Problem 2. Determine the perimeter of the scalene triangle $\mathrm{ABC}$ knowing that

$$
p^{2} \cdot \overrightarrow{G I}=(4 p-b-c) \cdot \overrightarrow{A I}+(4 p-c-a) \cdot \overrightarrow{B I}+(4 p-a-b) \cdot \overrightarrow{C I}
$$

where $\mathrm{G}$ is the centroid of $\triangle \mathrm{ABC}$, I is the incenter of $\triangle \mathrm{ABC}$, and $p$ is the semiperimeter of $\triangle \mathrm{ABC}$.
Solution:

$4 p-b-c=2 p+a, 4 p-c-a=2 p+b, 4 p-a-b=2 p+c$

The relationship from the statement becomes:

$p^{2} \cdot \overrightarrow{G I}=a \cdot \overrightarrow{A I}+b \cdot \overrightarrow{B I}+c \cdot \overrightarrow{C I}+2 p(\overrightarrow{A I}+\overrightarrow{B I}+\overrightarrow{C I})$

$1 \mathbf{p}$

$\overrightarrow{A I}=\frac{b \cdot \overrightarrow{A B}+c \cdot \overrightarrow{A C}}{a+b+c}, \overrightarrow{B I}=\frac{c \cdot \overrightarrow{B C}+a \cdot \overrightarrow{B A}}{a+b+c}, \overrightarrow{C I}=\frac{a \cdot \overrightarrow{C A}+b \cdot \overrightarrow{C B}}{a+b+c} \Rightarrow$

$a \cdot \overrightarrow{A I}+b \cdot \overrightarrow{B I}+c \cdot \overrightarrow{C I}=\overrightarrow{0}$

3 p

According to the Leibniz relation

$3 \overrightarrow{G I}=\overrightarrow{A I}+\overrightarrow{B I}+\overrightarrow{C I} \Rightarrow$

$1 p$

$p^{2} \cdot \overrightarrow{G I}=6 p \cdot \overrightarrow{G I} \Leftrightarrow\left(p^{2}-6 p\right) \cdot \overrightarrow{G I}=\overrightarrow{0}$

$\triangle \mathrm{ABC}$ non-equilateral $\Rightarrow G \neq I \Rightarrow\left(p^{2}-6 p\right)=0 \Rightarrow p=6 \Rightarrow P_{\triangle \mathrm{ABC}}=12$.

2p