Problem 1462

For any two Welsh darts boards that can be matched, if two of their superimposed sectors have the same color, we call it a *concordance*, and if two of their superimposed sectors have different colors, we call it a *non-concordance*.

On each of the two Welsh darts boards that can be matched, we write $+1$ on every red sector, and $-1$ on every white sector. At some matching of the boards, the product of the numbers written in the overlapping sectors equals $+1$ in the case of a concordance, respectively $-1$, in case of a non-concordance.

Moreover, if $t$ is the number of all the concordances (whites and reds), the sum $S$ of the products of the numbers written on the overlapping sectors represents the difference between the number of concordances and the number of non-concordances, therefore $S = t \cdot 1 + (2n - t) \cdot (-1) = 2(t - n)$.

At any matching, let $k$ be the number of concordances between white sectors and $j$ the number of concordances between red sectors. Consequently, $n-k$ white sectors of the first board (those for which we have non-concordances) overlap over $n-k$ red sectors of the second board, and $n-j$ red sectors of the first board overlap over $n-j$ white sectors of the second board.

Therefore, on the second board we have $k + (n-j)$ white sectors and $j + (n-k)$ red sectors, hence $k + (n-j) = j + (n-k) = n$, thus $k = j$. Consequently, the number $t = k + j$ of all concordances is even.

Consider two matched Welsh darts boards, such that they have more than $n$ non-concordances. Let $u_1, u_2, \dots, u_{2n} \in \{-1, +1\}$ be the numbers written (clockwise) on the sectors of the first board and $v_1, v_2, \dots, v_{2n} \in \{-1, +1\}$ the numbers written on the correspondent sectors of the second board.

By fixing the sector with the number $u_1$ and by rotating the second board, we obtain all the possible matchings, and the sums $S_1 = u_1v_1 + u_2v_2 + \dots + u_{2n}v_{2n}$, $S_2 = u_1v_2 + u_2v_3 + \dots + u_{2n}v_1$, ..., $S_{2n} = u_1v_{2n} + u_2v_1 + \dots + u_{2n}v_{2n-1}$.

Moreover, we have
$$S_1 + S_2 + \dots + S_{2n} = (u_1 + u_2 + \dots + u_n)(v_1 + v_2 + \dots + v_n) = 0.$$
Since initially there were more than $n$ non-concordances between the boards, we have $S_1 < 0$. Therefore, $j = \frac{1}{2n}$ exists, such that $S_j > 0$. If $t$ is the number of all the concordances of the sum $S_j$, then $2(t-n) > 0$, therefore $2(t-n) \ge 2$. We obtain $t \ge n+1$ and since $t$ is even, it follows that $t \ge 2\left\lfloor\frac{n}{2}\right\rfloor + 2$.