In the triangle $ABC$, we divided each side into $p$ equal parts, where $p$ is a prime number. Then, from each side, we connected one of the division points to the opposite vertex of the triangle in such a way that these three lines intersect at a single point. Determine the possible values of $p$.
Solution. Suppose that on the side $AB$, we connect the $c$-th division point from $A$, on the side $BC$, the $a$-th division point from $B$, and on the side $CA$, the $b$-th division point from $C$, to the vertices $C$, $A$, and $B$ respectively, where $a$, $b$, and $c$ are positive integers less than $p$. According to Ceva's theorem,

$$
\frac{a}{p-a} \cdot \frac{b}{p-b} \cdot \frac{c}{p-c}=1
$$

which means $a b c=(p-a)(p-b)(p-c)$. Expanding the right-hand side, the only term that does not contain $p$ as a factor is $-a b c$; therefore, by rearranging the equation to solve for $a b c$, we get that $2 a b c$ is divisible by the prime number $p$. This is only possible if one of the factors in the product is divisible by $p$. The numbers $a$, $b$, and $c$ cannot be such, as they are positive integers less than $p$; hence, $p$ must be a divisor of 2, i.e., $p=2$. In this case, the segments of the three lines within the triangle are the medians of the triangle, which intersect at the centroid.