Monsky's theorem: Difference between revisions

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#Conclude from the straight-line property that a tricolored triangle must also exist in every dissection of the square into triangles, not necessarily meeting edge-to-edge.
#Use Cartesian geometry to show that the 2-adic 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 2-adic valuation greater than 1.
#If ''n'' is odd then the 2-adic valuation of 1/''n'' is 1, so it is impossible to dissect the square into triangles all of which have area 1/''n''.<ref>{{Waybackwebarchive |url= |date=August 18, 2010 |title=Dissecting a square into triangles|date=20100818142143 }}</ref>