In the Mondrian Art Problem, a square is divided into non-congruent integer-sided rectangles so that the largest area and smallest area are as close as possible.

A lattice square can be divided into non-congruent lattice-vertex triangles so that all areas are identical. Pick's theorem, Area = 2 Interior + Boundary/2 - 1, is handy. Below, all triangles have area 3. Define a Perfect Pick as a dissection into non-congruent lattice-vertexed triangles of the same area.

Triangles of area 3

What is the largest possible smallest angle within a a Perfect Pick square dissection?

  • 1
    $\begingroup$ Is there any condition on the area of the square? Otherwise, there is a clear improvement to the above example with 4 triangle above and 4 triangles below the diagonal (using essentially the same method as your example). However, in order to satisfy the lattice vertex condition, one needs to work on a different sized square (I think 4x4 works, though I need to check. Otherwise, there should be a larger square that works). $\endgroup$ – Dan Rust Nov 16 '16 at 23:17
  • $\begingroup$ Sure, any sized square can be used. $\endgroup$ – Ed Pegg Nov 16 '16 at 23:27
  • 1
    $\begingroup$ By the way Ed, as you seem to be interested in this problem recently, I thought you might also be interested in a recent preprint of a colleague of mine. It's a similar flavour, but for the entire plane, rather than a square. $\endgroup$ – Dan Rust Nov 16 '16 at 23:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.