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I am looking for proofs of or counterexamples to the following assumptions:

if the corners of a triangle are chosen from a compact subset $\mathcal{S}$ of the Euclidean plane then all three corners of the following triangles are elements of the convex hull $CH(\mathcal{S})$ of $\mathcal{S}$

  • the triangles with corners $A,B,C\ \in\ \mathcal{S}$,$\ $ that have the largest incircle

  • the triangles with corners $A,B,C\ \in\ \mathcal{S}$,$\ $ that maximize $a+b-c$, where $\left(c:=\|B-A\|\right) \ge \left(b:=\|A-C\|\right)\ \ge\ \left(a:=\|C-B\|\right)$

My motivation is to identify "fat" triangles, for which it is guaranteed, that their corners are elements of the convex hull of some finite planar point sets.

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1) No. Consider the square $ABCD$. Then the triangle $ABC$ does not have the largest incircle. Indeed, it has smaller incircle than $ABM$, where $M$ is the midpoint of $CD$. This follows from the formula $r(\Delta)=S(\Delta)/p(\Delta)$, where $r,S,p$ denote the inradius, area and semi-perimeter of a triangle $\Delta$. Perimeter of $\triangle ABM$ is smaller than that of $\triangle ABC$, it is seen from considering the point $K$ symmetric to $B$ against $C$: $$2p(ABM)=AB+AM+BM=AB+AM+MK=AB+AK<AB+AC+CK=2p(ABC).$$

2) No, the same counterexample works.

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  • $\begingroup$ why is the answer "no"? In your example all three points are on the convex hull, albeit not a corner; using however the square and a 5th point inside, whose distance to your $M$ is $\epsilon$ provides a valid counter example. I will accept your answer as it provided the right idea. $\endgroup$
    – Manfred Weis
    Commented Sep 22, 2017 at 9:31
  • $\begingroup$ Ah, I was thinking about these points being extreme points. You are correct, of course. $\endgroup$
    – Fedor Petrov
    Commented Sep 22, 2017 at 10:11
  • $\begingroup$ The valid counter example mentioned in my comment is of course only valid, if the corners of triangles are restricted to be elements of non-convex $\mathcal{S}$; if the corners may be any elements of $CH(\mathcal{S})$ then the assumptions are true, as can be seen by the possibility of otherwise finding a larger triangle, that contains the candidate triangle and has larger inner circle or detour. $\endgroup$
    – Manfred Weis
    Commented Sep 22, 2017 at 11:18

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