A rectangular sheet of paper is folded so that two diagonally opposite corners come together. The crease thus formed is as long as the longer side of the rectangle.
What is the ratio of the longer side of the rectangle to the shorter?
Let ABCD be the rectangle with AB = a < b = BC. Let corners A and C touch to form crease EF ( E on AD and F on BC ). EF is the perpendicular bisector of diagonal AC. Let G be the intersection of AC and EF. Triangles ABC and FGC are similar so that BC/AB = GC/FG. Let k = b/a = [sqrt(a^2+b^2)/2]/[b/2] = sqrt(1+/k^2) or k^4-k^2-1 = 0. k = sqrt((1+sqrt(5))/2) ~= 1.272.
Edited on July 27, 2004, 5:09 pm
Posted by Bractals
on 2004-07-27 17:07:57