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#Conclude from the straightline property that a tricolored triangle must also exist in every dissection of the square into triangles, not necessarily meeting edgetoedge.
#Use Cartesian geometry to show that the 2adic valuation of the area of a triangle whose vertices have three different colours is greater than 1. So every dissection of the square into triangles must contain at least one triangle whose area has a 2adic valuation greater than 1.
#If ''n'' is odd then the 2adic valuation of 1/''n'' is 1, so it is impossible to dissect the square into triangles all of which have area 1/''n''.<ref>{{
==Generalizations==
