Problem 12 · The ten-team tournament
Maximising points = minimising draws under the constraints.
Integer answer, at most 4 digitsTen teams played a football tournament. Each team faced each other team exactly once; each match awarded three points to the winner and zero to the loser, or one point each in case of a draw. No team won all its matches, and none lost all of them; only one team lost no match, yet it scored fewer points than the top team. The ten teams' scores were added up. What is the largest possible value of this sum?
Reasoned solution
Key idea: there are $\binom{10}{2} = 45$ matches. Each match distributes $3$ points if decisive and $2$ if drawn: the total is $135 - (\text{number of draws})$. Maximising the total means minimising draws.
At least $2$ draws are needed. Let $U$ be the unique unbeaten team. $U$ didn't win everything, so it has some draw. If it had only one, $U$ would have $8$ wins and $1$ draw: $25$ points. But the top team is not $U$ and did lose some match, so it has at most $8$ wins and $1$ loss: $24 < 25$ points — contradiction. Hence $U$ has at least $2$ draws, and so does the tournament.
$2$ draws are achievable: $U$ draws with two teams and beats the other $7$ → $7\cdot 3 + 2 = 23$ points. Another team $T$ wins $8$ matches and loses only one, scoring $24 > 23$; all remaining matches are decisive, arranged so that no team loses everything (easy to set up).