Round Robin Tournament

First, WLOG, label the teams $1,2, \ldots,13$ . Then, to represent the problem in matrix (or table) form, see the picture below:

Matrix where $1$ represents a win for the team represented in the column and $0$ represents a loss for the team also in the column. For example, if Team $1$ wins against Team $2$ we place a $1$ in the cell $(1,2)$ and a $0$ at cell $(2,1)$ where cell $(x,y)$ is found at the intersection of column $x$ and row $y$ .

In the matrix, each column (row too) will have 6 $1$ s and $0$ s. Hence, there will two "symmetrically inverse" triangles (each with equal numbers of $1$ and $0$ as each other) in the matrix over the yellow coloured diagonals (where the team are matched against itself). Symmetric in the sense that if in one triangle has a cell $(x_1,y_1)$ with $1$ , then the cell in the other triangle reflected over the yellow diagonal will be $0$ or the opposite of $1$ . ( Can you explain why? )

As such, we only need to focus on one triangle, namely the blue one highlighted above. Notice that for a circular triangle to be found in a set of $3$ unordered and distinct teams say $(A,B,C)$ then, every team has to win $1$ game in that set.

• A wins B.
• B wins C. (forced since B already lost once.)
• C wins A.

In addition to satisfy the conditions for a circular triangle, we see that the triangle must have a $0$ and $1$ in the column with $2$ cells and $1$ in the other as shown - refer to the green triangle (sry the matrix as a whole doesn't make sense): ( Can you see why? - Hint: Try to plot the above ABC team above )

Example triangle

Thus, we just need to find number of triangles of the form above in our matrix to be done!

Lemma Realise that the matrix shows one example of a possible configuration of loss and wins among the teams. Now, I claim that the number of circular triangles for $13$ teams will be the same no matter how I rearrange the $1$ s and $0$ s in my matrix. I don't have a vigorous proof for this but you can see it as the teams are not ordered meaning that if I put Team $1$ won against Team $2$ here, it could jolly well mean the same thing as saying that Team $5$ in another matrix won against Team $7$ .

Solutions! So, I arrange the wins and losses in the matrix above to facilitate counting of the triangles of the form above. Notice that if the first cell I choose is $(1,2)$ , then, I can only have cells $(1,8)$ and cell $(2,8)$ to form a circular triangle. If I chose cell $(1,3)$ , then, I can have 2 choices: cells $(1,8)$ and $(3,8$ or cells $(1,9)$ and $(3,9)$ . Repeating this we can just get: - For column $C1$ : there are $1+2+3+4+5+6 = 21$ circular triangles formed.

• For $C2$ : $1+2+3+4+5+5 = 20$
• For $C3$ : $1+2+3+4+4+4 = 18$
• For $C4$ : $1+2+3+3+3+3 = 15$
• For $C5$ : $1+2+2+2+2+2 = 11$
• For $C6$ : $1+1+1+1+1+1 = 6$
• For $C7$ to $C13$ notice that there are no circular triangles because there simply isn't a $0$ in these columns! (this is unlike the green triangle example given).

Hence, adding up we get total number of circular triangles is $21+20+18+15+11+6 = \boxed{91}$ .

We say a player "blocks" a triangle when it loses to other 2 players or wins over 2 players:

Now, every player blocks binomial(6,2) triangles among its 6 loses and blocks binomial(6,2) triangles among its 6 wins.

So every player blocks 15+15=30 triangles. Therefore there are 30*13 = 390 blocked triangles

WAIT NO! Each triangle is blocked twice, so there are 390/2 = 195 blocked triangles.

Since the total of triangles is binomial(13,3) =286, we conclude there are 286 - 195 = 91 circular triangles.