On Sat, Feb 18, 2012 at 11:51 PM, Bill Gosper <billgosper@gmail.com> wrote:
Here <http://gosper.org/isos10.png>'s a 10 triangle solution (like with the square), while we await the wondrous 9.
Oops, my algebra was wrong, as was the picture, by one jaggy. Please refetch.
SA> http://mathworld.wolfram.com/SquareTiling.html M. Laczkovich has shown that there are exactly three shapes of non-right triangles that tile the square with similar copies, corresponding to angles (pi/8,pi/4,5pi/8), (pi/4,pi/3,5pi/12), and (pi/12,pi/4,2pi/3) (Stein and Szabó 1994). in degrees that is (22.5, 45, 112.5), (45,60,75), and (15, 45, 120) The Gosper-Hoggatt tilings use these angles. Stuart [Anderson] Perhaps no longer. --rwg Easy one: Everybody knows that you get a cube when you start with a regular
tetrahedron and perform the Snowflake recursion in 3D. But what do you get if you start with an octahedron instead, and add tetrahedra? Answer <http://gosper.org/cuboct.png>. (Simply typed in by Julian.) What percentage of true/false quiz takers would bite on "This correct name for this solid is "Cuboctahedron"." ?
In another, overlong thread I said
Trying the g.f. sum, I hit this weirdy: Sum[(c^n*QPochhammer[b, q, n])/(b^n*QPochhammer[c*q, q, n]), {n, 0, Infinity}] == (b*(-1 + c))/(-b + c)
independent of q. Plotting for b=2/3,c=1/3, it holds somewhat past q=1, and then develops a very interesting collection of poles.
And fake poles? ( http://gosper.org/poles.png)
All but one of the poles are completely spurious artifacts of insufficient WorkingPrecision. c=1,b=2 near q=1 is about the worst floating point noise I've ever seen. (Amplitude >> 10^4.) --rwg
On Wed, Feb 15, 2012 at 4:11 PM, Bill Gosper <billgosper@gmail.com> wrote:
Neil has just uncovered fq.math.ca/Scanned/6-6/hoggatt.pdf predating my http://mathworld.wolfram.com/SquareDissection.html of a square into ten acute isosceles triangles.
Neil also found in a Gardner book a reference to a Monthly paper (June-July 1962, pp550-552) claiming that *any* obtuse triangle can be cut into eight acute isosceles triangles, implying at most nine for dissecting a right isosceles, in contradiction of my round(tan(69)) solutions assertion. Has anyone a picture of this dissection? --rwg This is weird: Calculating with pen and paper for a change, it seems that sunlight falling on the diagrams and equations actually makes them *easier* to read.