Probability that the orthographic projection of a randomly oriented regular tetrahedron is a triangle I wanted to find the probability that, given a uniformly sampled rotation matrix, when applied to a regular tetrahedron, its orthographic projection is a triangle (instead of a quadrilateral) when viewed from a predetermined "camera angle".
Here is my approach. First, some definitions and convenient choices:


*

*The regular tetrahedron has side length $a$ be centered at the origin.

*We are looking at it from the positive $z$ axis. Thus we are projecting the tetrahedron onto the $xy$-plane.

*Given a triangular projection, let the triangle's vertices be $A, B, C$. The fourth vertex of the tetrahedron is $D$.

*Assume for now WLOG that $D$ is obscured by the face $\triangle ABC$.


I want to calculate how much area on the tetrahedron's circumsphere can be reached by $D$ through rotation without $D$'s projection ever exiting the projection of $\triangle ABC$. Dividing that by the total surface area would give the probability.
To start, let us start with the tetrahedron being oriented such that $\triangle ABC$ is parallel to the $xy$-plane, and $AB$ (the edge) is parallel to the $x$-axis. Let $\theta = \phi = 0$ (spherical coordinates) at this orientation, as this describes the position of $D$. What I want to do, is for each value of $\theta$, where $0 \leq \theta \leq \pi/3$ (I chose this range since I believe the math to be symmetric enough to do on 1/6 of a full rotation), rotate the tetrahedron about the $z$-axis by $\theta$, then determine how much I can rotate the tetrahedron clockwise (as seen from positive $x$) about the $x$ axis without having vertex $D$ exit $\triangle ABC$. To help explain the next step, I will use the figures below. Figures 1 through 4 are viewed from the positive $z$-axis, Figure 5 is viewed from the positive $x$-axis.

See Figure 4 for what the rotation by $\theta$ mentioned earlier looks like, and Figures 1-3 for concrete examples. In Figure 4, there is a point $P$ on the edge $AB$, and line segment $PD$. See Figure 5 for a partial view of Figure 4 from the positive $x$-axis. If we are rotating clockwise about the $x$-axis, then we cannot rotate further once $PD$ becomes parallel to the $z$-axis. That is, when the projections of $P$ and $D$ would coincide. Thus, the value $\alpha$ in Figure 5 is the most we can rotate about the $x$-axis for this particular value of $\alpha$. In particular, the value of $\phi$ ranges from $0$ to $\alpha$.
To get $\alpha$ as a function of $\theta$, note that $|EF| = \sqrt{3}a/6$ is the inradius of $\triangle ABC$, and so $|PF| = |EF| \sec \theta = (\sqrt{3}a \sec \theta)/6$. As $|DF| = \sqrt{6}a/3$ is the height of the tetrahedron and $\tan \alpha = |PF|/|DF|$, we have $$\alpha(\theta) = \tan^{-1}\left(\frac{|PF|}{|DF|}\right) = \tan^{-1}\left(\frac{\frac{\sqrt{3}}{6}a \sec \theta}{\frac{\sqrt{6}}{3}a}\right) = \tan^{-1}\left(\frac{\sec \theta}{2 \sqrt{2}}\right).$$
We now prepare to integrate in spherical coordinates. The differential for the surface area of the sphere is $dA = R^2 \cos \theta \; d\phi \; d\theta$ where $R$ is the circumradius of the tetrahedron. Using this information, the area that $D$ can reach when $\theta \in [0, \pi/3]$ is $$A_{\pi/3} = \int_0^{\pi/3}\int_0^{\tan^{-1}\left(\frac{\sec \theta}{2 \sqrt{2}}\right)} R^2 \cos \theta \; d \phi \; d \theta.$$
The total area reachable for $\theta \in [0, 2 \pi)$ is $6$ times this value due to symmetry. But, we need to multiply by an additional factor of $2$ to account for the case when $D$ is in front of $\triangle ABC$ (the math should be the same). As the surface area of the circumsphere is $4 \pi R^2$, the final probability is $$P = \frac{12A_{\pi/3}}{4 \pi R^2} = \frac{3}{\pi} \int_0^{\pi/3}\int_0^{\tan^{-1}\left(\frac{\sec \theta}{2 \sqrt{2}}\right)} \cos \theta \; d \phi \; d \theta \approx 0.33222.$$
The only problem is, from numerical results, I am getting a probability of about $0.35$, so there is a mistake here somewhere, probably with the procedure itself. But that is the approach I came up with, so what I would like to ask is 


*

*whether this approach solves the problem at all, and

*if you have a better idea, whether it be a correction/improvement to mine or something else entirely.

