Love this post from Matt Baker in which he explains the tropical / 2-adic proof (in fact the only proof!) that you can’t dissect a square into an odd number of triangles of equal area. In fact, his argument proves more, I think — you can’t dissect a square into triangles whose areas are all rational numbers with odd denominator!

- The space of quadrilaterals in R^2, up to the action of affine linear transformations, is basically just R^2, right? Because you can move three vertices to (0,0), (0,1), (1,0) and then you’re basically out of linear transformations. And the property “can be decomposed into n triangles of equal area” is invariant under those transformations. OK, so — for which choices of the “fourth vertex” do you get a quadrilateral that has a decomposition into an odd number of equal-area triangles? (I think once you’re not a parallelogram you lose the easy decomposition into 2 equal area triangles, so I suppose generically maybe there’s NO equal-area decomposition?) When do you have a decomposition into triangles whose area has odd denominator?
- What if you replace the square with the torus R^2 / Z^2; for which n can you decompose the torus into equal-area triangles? What about a Riemann surface with constant negative curvature? (Now a “triangle” is understood to be a geodesic triangle.) If I have this right, there are plenty of examples of such surfaces with equal-area triangulations — for instance, Voight gives lots of examples of Shimura curves corresponding to cocompact arithmetic subgroups which are finite index in triangle groups; I think that lets you decompose the Riemann surface into a union of fundamental domains each of which are geodesic triangles of the same area.

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