Show that any convex polygon of area 1 is contained in some parallelogram of area 2.
Solution
Let the vertices X, Y of the polygon be the two which are furthest apart. The polygon must lie between the lines through X and Y perpendicular to XY (for if a vertex Z lay outside the line through Y, then ZY > XY). Take two sides of a rectangle along these lines and the other two sides as close together as possible. There must be a vertices U and V on each of the other two sides. But now the area of the rectangle is twice the area of XUYV, which is at most the area of the polygon. [In the case of a triangle one side of the rectangle will be XY.]
© John Scholes
jscholes@kalva.demon.co.uk
31 July 2002