Solution
Triangles in a Hexagon
Show the problem again
A convex hexagon is drawn together with all 9 of its diagonals, and no three diagonals pass through a common interior point. How many triangles (of any size) appear in the figure, i.e. triangles whose three sides each lie along drawn segments? Give an integer.
Worked solution
The answer is 111. Classify each triangle by how many of its vertices are hexagon vertices; every other vertex is an interior crossing of two diagonals. Three polygon vertices: $\binom{6}{3} = 20$. Two polygon vertices: such triangles live inside a convex quadrilateral formed by 4 of the hexagon's vertices; its two diagonals cross at a point $P$, and each of the 4 sides of the quadrilateral with $P$ gives a triangle: $4\binom{6}{4} = 60$. One polygon vertex: such a triangle lives inside a convex pentagon on 5 of the hexagon's vertices. For each vertex $V$ of the pentagon, the diagonal joining $V$'s two pentagon-neighbours crosses the two pentagon diagonals from $V$, and these three segments bound a triangle with apex $V$; so each 5-subset yields 5 such triangles: $5\binom{6}{5} = 30$. No polygon vertex: the three long diagonals of the hexagon form 1 triangle in the centre: $\binom{6}{6} = 1$. Total $20 + 60 + 30 + 1 = 111$.
Source: Count of triangles formed by the sides and diagonals of a convex polygon in general position; classic in wide circulation. Statement written for AxiomIQ.