Problem 1427

1. Restate the problem: We need to show that the sum of vectors belonging to any proper subset of $M = \{ \overrightarrow{OA_j} \mid j = 1, 2, \ldots, 1987 \}$ is nonzero for a regular 1987-gon with center $O$.

2. Prime number property: We start by noting that 1987 is a prime number. We will prove that the result holds for any $n$-gon if $n$ is prime.

3. Non-prime case: If $n$ is not prime, then $n$ has a divisor $1 < d < n$. In this case, the vertices $A_d, A_{2d}, \ldots, A_{nd}$ form a regular $\frac{n}{d}$-gon. The sum of the vectors corresponding to these vertices is zero because they are symmetrically distributed around the center $O$.

4. Prime case: Now, suppose $n$ is prime. Let $\zeta = e^{\frac{2\pi i}{n}}$ be a primitive $n$-th root of unity. The problem reduces to showing that there is no proper subset $I \subset \{1, 2, \ldots, n\}$ such that $\sum_{i \in I} \zeta^i = 0$.

5. Minimal polynomial: The minimal polynomial of $\zeta$ over the rationals is the $n$-th cyclotomic polynomial $\Phi_n(x) = 1 + x + x^2 + \cdots + x^{n-1}$. Since $n$ is prime, $\Phi_n(x)$ is irreducible over the rationals.

6. **Polynomial $P(x)$**: Suppose there exists a subset $I \subset \{1, 2, \ldots, n\}$ such that $\sum_{i \in I} \zeta^i = 0$. Define the polynomial $P(x) = \sum_{i \in I} x^i$. If $P(\zeta) = 0$, then $\Phi_n(x)$ must divide $P(x)$.

7. Degree and coefficients: The polynomial $P(x)$ has degree at most $n-1$ and all coefficients are either 0 or 1. Since $\Phi_n(x)$ is irreducible and has degree $n-1$, the only way $\Phi_n(x)$ can divide $P(x)$ is if $P(x) = \Phi_n(x)$.

8. Contradiction: If $P(x) = \Phi_n(x)$, then $I = \{1, 2, \ldots, n\}$, which is not a proper subset. This is a contradiction, as we assumed $I$ was a proper subset.

9. Conclusion: Therefore, for a prime $n$, the sum of vectors belonging to any proper subset of $M$ is nonzero.

$\blacksquare$