## Task A-2.5.

Let $n$ be a natural number. The vertices of a regular $2n$-gon are alternately colored red and blue, and all its sides and diagonals are drawn. If the number of segments connecting vertices of the same color is equal to 3192, determine the number of segments connecting vertices of different colors.
## First Solution.

The number of edges connecting blue vertices is $\frac{n(n-1)}{2}$. The same applies to the edges connecting red vertices, so from the given condition we have

$$
2 \cdot \frac{n(n-1)}{2}=n(n-1)=3192
$$

By factoring $3192=2^{3} \cdot 3 \cdot 7 \cdot 19=56 \cdot 57$, we see that the solutions to the quadratic equation are $n=57$ and $n=-56$. The only positive solution is $n=57$.

Each edge connecting vertices of different colors connects one red vertex and one blue vertex. Therefore, the desired number is $n \cdot n=57^{2}$.

## Second Solution.

Each vertex is connected to $n-1$ other vertices of the same color, and each edge connects two vertices. Therefore, the number of edges connecting vertices of the same color is

$$
\frac{2 n(n-1)}{2}=n(n-1)=3192
$$

The quadratic equation has exactly one positive real solution (and exactly one negative solution). Since $57 \cdot 56=3192$, it follows that $n=57$.

Each vertex is connected to $n$ other vertices of a different color. The desired number is

$$
\frac{2 n \cdot n}{2}=n^{2}=57^{2}=3249
$$

Note: The student must solve the quadratic equation $n(n-1)=3192$ or otherwise explain why 57 is the only possible solution. Without this explanation, the student should receive 1 point less.