On Sunday 10 June 2007 02:11, George W. Hart wrote:
Gareth McCaughan wrote:
...Is there a better solution?
If you draw ABC properly within a tessellation of unit equilateral triangles, it is clear from the Pythagorean theorem that its edge length is sqrt[7]. See figure 5 of: http://www.georgehart.com/radiolaria/radiolaria.html
Very nice (and, completely irrelevantly, I like your ray-trace-olarian) but that only answers the question for p=1/3. ... No, wait. If you do the same construction but at a different angle (the picture you want to end up with is an equiangular hexagon with sides a,b,a,b,a,b with 3+3 of its diagonals drawn), you rapidly get that for a,b integers F(a/(a+b)) = (a-b)^2 / (a^2 + ab + b^2) or F(p) = (1-4pq) / (1-pq) which by continuity means that the latter is true for all p. In fact, forgetting about the underlying triangular grid, we can just draw the aforementioned equiangular hexagon and get the result for arbitrary p straight off. Very nice, and it does seem to me to give some indication of why the thing is true. Thanks! -- g