74th NMO Selection Tests for JBMO

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 $S_1=u_1{v}_1+u_2{v}_2+\cdots +u_{2n}{v}_{2n}$, $S_2=u_1{v}_2+u_2{v}_3+\cdots +u_{2n}{v}_1$, ..., $S_{2n}=u_1{v}_{2n}+u_2{v}_1+\cdots +u_{2n}{v}_{2n-1}$.

Moreover, we have $$S_1+S_2+\cdots +S_{2n}=(u_1+u_2+\cdots +u_n)(v_1+v_2+\cdots +v_n)=0.$$ 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 $t\ge 2\lfloor \frac{n}{2}\rfloor +2$.