Put three congruent triangles inside a unit square so that they don't overlap one another.
What is the maximum possible area of one of the triangles?
(In reply to
Possible solution by Jer)
Punching in the expression for your area of (6-4sqrt(2))/sqrt(3) into a calculator yields 0.198115. This is smaller than a trivial arrangement by drawing the diagonals of the square and taking three of the four triangles, each with area of 0.25.
Nothing in the problem states we need to be restricted to equilateral triangles; just that the triangles are congruent. Although if we were restricted to equilateral triangles then your figure would be best.
So far I can't find anything better than the trivial arrangement for triangles in general.
re: Possible solution
