Comment by bmacho
19 days ago
I like this reasoning. Define a probability distribution on all circles of (x,y,r>0) based on how likely a given circle is. Then we can just sum the good circles and all the circles.
And the probability distribution is simple: a given (x,y,r) is as likely as its circumference in the unit circle.
Reasoning: Let C:(x,y,r) a given circle. We want to know how likely is it that the circle on 3 random points are close to it, closer than a given value d. (A d wide ball or cube around C in (x,y,r) space. Different shapes lead to diffferent constants but same for every circle.) The set of good 3 points is more or less the same as the set of 3 points from the point set C(d): make C's circumference d thick, and pick the 3 points from this set. Now not any 3 points will suffice, but we can hope that the error goes to 0 as d goes to 0 and there is no systematic error.
Then we just have to integrate.
ChatGPT got me the result 2/3, so it's incorrect. I guess the circumference must not be the right distribution.
Maybe it should be the cube of the amount of circumference inside the unit circle, because 3 random points have to land on the circle for it to be chosen.
I think that's correct.
But that wouldn't work either. The problem is that there is an infinite volume of circles with ~2 as the circumference inside the unit circle. (Volume in x,y,r space, and, consequently in the proposed probability distribution as well.) So this proposed probability distribution gives infinity, when summed up.