I don't know how to work in latex yet, so I'm going to write the solution in normal text form, as it's not that long. During the solution, 'AB' will mean the vector form of segment AB. First, let x,y,z in the interval (0,1), such that 'AM'='AB'*x , 'BN'='BC'*y , 'CP'='CA'*z. Since I is the center of mass of triangle MNP and 'GA'+'GB'+'GC'='0', we say 'GI' = ('GM'+'GN'+'GP')/3 = ('GA'+'AM'+'GB'+'BN'+'GC'+'CP')/3 = ('AM'+'BN'+'CP')/3 and, by replacement: 'GI' = ('AB'*x+'BC'*y+'CA'*z)/3. 'CA' = 'CB'+'BA', so 'GI' = 'AB'*(x-z)/3+'BC'*(y-z)/3 (normally in latex i'd make this last equation bolded, but since I can't, I'll just say here that it's important, we will use it later). I is the incenter of triangle ABC, so: 'GI' = ('GA'*a+'GB'*b+'GC'*c)/(a+b+c). We then replace GA, GB, GC: 'GI' = 'AB'*(b+c-2*a)/(3*a+3*b+3*c)+'BC'*(2*c-a-b)/(3*a+3*b+3*c). Using this equation and the one I marked as important before, we find: x-z = (b+c-2*a)/(a+b+c) y-z = (2*c-a-b)/(a+b+c), so: z = x-(b+c-2*a)/(a+b+c) which, then, must be in the interval (0,1), for any x y = x+(a+c-2*b)/(a+b+c) which, too, must be in (0,1) , for any x WLOG, suppose a>=b>=c, or a>=c>=b, then b+c<=2*a. If b+c<2*a, we choose x = 1+(b+c-2*a)/(a+b+c), which leads to z = 1, false. Then, 2*a = b+c. Since we supposed a>=b>=c, or a>=c>=b, it follows that a = b = c, so triangle ABC is equilateral. I hope the solution was as clear as a solution that's not in latex can be. I took >= from informatics, I thought it is clear it means bigger or equal. Of course, there's probably a solution without vectors, but I find this one easier.

I will go ahead and LaTeX @above's solution. First, let $x,y,z$ be real numbers from the interval $(0,1),$ such that \[\overrightarrow{AM}=\overrightarrow{AB}\cdot x, \overrightarrow{BN}=\overrightarrow{BC}\cdot y\text{ and } \overrightarrow{CP}=\overrightarrow{CA}\cdot z.\] Since $I$ is the center of mass of $\triangle MNP$ and $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0},$ we have \[\overrightarrow{GI} = \frac{\overrightarrow{GM}+\overrightarrow{GN}+\overrightarrow{GP}}{3} = \frac{\overrightarrow{GA}+\overrightarrow{AM}+\overrightarrow{GB}+\overrightarrow{BN}+\overrightarrow{GC}+\overrightarrow{CP}}{3} = \frac{\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}}{3}\] and, by replacing, we get \[\overrightarrow{GI} = \frac{(\overrightarrow{AB}\cdot x+\overrightarrow{BC}\cdot y+\overrightarrow{CA}\cdot z)}{3}.\] Note that $\overrightarrow{CA} =\overrightarrow{CB}+\overrightarrow{BA},$ so \[\overrightarrow{GI} = \frac{\overrightarrow{AB}\cdot (x-z)}{3}+\frac{\overrightarrow{BC}\cdot (y-z)}{3}.\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }(\dagger)\] $I$ is the incenter of $\triangle ABC,$ so \[\overrightarrow{GI} = \frac{\overrightarrow{GA}\cdot a+\overrightarrow{GB}\cdot b+\overrightarrow{GC}\cdot c}{a+b+c}.\] We then replace $GA, GB$ and $GC$ and we get \[\overrightarrow{GI} = \overrightarrow{AB}\cdot\frac{b+c-2a}{3a+3b+3c}+\overrightarrow{BC}\cdot\frac{2c-a-b}{3a+3b+3c}.\] Using the above equation and $(\dagger)$, notice that \[x-z = \frac{b+c-2a}{a+b+c}\text{ and }y-z = \frac{2c-a-b}{a+b+c}\]which gives us $z = x-\frac{b+c-2a}{a+b+c}$ so $z$ must be in the interval $(0,1)$ for any $x$ and $y = x+\frac{a+c-2b}{a+b+c}$ so $y$ must be too in $(0,1)$ for any $x.$ WLOG, suppose that $a\geq b\geq c,$ or $a\geq c\geq b$, which gives us $b+c\leq 2a.$ If $b+c<2a,$ we choose $x=1+\frac{b+c-2*a}{a+b+c},$ which leads to $z=1,$ false. Then, $2a = b+c.$ Since we supposed $a\geq b\geq c,$ or $a\geq c\geq b,$ it follows that $a = b = c$, so $\triangle ABC$ is equilateral.