Expected area of the intersection of of triangles made up random points inside a circle, all the triangles must contain the origin How to find the expected area of the intersection of a set of triangles made up $N$ random points that are picked uniformly inside a circle? The triangles must contain the origin of the circle. If none of the trinagles contain the rigin we pick again $N$ random points until at least one triangle contain the origin and calculate the expected intersection area of all the triangles that contain the origin.
 A: Here's the answer for $N=3$, along with some thoughts on the general problem.
By ordering the points by angle and arbitrarily fixing the angle of the first point at $0$, the expected area of a triangle containing the origin can be obtained as
$$
\frac{\displaystyle3\int\mathrm d\triangle\frac12r_1r_2\sin\phi_2}{\displaystyle\int\mathrm d\triangle}\;,
$$
where the integrand in the numerator is the area of the triangle formed by the origin and the first two points, whose integral is by symmetry equal to that of the areas of the triangles formed by the origin with the other two pairs of points (hence the factor $3$), and where
$$
\int\mathrm d\triangle=
\int_0^\pi\mathrm d\phi_2\int_{\pi}^{\phi_2+\pi}\mathrm d\phi_3\int_0^1\mathrm dr_1r_1\int_0^1\mathrm dr_2r_2\int_0^1\mathrm dr_3r_3
$$
integrates over the polar coordinates for which the origin is contained in the triangle. The radial integrations yield $\left(\frac13\right)^2\cdot\frac12$ in the numerator and $\left(\frac12\right)^3$ in the denominator, so this is
$$
\begin{align}
&
\frac23\frac{\displaystyle\int_0^\pi\mathrm d\phi_2\int_{\pi}^{\phi_2+\pi}\mathrm d\phi_3\sin\phi_2}{\displaystyle\int_0^\pi\mathrm d\phi_2\int_{\pi}^{\phi_2+\pi}\mathrm d\phi_3}
\\
=&
\frac23\frac{\displaystyle\int_0^\pi\mathrm d\phi_2\phi_2\sin\phi_2}{\displaystyle\int_0^\pi\mathrm d\phi_2\phi_2}
\\
=&
\frac23\frac\pi{\frac12\pi^2}
\\
=&
\frac4{3\pi}
\approx
0.4244\;.
\end{align}
$$
This is to be compared to the average area $\dfrac{35}{48\pi}\approx0.2321$ without the condition that the triangle contains the origin (which is slightly more difficult to calculate).
The problem for general $N$ can be reformulated as finding the probability that none of the segments joining $N$ random points whose convex hull contains the origin lies between a further random point and the origin. What makes this seem difficult is the fact that the events for the various segments aren't independent.
Numerical simulations indicate that the answer for $N=4$ might be $\dfrac9{10\pi}$.
