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This is the same question as this question except that the random triangles do not need to necessarily touch the circle and I am only interested in the distribution for the perimeter. And I am interested in the exact closed form for the PDF, if that is possible even though it would be complicated.

Here are 4 random examples:

These are 4 equivalent circles each with a different triangle in it.

For the case when the $r=1$, the maximum perimeter is still the same, namely $3\sqrt{3}$

When I ran $100000$ trials I got a distribution that looks like:

This is the estimated PDF from 1000 trials with the correct code

With the corrected code, the expected value is simply the same as what was mentioned in the comments, namely, $\frac{128}{15\pi}$. The empirical results matched it.

For all the related problems of finding the geometric probabilities, the pdf's have all involved the Beta distribution somehow. I don't know why.

The Mathematica code that I had used was:

 (* Clear all variables and global definitions *)
ClearAll["Global`*"]

data=Table[Perimeter[Polygon[RandomPoint[Disk[],3]]],{i,1,100000}];
Mean[data]
SmoothHistogram[data,Automatic,"PDF",Filling->Axis]
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    $\begingroup$ The way you get $p_3$ is curiously intricate. Is it supposed to be uniformly distributed within the circle? Shouldn't the answer be three times the length of a random segment within the circle? We souldn't bother with three segments. $\endgroup$
    – Claude Chaunier
    Commented Jul 5 at 23:41
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    $\begingroup$ Which requirements? Three points always make a triangle, don't they? $\endgroup$
    – Claude Chaunier
    Commented Jul 5 at 23:47
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    $\begingroup$ @TegLouis Three points always make a triangle if they are not collinear. With probability 1, three random points in the unit disk are not collinear. Also, by linearity of expectation, the expected value of the perimeter is equal to three times the expected length of a random segment. $\endgroup$
    – mathworker21
    Commented Jul 6 at 0:02
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    $\begingroup$ I believe the reason this doesn't match your simulation is that the answer there is for three points uniformly chosen in the circle, and your third point isn't uniform. $\endgroup$
    – Kevin P. Costello
    Commented Jul 6 at 0:31
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    $\begingroup$ I can't see what use can this have. Maybe it's a part of another problem you work on. But if it's just for fun then there is a more beautiful similar problem: to find the PDF of a random triangle area on lobachevsky plane $\endgroup$
    – Марат Рамазанов
    Commented Jul 7 at 18:40

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