Two diagonals - right angle

Question

For which natural number $n$ there exists a regular $n$-angle that has two diagonals that intersect at right angles? I tired to consider various cases, but I have problem with $n=5,7,9,11...$. Do you have any hint?

Answer 1

For a given $n$-gon, it appears that the size of the angles between the diagonals are of the form $\frac{180k}{n}$, where $k$ ranges from $1$ to $(n-4)$.

A $9$-gon, for example, has angles $\frac{180}{9}=20$, $\frac{360}{9}=40$, ..., $\frac{900}{9}=100$.

To answer your question, we are searching for when $\frac{180k}{n}=90$, subject to the constraint that $1\leq k\leq (n-4)$.

Rearrange to get $180k=90n$ and you can see that we need $n=2k$. That is, $n$ must be even, which explains why you were unable to find examples for polygons with an odd number of sides.

For even-sided polygons, when we consider $k=\frac{n}{2}$ it is evident that this value is in the range $1\leq k\leq (n-4)$ for all polygons with number of sides $n\geq 8$.

Note: I don't have a rigorous proof that the angles between diagonals are of the form $\frac{180k}{n}$. This is just something I observed by playing around with a few examples.