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.

the key to this interesting problem is the 30-60-90 triangle.
when the short side is 1, the hypotenuse is 2 and the adjacent side is sqrt3.

so to make this problem easy for myself, i designated the square paper 3x3, or 9 square units, thus our problem is to make a square which is 3 square units.

first i made two folds, with the creases originating in the upper left corner, such that the paper is divided into three wedges of equal angle: 30 degrees.

so you have the upper right and the lower left corners folded over into flaps, which are 30-60-90 triangles with 'adjacent' side of 3.

unfold the lower left flap, but make sure there is a crease. now you're left with the original square paper with the upper right corner folded over.

that flap is your triangle, the short side of which is the first complete side of your square, since it is the adjacent side's length, 3, divided by sqrt3, which is: sqrt3.

by folding and unfolding and refolding, it is pretty simple to create the square. what IS difficult is explaining it in words, but if you make it this far, you're pretty much there, so i'll leave my answer incomplete.

nice problem.

Posted by rixar on 2004-09-22 21:25:15