## Task 4 - 270714

It is known that every quadrilateral has exactly two diagonals.

a) Determine the number of diagonals in a pentagon and a hexagon!

b) Find a formula for the number of diagonals of a polygon depending on the number of vertices $n$ of the polygon! The formula should be valid for all natural numbers $n \geq 4$. Justify this formula!

c) What value does this formula give when applied for $n=3$? Can this value also be expressed in a geometrically intuitive statement?
a) The pentagon has exactly five and the hexagon exactly nine diagonals, as can be seen from the illustration, for example.

b) To capture all diagonals in an n-gon, one can consider for each of the $n$ vertices the line segment to every other vertex except the two adjacent vertices. Thus, one has considered a total of $n \cdot (n-3)$ segments, and each diagonal of the n-gon exactly twice. If the number of diagonals is denoted by $x$, it follows that $x = \frac{1}{2} n \cdot (n-3)$.

c) For $n=3$, this formula yields the value $x=0$. This can be stated as the fact that for every triangle, the number of diagonals is zero.

Solutions of the 1st Round 1987 adopted from [5]

# 4.29.2 2nd Round 1987, Class 7