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This question already has an answer here:

Consider a unit circle C. The goal is to find a curve L such that:

  1. all secant lines of C intersect L;
  2. the length of L is minimal among those with property 1 above.

Any closed curve containing C (for example a circle with the same center as C, but larger) clearly satisfies property 1, but is not minimal.

The curve L does not need to be connected.

A proof of minimality of the length of L is required.

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marked as duplicate by Oray, Glorfindel, Community♦ Feb 1 '18 at 7:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ If L skips even a tiny epsilon-region around a point on C, we can slice a chord through there (almost a tangent, but still a secant), so isn't it obvious that L coincides with C? I must be missing something... $\endgroup$ – ngn Jan 31 '18 at 22:11
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    $\begingroup$ This seems like a question for Mathematics $\endgroup$ – Kieran Moynihan Jan 31 '18 at 22:49
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    $\begingroup$ @KieranMoynihan I was in doubt whether to post it here because of its mathematical flavor, but I've seen quite tough math problems on this site $\endgroup$ – Maiaux Jan 31 '18 at 23:01
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    $\begingroup$ @ngn That was my first thought too, but a "plus sign" centered around the center of C would intersect all secants providing each arm was sqrt(2)*r units long. This would have a total length of 4*sqrt(2)*r units, which is better than 2*pi*r. $\endgroup$ – DqwertyC Jan 31 '18 at 23:02
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    $\begingroup$ So it seems that (1) this is an open problem, (2) the best known answers aren't particularly elegant, and (3) most likely you can always do better by adding more separate components. All in all, this is looking less and less like a puzzle and more and more like a highly nontrivial mathematical problem... $\endgroup$ – Gareth McCaughan♦ Feb 1 '18 at 7:43
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No proof, but another example of a solution better than the full circle:

enter image description here

The total length is $\pi + 2 \approx 5.1416$.

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    $\begingroup$ If you cut along the dotted line, you can shorten one of the "horns" by moving the lower end point a bit more towards the center: it only needs to reach the line between the other horn's tip and the cutting point. $\endgroup$ – Bass Feb 1 '18 at 1:20
  • $\begingroup$ @Bass I don't get what you mean at all. $\endgroup$ – ffao Feb 1 '18 at 3:58
  • $\begingroup$ Separate the right side vertical portion. Keeping the upper end in the same place, rotate it a couple of degrees (30?) clockwise. Notice how a portion of the rotated piece becomes unnecessary. Cut the unnecessary part off for a better solution. $\endgroup$ – Bass Feb 1 '18 at 5:09
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Circumscribing a square and connecting the corners looks like a very promising solution. The shortest way to connect the corners of a square looks something like this, IIRC (pardon my art):

enter image description here

The total length of that thingy seems to be $2(1+\sqrt{3}) \approx 5.464$, which was handily provided by this answer.

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  • $\begingroup$ If I squint really hard, the angle "lower right corner" - "lower middle point" - "upper right corner" is 90 degrees. But it probably is a bit less than that. Therefore, the lower legs can be made shorter by aiming them a bit higher, without any secants escaping. Oh well, at least this serves as an upper bound. $\endgroup$ – Bass Feb 1 '18 at 0:31

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