A familiar geometry problem is to consider an isosceles triangle APB with vertex P 20 degrees, and draw line BM with angle ABM 60 degrees, and M lying on AP, and line AN with angle BAN 50 degrees and N lying on BP. Then connect M and N, and the problem is to find angle ANM.
There are several clever ways to solve this involving drawing two or more lines and reasoning about various similar triangles, and all the angles nicely come out to be multiples of 10 degrees.
But ever since solving it for myself in 1969, after several months of futile struggling with the analogous problem where vertex P is 40 degrees instead of 20, I have been wondering about the following extension:
Find all sets of three distinct angles APB, ABM and BAN, all rational multiples of $\pi$, such that angle ANM (and hence all the other angles in the diagram) is also a rational multiple of $\pi$. This may be too hard -- then find any other such set of angles.
[Change after Josh B's comment -- any other such set of angles with the vertex not 20 degees. Because evidently 20, 70, 60 works.]
I have approached this using analytic geometry, and after a prodigious amount of algebra and trig identity shuffling, managed to replicate the answer for the 20/60/50 problem (by the way, Mathematica is wonderful, but it is quite lame at applying trig identities so this was done mostly by hand). But this approach gave me less than no insight as to why these magic angles work together.
