Geometry and Permutation and Combination problem

Question

In an n sided regular polygon, the formula for the number of diagonals is $\frac{n(n-3)}{2}$. This formula is derived by selecting $2$ out of $n$ sides i.e. $n\choose 2$ and then subtracting the number of sides. Among these diagonals how many passes through the centre?

Answer 1

I believe that in such questions, It is best to observe the characteristics for a smaller case, and then see how we can generalize.

Lets start with a pentagon(or even a triangle). You'll see that no diagonal passes through the center. Which is not really helpful to generalize.(Same is the case with a heptagon). There is something weird going on with the odd polygons, And so We must realize that it is helpful to work with only even polygons first. We might see then,Why exactly are the odd polygons causing a problem.

We now move on to a square: Label the vertices $A1,A2,A3,A4$ in order. We see that the diagonals $A1A3$ and $A2A4$ pass through the center. Observe that $13$ and $24$ have a common link: $3-1 = 4-2 =2= 4/2$.

Consider a hexagon next. Label it $A1,A2,...A6$. The diagonals that pass through the center are $A1A4, A2A5$ and $A3A6$. Again, $4-1=5-2=6-3=3=(6/2)$.

It now becomes easy to see that for a polygon with $n=2k$ sides, the no. of solutions to the equation $(x-y)=k$ give us the diagonals passing through the center. $x=y+k$ and both $x,y<=k$. Its easy to see that exactly k solutions exist, which is simply $ n/2$ as Pi_die_die claimed.

While working through this solution, You might also observe something: a "diagonal through the center" is simply a line of symmetry for the polygon : The no. Of sides on either side is the same. This realization makes it obvious that Such diagonals cannot exist for odd polygons, and will be $n/2$ in number for even polygons.

Answer 2

Not all n-sided polygons will have such diagonals.

Only for n=2k will you have n/2 such diagonals. No need for a formula to think this through in my opinion.