Motivated by the discussion in The expected area of a triangle formed by three points randomly chosen from the unit square I tried to find an expression for the expected area of a triangle formed by three randomly chosen points inside the unit circle. I could not find one. Perhaps this is one of those problems which look easy but are difficult to solve analytically.

  • $\begingroup$ It doesn't look easy to me! $\endgroup$ – TonyK Apr 20 '15 at 9:45
  • $\begingroup$ @TonyK: central symmetry is easily exploitable! $\endgroup$ – Jack D'Aurizio Apr 20 '15 at 9:56
  • $\begingroup$ So what is the distribution for selecting the points? These kinds of questions can often be ill defined if not being careful about defining "random". $\endgroup$ – mathreadler Apr 23 '15 at 15:41

Preliminary remark: @Jack D'Aurizio 's answer left me unconvinced. Therefore I ran a few simulations. They revealed that the expected value is about $0.23$, which is definitely larger than Aurizio's value ${2\over3\pi}=0.212$. On the other hand I found out that the problem had been treated in a 1996 paper by B. L. Burrows and R. F. Talbot. Since their paper resides behind a paywall I decided to solve the problem anew from scratch.

We start with the simpler situation that the vertex $C$ of $T$ lies on the rim of the disc $D$. We momentarily put the center of $D$ at $(0,1)$ and fix the vertex $C$ at the origin. The random vertex $A\in D$ is then given by $A=(r\cos\phi,r\sin\phi)$, whereby $$0\leq r\leq 2\sin\phi,\quad 0\leq \phi\leq\pi,\qquad{\rm d}(x,y)=r\>{\rm d}(r,\phi)\ .$$ Let $B=(s\cos\psi,s\sin\psi)$ be the second random vertex, whereby we restrict to $\phi\leq\psi\leq\pi$ and multiply by $2$ at the end. We then get the following expression for the expected area of $T$: $$E_1={2\over\pi^2}\int_0^\pi d\phi\>\int_\phi^\pi d\psi\>\int_0^{2\sin\phi}dr\>\int_0^{2\sin\psi} ds\left({1\over2}rs\sin(\psi-\phi)\>r\>s\right)\ .$$ Integrating first with respect to $r$ and $s$ we obtain $$E_1={1\over\pi^2}\int_0^\pi \int_\phi^\pi {64\over9} \sin(\psi-\phi)\sin^3\phi\sin^3\psi\>d\psi d\phi ={35\over36\pi}\ .$$ We now move $D$ back to the origin and put $Q:=\max\bigl\{|A|, |B|, |C|\bigr\}$. The expected area of $T$, conditioned on the value $q$ of $Q$, is then given by $$E_q={35\over36\pi}q^2\ .$$ The probability density of each of the three radii $|A|$, $|B|$, and $|C|$ is given by $f_R(x)=2x$. It is then an easy exercice to prove that the probability density of $Q$ is $f_Q(q)=3q^2\cdot2q=6q^5$. Therefore the expectation of $Q^2$ comes to $$E(Q^2)=\int_0^1 q^2\cdot 6q^5\>dq={3\over4}\ ,$$ and the final result is $$E={35\over36\pi}E(Q^2)={35\over48\pi}=0.232101\ .$$

  • $\begingroup$ (+1) Really nice. So, where my argument is flawed? $\endgroup$ – Jack D'Aurizio Apr 23 '15 at 9:07
  • $\begingroup$ @Jack D'Aurizio: I guess in the handling of the conditional expectations. $\endgroup$ – Christian Blatter Apr 23 '15 at 9:14
  • $\begingroup$ "Since their paper resides behind a paywall I decided to solve the problem anew from scratch." See? At least one advantage to promote closed access! $\endgroup$ – Jill-Jênn Vie Jun 14 at 19:13

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