# What is the probability that the triangle formed by three uniformly random points on the sphere contains its circumcentre?

In answering Conjecture: If $A,B,C$ are random points on a sphere, then $E\left(\frac{\text{Area}_{\triangle ABC}}{\text{Area}_{\bigcirc ABC}}\right)=\frac14$. it turned out that if you choose three points uniformly randomly on the sphere, the density distribution of their angles on the circle they form is proportional to the area of the triangle they form. So these points tend to be spaced further apart on the circle than points randomly uniformly chosen on the circle. (That makes sense, since on the sphere the probability distribution of the distance between two points vanishes at distance $$0$$.)

Since that seemed like a nice result, I wanted to do something else with it. A question often asked about random points on a circle is about the probability that the triangle they form contains the centre of the circle. If the points are uniformly distributed on the circle, that probability is $$\frac14$$ (see e.g. Probability that n points on a circle are in one semicircle). If the points are spaced further apart, this probability should be higher. So what is it for points uniformly randomly chosen on a sphere?

I’m posting this as a self-answered question, but I merely worked out the integrals and would be happy to see a more elegant solution that explains the simplicity of the answer with a symmetry argument.

As worked out for the other question, the joint density for the angles $$\phi_i$$ of the three points on the circle they form is proportional to the area of the triangle they form, which is proportional to $$|\sin\alpha+\sin\beta+\sin\gamma\,|$$, where $$\alpha=\varphi_2-\varphi_1$$, $$\beta=\varphi_3-\varphi_2$$ and $$\gamma=\varphi_1-\varphi_3$$. By symmetry, we can restrict to $$\phi_i$$ ordered clockwise (and thus $$\alpha$$, $$\beta$$, $$\gamma$$ positive) to get rid of the absolute value. Then the integral over that entire half of the space is

$$\begin{eqnarray*} 3\cdot2\pi\int_0^{2\pi}\int_0^{2\pi-\alpha}\sin\alpha\,\mathrm d\beta\,\mathrm d\alpha &=& 3\cdot2\pi\int_0^{2\pi}(2\pi-\alpha)\sin\alpha\,\mathrm d\alpha \\ &=& 3\cdot2\pi\cdot2\pi\;. \end{eqnarray*}$$

The integral over the region where the triangle doesn’t contain the centre of the circle has three disjoint parts, each where one of $$\alpha$$, $$\beta$$, $$\gamma$$ is reflex, and is thus

$$\begin{eqnarray*} 3\cdot2\pi\int_\pi^{2\pi}\int_0^{2\pi-\alpha}(\sin\alpha+\sin\beta+\sin\gamma)\,\mathrm d\beta\,\mathrm d\alpha &=& 3\cdot2\pi\int_\pi^{2\pi}\int_0^{2\pi-\alpha}(\sin\alpha+\sin\beta-\sin(\alpha+\beta))\,\mathrm d\beta\,\mathrm d\alpha \\ &=& 3\cdot2\pi\int_\pi^{2\pi}((2\pi-x)\sin x+2(1-\cos x))\,\mathrm d\alpha \\ &=& 3\cdot2\pi\cdot\pi\;. \end{eqnarray*}$$

(Note that the factor $$3$$ arises for different reasons: In the first case, it’s from the symmetry among the three summands in the density; in the second case, it’s from the symmetry among the three integration regions.)

Thus, the probability for the triangle not to contain the centre (and thus also for the triangle to contain the centre) is $$\frac12$$.

Here’s Java code that checks the result by simulation.

• I've asked another question about a geometrical probability of $1/2$, seeking an intuitive explanation.
– Dan
Jan 14 at 1:51

I haven't found a symmetry argument, but here is perhaps a simpler approach using integration.

The first chosen point, $$P$$, is fixed. The second point chosen, $$Q$$, is variable. The triangle will not contain its circumcentre (i.e. will be obtuse) if and only if the third chosen point is in the red or green regions shown below (the diagram shows a sphere, not a circle).

Let $$A$$ be the area of the red region. Let $$x$$ be the area of the blue region shown below (the blue region's axis of symmetry goes through $$P$$).

Assume that the surface area of the sphere is $$1$$. It is easy to show that $$A=\frac12(1-\sqrt x)$$. So the probability that the third point lies in the red region is $$\int_0^1A\mathrm dx=\frac16$$. The red and each green region have the same areas averaged over $$x$$, so the probability that the third point lies in the red or green regions is $$3\times\frac16=\frac12$$.

• Thanks, this is great (and I like the colours :-). It's perhaps worthwhile to mention that the reason that the red and each green region have the same average area is that they each correspond to one of the three angles of the triangle being obtuse; and that (since the area of a slice of a sphere is proportional to its height) $2x-1$ is proportional to the $z$ coordinate of $Q$. Jan 12 at 10:11