The approximate method can be described as follows:
Start with a square. Let A, B be the middle points of two adjacent sides and C the vertex farthest away from these two points. Let M be the middle of the segment BC. Consider the shaded rectangle whose diagonal is MC and rotate it 90 degrees counterclockwise so C goes to C1. Pick the point K so that KC1:KM = 1:3. The "pentagon" has vertices A, B, C, K and the symmetric of K.To see that it's not a regular pentagon, let the side of the original square have length 8u where u is the square root of 2 (for no good reason other than that I wanted to end up with integers). Then, using the Pythagorean theorem several times, we can see that AB equals sqrt(256)=16, but BK equals sqrt(250) which equals 15.81... (Pretty good approximation.) Notice that MK passes through the middle of BC and is perpendicular to it, so BK equals CK. By symmetry, 4 of the 5 sides of the pentagon have length 15.81... and only one has length 16.
Does anyone know an exact (and simple...) origami method for constructing regular pentagons?



http://origami.oschene.com/cp/Decagon%20SCP.pdf
ReplyDeleteI saw this on a calendar last year, I lost it for a while and found it again on the net. It works!