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Anonymous01/27/26, 22:36No.16900433
You choose three uniformly random points inside of a circle. Then draw another circle through the three points. What is the probability that the circles won't intersect?
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Replies
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Anon01/27/26, 22:58No.16900440
50%. It either does or it doesn't.
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Anon01/27/26, 23:39No.16900452
0.4this figure illustrates the non intersection case geometrically. the outer curve is the original unit circle centered at the origin, the three marked points are a random triangle chosen inside it, and the inner curve is the circumcircle determined by those three points. the dot at the center of the inner circle is the circumcenter u.the non intersection condition shown is
[math]|OU|+R\le1[/math]
where R is the circumradius. for points a=(a_x,a_y), b=(b_x,b_y), c=(c_x,c_y), define
[math]D=2\big(a_x(b_y-c_y)+b_x(c_y-a_y)+c_x(a_y-b_y)\big)[/math]
then the circumcenter coordinates are
[eqn]
u_x=\frac{(a_x^2+a_y^2)(b_y-c_y)+(b_x^2+b_y^2)(c_y-a_y)+(c_x^2+c_y^2)(a_y-b_y)}{D}
[/eqn]
[eqn]
u_y=\frac{(a_x^2+a_y^2)(c_x-b_x)+(b_x^2+b_y^2)(a_x-c_x)+(c_x^2+c_y^2)(b_x-a_x)}{D}
[/eqn]
and the circumradius is
[math]R=|u-a|[/math]this configuration contributes to the roughly forty percent probability because a large fraction of random triangles still satisfy [math]|OU|+R\le1[/math], meaning their circumcircles sit fully inside the original circle and therefore do not intersect it.