 A: A good way to model this, I think, is to fix the orientation of the tetrahedron but let the projection plane be randomly oriented.
The direction from any vertex to the projection plane is then uniformly distributed over the unit sphere around that vertex.
To convert this back into the original problem, after choosing a random orientation of the projection plane we rotate everything so that the projection direction becomes the original fixed projection direction.
The probability that vertex $D$ is obscured by face $ABC$ is then 
$\frac{2\Omega}{4\pi} = \frac{\Omega}{2\pi},$
where $\Omega$ is the solid angle subtended by face $ABC$ as seen from vertex $D$ and $4\pi$ is the total solid angle around the vertex.
(Specifically, there is probability $\frac{\Omega}{4\pi}$ that the direction from the vertex to the projection plane will pass through the opposite face, and probability $\frac{\Omega}{4\pi}$ that the direction away from the projection plane will pass through the opposite face. The point on the opposite face in that direction may be on either side of the projection plane; that does not matter.)
From multiple sources (for example http://mathworld.wolfram.com/RegularTetrahedron.html),
$$ \Omega = 3 \arccos\left(\frac13\right) - \pi
= \arccos\left(\frac{23}{27}\right) \approx 0.55129.$$
Moreover only one vertex can be obscured in any projection.
Therefore the events that $A,$ $B,$ $C,$ or $D$ respectively are obscured are disjoint events, and their total probability is $\frac{8\Omega}{4\pi}.$
The probability that the tetrahedron projects as a triangle is therefore
$$
\frac{8\Omega}{4\pi} = \frac 2\pi \arccos\left(\frac{23}{27}\right) \approx 
0.350959,
$$
which agrees with your numerical results.

As an aside, the formula $3 \arccos\left(\frac13\right) - \pi$
can be obtained by letting the angles between the three pairs of edges meeting at $D$ be $\alpha,$ $\beta,$ and $\gamma,$ so that
$\alpha = \beta = \gamma = \frac\pi3,$
and then applying 
the formulas I derived in another answer.
Also note that by the triple angle formula for cosine
(that is, $\cos(3\theta)=4\cos^3\theta - 3\cos\theta$),
\begin{align}
\cos\left(3 \arccos\left(\frac13\right)\right)
&= 4\cos^3\left(\arccos\left(\frac13\right)\right)
   - 3\cos\left(\arccos\left(\frac13\right)\right) \\
&= 4\left(\frac13\right)^3 - 3\left(\frac13\right) \\
&= \frac{4}{27} - 1 \\
&= - \frac{23}{27},
\end{align}
so
$$
\cos\left(3 \arccos\left(\frac13\right) - \pi\right)
= -\cos\left(3 \arccos\left(\frac13\right)\right)
= \frac{23}{27}.
$$
Then, since $0 \leq 3 \arccos\left(\frac13\right) - \pi \leq \pi$
and  $0 \leq \frac{23}{27} \leq \pi,$
the arc cosines of both sides of the equation above are equal:
$$
3 \arccos\left(\frac13\right) - \pi = \arccos\left(\frac{23}{27}\right).
$$

Another related fact is that you can place a tetrahedron of edge length $\sqrt2$ in a cube of edge length $1$ so that the vertices of the tetrahedron are all vertices of the cube and the edges of the tetrahedron are face diagonals of the cube.
Now arrange eight such figures inside a cube of side $2$ so that all eight tetrahedra share a vertex at the center of the large cube.
The edges of the tetrahedra that lie along the faces of the large cube now form the edges of a cuboctahedron and the triangular faces of that cuboctahedron are faces of the tetrahedra. The solid angle subtended at a vertex of a tetrahedron by the opposite face is therefore the solid angle subtended at the center of a cuboctahedron by one of its triangular faces, which is featured in 
OEIS sequence A236555
as mentioned in a comment below.
It follows that the probability that the projection will be a triangle is equal to the total fraction of the solid angle around the center of a cuboctahedron that is subtended by its triangular faces.
