Consider a chess tournament involving humans and aliens where each player plays against all others exactly once. At the end of the tournament, each player obtained half of the points they scored playing against humans and half playing against aliens. Let \( t \) and \( a \) be the number of humans and aliens, respectively. Answer the following:

a) What is the total number of games in this tournament, in terms of \( t \) and \( a \)?

b) In this tournament, each win is worth 1 point, each draw is worth 0.5 points, and each loss is worth zero points. What is the total number of points for the humans in terms of \( t \) and \( a \)?

c) Verify that the total number of players in the tournament is a perfect square.
Solution

a) Let $n=t+a$ be the total number of chess players present, with each pair facing off in unique matches, the tournament had

$$
\frac{n(n-1)}{2}=\frac{(t+a)(t+a-1)}{2} \text { matches. }
$$

b) If we sum the points each athlete received per game (regardless of the outcome), we will always get a sum of 1 per game. Thus, there is a direct relationship between the number of games and the points distributed. Since each Earthling won half of their points playing against opponents of the same race and the other half against aliens, the total points that the Earthlings scored against Earthlings is equal to the number of points that the Earthlings scored against the aliens. Each Earthling played $\frac{t(t-1)}{2}$ matches against Earthlings (total points of Earthlings against Earthlings) and $\frac{a(a-1)}{2}$ against aliens (total points of Earthlings against aliens). The same reasoning applies to the aliens. Therefore, the total points of the Earthlings were

$$
\frac{t(t-1)}{2}+\frac{a(a-1)}{2}
$$

c) In light of the previous items, the total number of points distributed coincides with the total number of games $\frac{n(n-1)}{2}$, and is also equal to the sum of the points of all Earthlings against their own and against those of another race, plus the sum of all points obtained by the aliens among themselves and against the Earthlings. Therefore, we can write that:

$$
\begin{aligned}
\frac{(t+a)(t+a-1)}{2} & =2 \cdot \frac{t(t-1)}{2}+2 \cdot \frac{a(a-1)}{2} \\
\left(t^{2}+t a-t+a t+a^{2}-a\right) & =2\left(t^{2}-t+a^{2}-a\right) \\
a^{2}-2 a t+t^{2} & =t+a \\
(a-t)^{2} & =n
\end{aligned}
$$

Thus, we can express $n$ as a perfect square.