Distance between triangle's centroid and incenter, given coordinates of vertices 
If $G$ is the centroid and $I$ is the in-center of the triangle, with vertices $A(-36,7)$, $B(20,7)$, and $C(0,-8)$, then find the length of $GI$.

Well the obvious way to approach this problem would be to centroid of the triangle and then the incenter of the triangle, and then find the distance. Is there an easier method to do this problem? Doing it that way would get a bit lengthy.
 A: The side lengths are $a=25,b=39,c=56$ by the pythagorean theorem and the area is $\Delta=420$ by the Heron's formula or the shoelace formula, hence $r=7$. Since $h_c=\frac{2\Delta}{c}$, the length of the height relative to $C$ is $15$, hence the distance of $G$ from $AB$ is just $5$. 
It is well known that $I=\frac{aA+bB+cC}{a+b+c}$ and $AI^2=bc-4rR=bc-\frac{2abc}{a+b+c}$, hence $IG^2$ can also be computed through the parallel axis theorem. Let we assume that $A$ has mass $a$, $B$ has mass $b$ and $C$ has mass $c$. With such assumptions, $I$ is the center of mass of $S=\{A,B,C\}$ and the moment of inertia of $S$ around $I$ is given by:
$$ M_I = a AI^2 + b BI^2 + c CI^2 = 3abc - 4(a+b+c) rR = abc $$
while the moment of inertia of $S$ around $G$ is given by:
$$ M_G = a AG^2 + b BG^2 + c CG^2 = \frac{1}{9}\sum_{cyc}a(2b^2+2c^2-a^2) $$
and by the parallel axis theorem $ M_G = (a+b+c) IG^2 + M_I$, hence:

$$ IG^2 = \frac{M_G-M_I}{a+b+c} = \boxed{\frac{\frac{2}{3}(a+b+c)^3-\frac{5}{3}(a^3+b^3+c^3)-13abc}{9(a+b+c)}}.$$


A difficult problem has just born:
 Prove that if $a,b,c$ are the side lengths of a triangle,
 $$ 2(a+b+c)^3 \geq 5 (a^3+b^3+c^3) + 39abc. $$
Tricky solution: $IG^2\geq 0$, and equality is achieved only when $I\equiv G$, i.e. when $a=b=c$.
