... where "rights" is a clever allusion to the clever pun I made in my last title. I actually do deserve rights.
An astute mathematician pointed out to me that there is a fatal error in my reasoning in my last post, so ridiculously terrible that I am thinking about deleting the whole post and pretending it never happened. But I'm honest, so I won't delete it; I'll leave it for historical interest. The key is this: Triangles actually have to be on the plane x+y+z=180. Inside the pyramid, there are points like (1,2,3) which clearly are not the angles of any triangle. So, welcome to the plane.
The plane intersects the first octant in an equilateral triangle with vertices at (180,0,0), (0,180,0), and (0,0,180). An equilateral triangle appears at the centroid (60,60,60), and isosceles triangles appear along the medians, which I am not going to bother expressing in parametric form. I think that right triangles appear along the midlines (connect the midpoints to make a small equilateral triangle).
Here is what I propose: Draw in all the medians. This creates six 30-60-90 triangles. I assert that each one carries exactly the same triangles, just with the angle coordinates in a different order. There are six triangles, which corresponds to the 3!=6 ways to rearrange three different numbers. There are three different hypotenuses and three different short legs, which correspond to the isosceles triangles. The 3 corresponds to the 3!/2! = 3 ways to rearrange three numbers when two of them are the same. And there is one center point, because there is only one way to "rearrange" 60-60-60.
Thus, we need only inspect one of these small 30-60-90 triangles. The coordinates of one of them, for example, are (60,60,60), (0,0,180), and (0,90,90). In my picture, this is the one on the top right.
Now look at the midline that cuts across each small triangle (from (45,45,90) to (0,90,90) in the example). The midline acts as an altitude drawn from the right angle to the hypotenuse. This cuts the 30-60-90 triangle into two smaller, similar triangles, one of which is 1/4 the area and one of which is 3/4 of the area of the original. The smaller one, whose vertices are (60,60,60), (45,45,90), and (0,90,90) in the example, contains all of the acute triangles. The larger triangle contains the obtuse triangles, and the right triangles are on the dividing line.
So, in conclusion, 1/4 of triangles are acute, 3/4 are obtuse, and none of them are right. QED.
10/16/17 PHD comic: 'Confusing Malaise'
15 hours ago