
4. Consider a 20 -sided convex polygon $K$, with vertices $A_{1}, A_{2}, \ldots, A_{20}$ in that order. Find the number of ways in which three sides of $K$ can be chosen so that every pair among them has at least two sides of $K$ between them. (For example $\left(A_{1} A_{2}, A_{4} A_{5}, A_{11} A_{12}\right)$ is an admissible triple while $\left(A_{1} A_{2}, A_{4} A_{5}, A_{19} A_{20}\right)$ is not.)


Solution: First let us count all

![](https://cdn.mathpix.com/cropped/2024_06_05_c77f8a1613e773dc8337g-2.jpg?height=392&width=717&top_left_y=615&top_left_x=424)
the admissible triples having $A_{1} A_{2}$ as one of the sides. Having chosen $A_{1} A_{2}$, we cannot choose $A_{2} A_{3}$, $A_{3} A_{4}, A_{20} A_{1}$ nor $A_{19} A_{20}$. Thus we have to choose two sides separated by 2 sides among 15 sides $A_{4} A_{5}, A_{5} A_{6}, \ldots, A_{18} A_{19}$. If $A_{4} A_{5}$ is one of them, the choice for the remaining side is only from 12 sides

$A_{7} A_{8}, A_{8} A_{9}, \ldots, A_{18} A_{19}$. If we choose $A_{5} A_{6}$ after $A_{1} A_{2}$, the choice for the third side is now only from $A_{8} A_{9}, A_{9} A_{10}, \ldots, A_{18} A_{19}$ (11 sides). Thus the number of choices progressively decreases and finally for the side $A_{15} A_{16}$ there is only one choice, namely, $A_{18} A_{19}$. Hence the number of triples with $A_{1} A_{2}$ as one of the sides is

$$
12+11+10+\cdots+1=\frac{12 \times 13}{2}=78
$$

Hence the number of triples then would be $(78 \times 20) / 3=520$.

Remark: For an $n$-sided polygon, the number of such triples is $\frac{n(n-7)(n-8)}{6}$, for $n \geq 9$. We may check that for $n=20$, this gives $(20 \times 13 \times 12) / 6=520$.
