60th Belarusian Mathematical Olympiad

We number chess players in accordance with their power: $P_1,P_2,P_3,P_4,P_5,P_6$ (if the $i$'th player is stronger than the $j$'th one, then $i<j$). Suppose, contrary to our claim, that the tournament can be organized. Let chess player $P_i$ get $a_i$ points ($i=1,\dots ,6$). By conditions, $a_i-a_{i+1}\ge 0.5$ for $i=1,\dots ,5$. In particular $$a_i-a_{i+3}=(a_i-a_{i+1})+(a_{i+1}-a_{i+2})+(a_{i+2}-a_{i+3})\ge 0.5+0.5+0.5=\mathrm{1.5.}(1)$$ Let $A=\{P_1,P_2,P_3\}$, $B=\{P_4,P_5,P_6\}$, $S_A=a_1+a_2+a_3$, $S_B=a_4+a_5+a_6$. It is easy to see that $S_A+S_B=15$. Moreover $S_A-S_B\ge 4.5$. Indeed $$S_A-S_B=(a_1+a_2+a_3)-(a_4+a_5+a_6)=(a_1-a_4)+(a_2-a_5)+(a_3-a_6)\ge 1.5+1.5+1.5=\mathrm{4.5.}(2)$$ Let $S_Y(X)$ denote the total number of points got by the chess players of group X in the games with the chess players of group Y (here $X,Y\in \{A,B\}$). It is evident that $S_A(A)=S_B(B)=3$. It follows that $S_B(A)-S_A(B)=(S_A(A)+S_B(A))-(S_A(B)+S_B(B))=S_A-S_B$. The total number of the games between the players of A and B is equal to $3\cdot 3=9$, so $9\cdot 1=9$ points are distributed in these games. So for $S_A-S_B\ge 4.5$ it is necessary to have $S_B(A)\ge 7$. Otherwise, $S_B(A)\le 6.5$ gives $S_A(B)=9-S_B(A)\ge 2.5$, which implies $S_A-S_B\le 6.5-2.5=4$, contrary to (2). For $S_B(A)\ge 7$ it is necessary to use the fair board at least 5 times, otherwise, since only using the fair board gives 1 point to a stronger player, $S_B(A)\le 4\cdot 1+5\cdot 0.5=6.5$, which is impossible. Since the fair board is used exactly 5 times, the players from A must use the peaceful board for the remaining $4=9-5$ games with the players from B to obtain $S_B(A)\ge 7=5+4\cdot 0.5$. Therefore the peaceful board is used at most 1 time for the games between the players of the same group. It follows that $a_6\ge 1+0.5=1.5$. Then $$15=S_A+S_B=a_6+a_5+a_4+a_3+a_2+a_1\ge 1.5+2+2.5+3+3.5+4=16.5,$$ a contradiction. Hence it is impossible to organize the tournament satisfying the problem conditions.