One of my favorite teacher trollings is: Construct the tangent to a given point on a circle, without a compass. The first step is to choose four more points on the circumference at random. gosper.org/tanimation.gif This was shown me by PDP-1 hacker Dick Howell. It's a consequence of https://en.wikipedia.org/wiki/Pascal%27s_theorem about hexagons inscribed in conics. A completely other connection is with hexagons in Pascal's Triangle (GKP 5.1, p155): For example, there's a curious "hexagon property," illustrated by the six numbers 56, 28, 36, 120, 210, 126 that surround 84 in the lower right portion of Table 155. Both ways of multiplying alternate numbers from this hexagon give the same product: 56·36·210 = 28·120·126 = 423360. The same thing holds if we extract such a hexagon from any other part of Pascal's triangle. Talk about understatement. It even holds for hexagons of any integer "radius" r, anywhere in q-Pascal's Triangle! QBinomial[n, k + r, q] QBinomial[n - r, k - r, q] QBinomial[n + r, k,q] = QBinomial[n, k - r, q] QBinomial[n - r, k, q] QBinomial[n + r, k + r, q] Furthermore, if QBinomial is defined to return its q (Laurent) polynomials in factored form, this identity is syntactic. E.g., for radius 2 around {9,6}, In[521]:= pashexq[{9, 6}, 2] (1-q+q^2) (1+q+q^2) (1+q^4) (1-q+q^2-q^3+q^4) (1+q+q^2+q^3+q^4) (1+q^3+q^6)^2 (1+q+q^2+q^3+q^4+q^5+q^6)^2 (1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10) = (1-q+q^2) (1+q+q^2) (1+q^4) (1-q+q^2-q^3+q^4) (1+q+q^2+q^3+q^4) (1+q^3+q^6)^2 (1+q+q^2+q^3+q^4+q^5+q^6)^2 (1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10) Likewise for the formal expression in Γ functions for complex n, r, and k. (E.g., Fred's <https://www.dropbox.com/s/9vvl6l6ym1zkret/binomial.pdf?dl=1> crazy surfaces.) The property even holds for Mathematica's (and Maple's) broken "Bermuda" definition of Binomial. --rwg