Solution
Dissecting a Heptagon
Show the problem again
A convex heptagon has 14 diagonals. How many sets of diagonals (including the empty set) have the property that no two diagonals in the set cross at a point interior to the heptagon? Diagonals that share an endpoint do not count as crossing. Give an integer.
Worked solution
**Answer: 197.** Such a set is a dissection of the heptagon into smaller convex polygons. The Kirkman-Cayley formula counts dissections of a convex n-gon by k non-crossing diagonals as D(n, k) = (1 / (k + 1)) · C(n − 3, k) · C(n + k − 1, k). For n = 7 and k = 0, 1, 2, 3, 4: 1, 14, 56, 84, 42. (The k = 4 term is the 42 triangulations, a Catalan number, and k = 1 is simply the 14 diagonals.) The total is 1 + 14 + 56 + 84 + 42 = 197. These totals 1, 3, 11, 45, 197, ... for the triangle, square, pentagon, hexagon, heptagon, ... are the little Schröder (super-Catalan) numbers. An exhaustive check of all 2^14 subsets confirms 197.
Source: The count of dissections by number of diagonals is the Kirkman-Cayley formula; the totals are the little Schröder numbers. Statement written for AxiomIQ.