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This post adds a little to To find the longest circular arc that can lie inside a given convex polygon

A circular segment is formed by a chord of a circle and the line segment connecting its endpoints.

Is the following claim valid?

  • Assume that the longest connected subset of a circle (this subset could be an arc or a full circle) that can be drawn inside a planar convex region C has been found. The circular segment determined by the subset (this segment could be a full circle in some cases) is also the largest area circular segment lying inside C.

If the claim is invalid, one could ask for bounds on how much it is off.

A weaker version of the claim: the circular segment inscribed in C and with highest perimeter (perimeter of circular segment is sum of arc length and the chord between its endpoints; in extreme case, segment could be a full circle) is also the inscribed circular segment with max area.

Note added on March 3rd 2024: Numerical experiments - done following the suggestion from Pietro Majer (comment below) - have had difficulties with tolerance issues but whatever one could do clearly indicates that neither claim is valid in general even when C is restricted to be some triangle - indeed for most triangles (including the equilateral triangle), botch claims seem invalid. What is now not clear to me is whether there are 'special' triangles where either claim holds - ie for which the max area contained circular segment is the one determined by the longest contained arc (or circular segment with max perimeter).

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  • $\begingroup$ Is "The circular segment determined by the subset" the convex hull of the chosen arc? $\endgroup$
    – RavenclawPrefect
    Commented Feb 4 at 6:44
  • $\begingroup$ yes. that's right. $\endgroup$
    – Nandakumar R
    Commented Feb 4 at 6:52
  • $\begingroup$ The case of a triangle (especially a regular one) should be easy to treat for both problems. Do you know if your conjecture holds there? $\endgroup$
    – Pietro Majer
    Commented Feb 4 at 17:31
  • $\begingroup$ Thanks. shall try. just added a weaker version of the claim which is of course likelier to be true. $\endgroup$
    – Nandakumar R
    Commented Feb 5 at 10:56

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