Inside this regular octagon sit two squares of area 8. What’s the area of the shaded rectangle?pic.twitter.com/FgdfcoY1aj
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Inside this regular octagon sit two squares of area 8. What’s the area of the shaded rectangle?pic.twitter.com/FgdfcoY1aj
I think a dissection argument is simplest, and perhaps also the most elegant.pic.twitter.com/fbABrw2jzC
That's beautiful!
And how about these amber parts?
cc @mathhombrepic.twitter.com/wweOoVqsx3
I really like the way these A-size rectangles and half-squares can tile together. Lots of interesting things to see. What size regular octagons can we make?pic.twitter.com/Esi7y1YSHb
Thinking about this was so fun that I made an @observablehq notebook illustrating the two substitution tilings of the octagon from this thread! https://beta.observablehq.com/@bryangingechen/tiling-an-octagon-with-half-squares-and-a-size-rectangles … Here's an animation of the more interesting one: in the limit, the purple and green areas become equal, but why?pic.twitter.com/o2YM2BtnoV
That fact follows from the messy linear algebra I sketch in the notebook, but is there a nicer, more conceptual proof?
I love this! So would equal area also be the limit of an octagon-square tessellation of the plane? And how can a hexagon-square tiling be equivalent to this?
For the first question, I think so, assuming that you first subdivide the octagon-square tessellation into half-squares and A-size rectangles and then perform rule II repeatedly. For the second question, Cockayne's hexagon-square tiling looks like this (continued)pic.twitter.com/DoxqYmIGcq
In particular, his hexagons aren't regular and can be arbitrarily long (the previous image is fig 3 from https://doi.org/10.1088/0305-4470/27/18/019 … and only shows length rt(2)+1 and rt(2)+2 if square side=1). De Gier and Nienhuis cut up Cockayne's tiles like this (fig 1 of https://doi.org/10.1103/PhysRevLett.76.2918 …).pic.twitter.com/ngPHhCfeCA
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