Question

Let n be a natural number. Let $A_1{A}_2...,A_n$ be a regular polygon and X = 1, 2, . . , n}. A subset { $i_1,i_2,...,i_k$ } of $X$, with $k\ge 3$ and $i_1<i_2<...<i_k$ , is called a good subset if the angles of the polygon $A_{i1}{A}_{i2}...A_{ik}$ , when arranged in the increasing order, are in an arithmetic progression. If n is a prime, find number of elements that a proper good subset of $X$ contains?

Answer 1

We note that every angle of $A_{i1}{A}_{i2}...A_{ik}$ is a multiple of $\pi =n$ .
Suppose that these angles are in an arithmetic progression.
Let $r$ and $s$ be non-negative integers such that $r=\pi /n$ is the smallest angle in this progression and $\pi s/n$ is the common difference.
Then we have $\frac{\pi }{n}(rk+sk(k-1\left)/2\right)=(k-2)\pi$
$\therefore rk+sk(k-1)=2=(k-2)n$ .
Suppose that $k$ is odd.
Then $k$ divides the left-hand side and $k$ is coprime to $k-2$ .
$\therefore k$ divides $n$ .
On the other hand if $k$ is even then $k/2$ is coprime to $(k-2)/2$ and hence $k$ divides $4n$ .
If $n$ is prime and $k<n$ then it follows that $k$ divides 4.
Since $k>2$ , we have proved that $k=4$ .