Question

Let x be a point inside a convex quadrilateral q. Find the probability that x is neither on the boundary nor inside any of the circles drawn with the sides of the quadrilateral q as their diameters.

Answer 1

Here is a different approach to the problem. Consider a random triangle illustrated below. Regions are labelled near (NiNi) and far (FiFi) from point ii. TT is the triangle.

If the 4th point falls in regions F1F1, F2F2, or F3F3 a convex quadrilateral will result. The probability that 4 points produce a convex shape is equivalent to the probability that the 4th point will be in one of those regions. That is, p=E[F1+F2+F3]p=E[F1+F2+F3]where FiFi is the random variable expressing the area of that region.

To simplify the calculations, consider the shaded region labelled A1A1 and the counterparts A2A2 and A3A3.

F1+N2+N3=A1F1+N2+N3=A1

N1+F2+F3+T=1−A1N1+F2+F3+T=1−A1

Combining these one finds,

F1+F2+F3=2−2T−A1−A2−A3F1+F2+F3=2−2T−A1−A2−A3

and therefore,

p=E[F1+F2+F3]=2−2E[T]−3E[A]p=E[F1+F2+F3]=2−2E[T]−3E[A]

Fortunately others have worked out E[T]=11144E[T]=11144

To work out E[A]E[A], I considered separately the case where the unit square is bisected across opposite sides (as is the case for A1A1). You can show that this occurs with probability 2323, given two uniform random points. The smallest of the two areas AminAmin follows the distribution

f(amin)=4aminf(amin)=4amin

for 0<amin<120<amin<12. Given the two points (2,3)(2,3) point 11 point will fall in the smaller of the two areas with probability aminamin, in which case A1=1−aminA1=1−amin.

E[Aopp]=16∫120a2min(1−amin)damin=512E[Aopp]=16∫012amin2(1−amin)damin=512

For the case where the unit square is cut across a corner (as is the case for A2A2), Amin=12X0Y0Amin=12X0Y0 where X0X0 and Y0Y0 are independent random variables with the identical distribution, f(x)=2xf(x)=2x for 0<x<10<x<1. This gives the distribution for AminAmin,

f(amin)=−16aminlog2aminf(amin)=−16aminlog⁡2amin

for 0<amin<120<amin<12.

E[Acorner]=−32∫120a2min(1−amin)log2amindamin=2372E[Acorner]=−32∫012amin2(1−amin)log⁡2amindamin=2372

And so, overall

E[A]=P(Aopp)E[Aopp]+P(Acorner)E[Acorner]=23512+132372=83216E[A]=P(Aopp)E[Aopp]+P(Acorner)E[Acorner]=23512+132372=83216

Finally,

p=2−211144−383216=2536=(56)2=0.694¯p=2−211144−383216=2536=(56)2=0.694¯

which is close to your simulation result. A very simple form!