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A triangle with all side lengths rational is said to be a rational triangle. It is known - for example, Cutting the unit square into pieces with rational length sides - that the unit square allows partition into triangles all of which are rational. It follows from the above-linked discussion that any rectangle with length and width rational (diagonal whatever) can be cut into rational rectangles by a recursive application of the unit square method (thanks to Peter Taylor for clarifying this point I had overlooked in an earlier post).

Question: for a general polygon P to be divisible into some finite number of rational triangles, it is obviously necessary that all sides of P are rational in length. What could one say about sufficiency conditions? As a special case, how could one characterize a quadrilateral that can be cut into some finite number of rational triangles?

Nandakumar R
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    There are only countably many angles that occur in rational triangles. So only countably many polygons can be cut into finitely many rational triangles. – Robert Israel Dec 12 '23 at 16:03
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    Thanks. Isn't the set of all polygons with all edges of rational length (polygons satisfying just the necessary condition mentioned in question) already countable? – Nandakumar R Dec 13 '23 at 10:56
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    No. The set of side length sequences is countable, but varying the angles between them may give an uncountable set for a single side length sequence. E.g. there's an uncountable set of unit rhombi. – Peter Taylor Dec 14 '23 at 07:48
  • Thanks for pointing out the freedom - and uncountability - in angles. So, could one say that (say) in addition to all sides being rational, the cosines of all angles of P being rational would be a better (stronger) necessary condition? – Nandakumar R Dec 14 '23 at 11:04
  • Yes, that would be a necessary condition. – Robert Israel Dec 17 '23 at 04:38

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