Let $n \ge 4$ be a natural number. Let $A_1A_2 \cdots A_n$ be a regular polygon and $X = \{ 1,2,3....,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.
Problem
Source: RMO - Problem 6
Tags: calculus, integration, arithmetic sequence, number theory unsolved, number theory
dibyo_99
11.12.2013 09:58
Firstly, note that the angle between any $3$ vertices of the polygon is an integral multiple of $\pi/n$. This is because twice this angle is an angle subtended at the centre by a chord joining $2$ vertices of the polygon and therefore is a multiple of $\2\pi/n$ Let the arithmetic progression have $\frac{a\pi}{n}$ as the first term and $\frac{d\pi}{n}$ as common difference. The sum of the angles of this polygon is therefore $\frac{k\pi(2a+(k-1)d)}{2n}$. But, it is well know that the sum of the angles of a $k$-sided polygon is $(k-2)\pi$. We equate these sums and get $2n(k-2)=k(2a+(k-1)d)$. Therefore, $k|2n(k-2) \implies k|4n$. Obviously $n$ and $k$ are both $>2$. Therefore, the only factors of $4n$ are $1,2,4,n,2n,4n$. Since $k <n$ and $k>2$, we must have $k=4$.
guptaamitu1
13.10.2020 17:04
I have an alternative way(probably an easier way) to proceed after we get the shaded equation (I am talking about the official solution and I have also attached that here). So here goes the alternative way: the shaded equation is equivalent to k(2r + s(k-1)) = (2k-4)n ... (1) now the RHS is divisible by n, so the LHS must be divisible by n. since gcd (k,n) = 1 (as k < n, n is a prime), therefore n divides 2r + s(k-1). Now if 2r + s(k-1) >= 2n, then (1) implies that 2k-4 >= 2k. A contradiction. Also, as 2r + s(k-1) > 0 (as r > 0), therefore 2r + s(k-1) = n, so (1) is equivalent to k = 2k - 4. Hence, k=4
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