and $\Gamma_2$ - otherwise, you can "fill" the space between $\Gamma_1$ and $\Gamma_2$ with a circle. But for the same reason, if all the space is took, there exists $\Gamma_3$ arbitrarily close to $\Gamma_1$ and $\Gamma_2$ - otherwise, you can fill the gap between $\Gamma_1$ and $\Gamma_3$ or the space between $\Gamma_2$ and $\Gamma_3$ with a circle. So we can assume without loss of generality that the depicted configuration occurs. – Jack D'Aurizio Aug 18 '14 at 21:31
  • Yes, there must be some $\Gamma_1$ and $\Gamma_2$ that are as close as you want. And yes, there must be some $\Gamma_3$ as close to $\Gamma_2$ as you want. But there is no reason that any such $\Gamma_3$ would also be as close to $\Gamma_1$ as you want. Any gap between $\Gamma_3$ and $\Gamma_1$ can after all be filled by other ellipses. – Matt Aug 19 '14 at 02:33