By "rational angles" I mean a rational degree measure (equivalently, angle a rational multiple of $\pi$). Obviously similar triangles should be counted once.

Off the top of my head we have: 30-60-90, equilateral triangles, 45-45-90, and 36-36-72.

  • $\begingroup$ Note that while wendy kriger's answer covers your specific question (values of the form $r+s\sqrt{n}$ with $r$ and $s$ rational and $n$ integer), there are other rational angles whose sines and cosines can be expressed using nestded square roots whose 'leaf values' are rational numbers (or equivalently, which are constructible with ruler and compass). $\endgroup$ Aug 15 '15 at 4:14

The triangles would exist in a regular polygon of 4, 5, 6, sides. So there's also 36-72-72 and 108-36-36 in the pentagon, 30-30-120, 30-60-90 and 60-60-60 in the hexagon, and 45-45-90 in the square. So there's six of them.

The angle at the circumcentre of the triangle for each side, is twice the angle opposite angle. Since the angles make rational angles, the vertices must fall in the a polygon representing the least common denominator.

Since the chords of polygons are solutions to a polynomial of order ½ totient(2n), it's simply a matter of satisfying this for this=2, gives n=4, 5, 6 as the only solutions. 7 and 9 solve cubics, 8, 10, 12, 15 solve biquadratics, 11 solve quintics, 13, 14, 18, 21 solve hexics, and so forth.

  • 2
    $\begingroup$ Can you explain why? $\endgroup$
    – Alex Zorn
    Jul 10 '15 at 2:01
  • $\begingroup$ added comment for the proof of this. $\endgroup$ Jul 10 '15 at 4:17

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