[math-fun] name for golden ratio triangle tiling?
If you can divide a triangle into two similar triangles, then the small triangles are right triangles. Which means the small triangles are similar to the original triangle. If you want the small triangles' areas to be 1 : phi, then the sides of the triangles are 1 : r : r**2, where r = sqrt(1/phi) ~= .7861513777574233. r = sqrt( (sqrt(5) - 1) / 2 ) I don't see a simpler expression for that. As far as I can tell, this triangle is not related to Penrose tiles. But, googling for "golden ratio tiling" finds only mentions of Penrose tiles. -> Is this a familiar construction with a name? <- It seems about tied for simplicity with these other ways of recursively dividing the plane into similar shapes: square -> four squares 1 : sqrt(2) rectangle -> two of them (European paper sizes). 90-45-45 triangle -> two of them. From this you can get "gradual" tiling (see below) "gradual" Hilbert-like curve "gradual" sampling of a 2D space, "peppering." By gradual tiling, I mean that you can divide the original triangle into N smaller triangles by splitting in a systematic order. If N is a Fibonacci number, you only need two sizes, else three. Similarly you can evolve a path by putting a dot in the center of the original triangle, and when you split a triangle, replace the dot by a line segment between the two new centers (rubber-banding any segments that were already attached). By peppering I mean, as you split a triangle, put a dot at the middle of the split line. Gradually, the accumulating dots sample the triangle at finer and finer resolutions. Something like the way you (or a sunflower) can sample the unit interval using phi. Conway's tessellation / Charles Radin's "pinwheel tiling" divides a sqrt(5) : 2 : 1 triangle into five similar triangles. With the original triangle oriented correctly, all the vertices of the little triangles have rational coordinates! With this 1 : r : r**2 tiling, all of the coordinates are irrational. --Steve P.S. Strange, I had to add Penrose, Conway and Fibonacci to my spelling dictionary for the first time, all in the same email. And sqrt.
I forget if I ever mentioned this puzzle on math-fun, so just in case, this is aimed only at people who haven't seen it. ------- PUZZLE: Find all right triangles that can be tiled by congruent* triangles similar to itself. ------- --Dan ____________________________________________________________________________________ * Two shapes are congruent if there is a distance-preserving bijection between them. On 2014-02-09, at 11:56 PM, Steve Witham wrote:
If you can divide a triangle into two similar triangles, [followed by much interesting stuff]
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Steve Witham