Find the least positive integer n such that there are at least $1000$ unordered pairs of diagonals in a regular polygon with n vertices that intersect at a right angle in the interior of the polygon.
Here, I thought that for even $n$, starting at any vertex, I draw the diagonal joining the opposite vertex. Then we get $\dfrac{n-2}{2}$ diagonals perpendicular to this diagonal. We can draw the longest diagonal $\dfrac{n}{2}$ times and keep getting unique cases. Based on this, I get the inequality,
$$\frac{n}{2}\cdot \frac{n-2}{2}\ge1000 \Rightarrow n \ge 65$$
My argument after this is that $n$ can only be even, giving me the answer as $66$.
However, how do I prove that $n$ cannot be odd? Or am I making a mistake in my argument and an odd $n$ can give valid answer?
Best Answer
Claim : $\;$ Inside a regular polygon with an odd number of sides, there are no perpendicular diagonals.
Proof: $\;$ Imagine a regular $n$-gon is inscribed in a circle. Suppose four of its vertices $A,B,C,D$ give a pair of perpendicular diagonals.
This means $m(\overset{\frown}{AB})+m(\overset{\frown}{CD})=m(\overset{\frown}{BC})+m(\overset{\frown}{AD})$ implying that $l(\overset{\frown}{AB})+l(\overset{\frown}{CD})=l(\overset{\frown}{BC})+l(\overset{\frown}{AD})=$ half the circumference.
Then arcs $\overset{\frown}{AB},\overset{\frown}{CD}$ contain one half of remaining $(n-4)$ vertices while other half lies on arcs $\overset{\frown}{AD},\overset{\frown}{BC}$. Clearly $(n-4)$, hence $n$, must be even.
The formula $n(n-2)/4$ for even $n$, undercounts pairs of orthogonal diagonals because only longest diagonals (passing through center) have been taken into account. Shorter diagonals (not passing through center) can also intersect at right angles. The correct formula can be found in the OEIS link shared by @DanielMathias in comments. To be explicit,
Using this formula in the inequality, one gets $n=14$ i.e., a regular $28$-gon is first one to satisfy the given condition.