QUESTION: Let $n \ge 4$ be a natural number. Let $A_1A_2 \cdots A_n$ be a regular polygon and $X = \{ 1,2,3,\cdots,n \} $. A subset $\{ i_1, i_2,\cdots, i_k \} $ of $X$, with $k \ge 3$ and $i_1 < i_2 < \cdots < i_k$, is called a good subset if the angles of the polygon $A_{i_1}A_{i_2}\cdots A_{i_k}$ , when arranged in the increasing order, are in an arithmetic progression. If $n$ is a prime, show that a proper good subset of $X$ contains exactly four elements.
MY APPROACH: Practically, I could not proceed with this particular sum.. I have tried a lot, but in vain.. But I came up with a counterexample though.. If the polygon is regular then all the angles will be equal and they will form a constant AP.. But I think, according to the question they have meant only non-constant AP's and therefore such a counter example is not valid.. I don't know if my question will be closed or not :P.. but honestly, I could not connect why should $n$ be prime, how do we use that information, or how do I suddenly start off with angles with any arbitrary polygon...
Will be very happy and much obliged if someone may help me out..
Thank you so much.
Here is a rough sketch of how you can prove this.
Let $\alpha=\frac\pi{n}$. Show that all the angles involved will be integer multiples of $\alpha$. In this way the problem has changed from something to do with angles into something to do with integers. You can use the inscribed angle theorem for this.
Let $a\cdot\alpha$ be the smallest angle of the $k$-gon, and $b\cdot\alpha$ the largest. We are given that the angles form an arithmetic sequence, so their sum will be $k\cdot\frac{a\alpha+b\alpha}2$. It is also true that the sum of the angles of a $k$-gon is $\pi(k-2) = n\alpha(k-2)$.
From this it follows that $k$ divides $4n$. Since $2<k<n$ and using the fact that $n$ is prime, deduce that $k=4$.