  Given a right-angled triangle of squares in a grid whose horizontal and
vertical sides are $n$ squares long, let N(n) denote the maximum number of dots
that can be placed into the cells of the triangle such that each row, each
column, and each diagonal parallel to the long side of the triangle contains at
most one dot. It has been proven that
  $N(n) = \lfloor \frac{2n+1}{3} \rfloor$.
  In this note, we give a new proof of this result using linear programming
techniques.
