random
Yesterday there was a little festival sort of thing in Cambridge, where we saw (inter alia) a Rastafarian sheep and a business called "Nutty Tarts" which disappointingly sells nutty tarts.
Elsevier have finally decided to get out of the arms trade.
I wrote up some rambling notes on the texts for my choir's forthcoming concert, in the hope that they might be useful to other choir members; but who knows, they might be of interest to someone else, hence the link here.
(For geeks only.) Draw a triangle (any triangle will do, but it'll be prettiest with an equilateral one). Divide each side in the ratio 1:2. Join each division point to the opposite vertex of the triangle. You'll get a smaller triangle in the middle. What's the ratio of its area to that of the original triangle? What if you divide the sides in some other ratio? This is an old question, and it's easy to solve ploddingly with coordinate geometry. Someone pointed me at figure 5 in George Hart's explanation of his lovely "artificial radiolarian reticulum" (actually, the someone was George Hart himself), which provides a near-instant proof once you check the side-length of the outer triangle. Well, it turns out that this generalizes nicely. (I expect George knew this when he drew that diagram.) So: draw an equiangular hexagon (all interior angles 120 degrees) with sides p, q, p, q, p, q. Label the vertices ABCDEF. Join AC, CE, EA, giving an equilateral triangle of side sqrt(p2+pq+q2). Join AD, CF, EB, producing (1) an equilateral triangle of sides |p-q| in the middle and (2) a subdivision of each edge of ACE in the ratio p:q. Conclusion: The ratio of areas is
![\[\frac{(p-q)^2}{p^2+pq+q^2}.\]](http://www.mccaughan.org.uk/g/tex-images/6f170d0fc5d4d6e85551efe0f73b0303.png){kind=small}