Given a square piece of paper, show how by creasing and folding only, a square of one third the area of the original can be obtained.
Assume the square to be a unit square. Let the corners of the square be ABCD (in clockwise order). Fold the square in half to form a crease with endpoints M and N, the midpoints of AB and CD, respectively. Then unfold the paper.
Next, fold along a crease from B to a point on CD in a way such that C is on crease MN (is this possible?). Let the point on this crease be P; mark it by a crease. Note then that triangle ABP is equilateral, so that angle PBC is 30.
Fold along line BP to make a crease that hits side CD at a point Q. Note then that BQC is a 30-60-90 triangle with the side opposite the 60 having length 1. Thus CQ=1/sqrt(3). Perform the necessary folds to create a square with side CQ, which will have area (1/sqrt(3))^2=1/3, as desired.
I know there must be more elegant ways of solving this problem...be happy to hear them.
Solution
