3.- Let $p \geq 3$ be a prime number. Each side of a triangle is divided into $p$ equal parts, and each of the division points is connected to the opposite vertex. Calculate the maximum number of regions, disjoint from each other, into which the triangle is divided.

## SOLUTION:

## SOLUTION:
First, we will see that three of these segments (cevians) cannot be concurrent. Let the triangle $ABC$ and $X, Y, Z$ be points of the internal divisions of the sides $BC, AC, AB$ respectively. If $AX, BY$, and $CZ$ were concurrent, applying Ceva's theorem we would have

$$
\frac{AZ}{ZB} \cdot \frac{BX}{XC} \cdot \frac{CY}{YA}=1
$$

On the other hand, by the way we have constructed the division points, there exist positive integers $k, l, m \in\{1,2, \ldots, p-1\}$, such that

$$
\frac{AZ}{ZB}=\frac{k}{p-k}, \quad \frac{BX}{XC}=\frac{l}{p-l}, \quad \frac{CY}{YA}=\frac{m}{p-m}
$$

Substituting in the above expression

$$
klm=(p-k)(p-l)(p-m)
$$

or equivalently,

$$
2klm + pm(p-k-l) = p(p-k)(p-l)
$$

From this, it follows that $p$ divides the product $klm$, which is impossible, and our initial statement is proven.

Drawing the cevians from vertex $A$, the triangle $ABC$ is divided into $p$ triangles. The cevians drawn from $B$ divide each of the $p$ triangles into $p$ disjoint parts, totaling $p^2$ regions. Each cevian drawn from vertex $C$ increases the number of regions by a number exactly equal to its number of intersections with the lines it encounters (including side $AB$). That is, $2(p-1)+1=2p-1$. Therefore, the maximum number of pairwise disjoint regions into which the triangle $ABC$ is divided is $p^2 + (p-1)(2p-1) = 3p^2 - 3p + 1$.