Nine!
Choose two points--for example, the 2 blue points in the diagram--inside a square with area 9, such that drawing a line segment between each of these 2 points and each vertex of the square divides the square into 9 parts.
Is it possible that all of the 9 parts have an equal area of 1?
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1 solution
Suppose yes. Then [ A B E ] + [ E C D ] = 2 + 2 = 4 . If we draw parallels to the sides of the square through E , then we have four rectangles, and the A B E , E C D triangles are made from the four rectangle. From each rectangle exactly the half is a part of the A B E or the E C D triangle (because A E , B E , C E , D E are diagonals). So [ A B E ] + [ E C D ] = 2 [ A B C D ] , but 4 = 2 9 .
So the answer is: it isn't possible.