Question

In a regular hexagon, ropes are tied to connect every pair of vertices (all sides and all diagonals). How many distinct intersection points do the ropes create?

• A. 16
• B. 17
• C. 19
• D. 20

Hint:

For a complete graph on a convex $n$-gon, interior intersections start as $\binom{n}{4}$; in regular polygons, adjust for any special concurrencies (like opposite-vertex diagonals meeting at the centre in a hexagon).

Answer 1

The Correct Option is C

Step 1: Intersections formed by diagonals inside the polygon.
For a convex $n$-gon with all diagonals drawn, each \emph{interior} intersection is determined by a choice of $4$ vertices (the two diagonals joining opposite pairs of those $4$ vertices). If every such pair met at a distinct point, the count would be $\binom{6}{4}=15$. Step 2: Adjust for concurrency at the centre.
In a \emph{regular} hexagon, the three long diagonals joining opposite vertices, namely $(1,4)$, $(2,5)$, and $(3,6)$, all pass through the centre. The “$\binom{6}{4}$ rule” counts the centre three times (from the pairs $(1,4)\& (2,5)$, $(1,4)\& (3,6)$, $(2,5)\& (3,6)$), but these are the \emph{same} point. So distinct interior intersection points $=15-2=13$. Step 3: Include intersections at vertices.
Every vertex is also an intersection of ropes (sides and diagonals). There are $6$ vertices. Total distinct intersection points $=13+6=19$. \[ \boxed{19} \]