Central Region Perimeter

The locus of points which are equidistance from the centroid and an edge straight line is a parabola, where the centroid is the focus point $F (x_f, y_f)$ and the edge straight line $ax+by+c = 0$ is the directrix. And the equation of the parabola is given by ( reference ):

$\frac {(ax+by+c)^2}{a^2+b^2} = (x-x_f)^2 + (y-y_f)^2$

To make the calculations easy, we set $(x_f, y_f) = (0,0)$ or we move $G(x_g, y_g)$ to $G'(0,0)$ . Since $x_g = \dfrac {x_a+x_b+x_c}3 = \dfrac {0+2+1}3 = 1$ and $y_g = \dfrac {y_a+y_b+y_c}3 = \dfrac {0+0+1}3 = \dfrac 13$ , then $A' = \left(x_a-1, y_a - \dfrac 13\right) = \left(-1,- \dfrac 13\right)$ , $B'= \left(1,- \dfrac 13\right)$ , and $C' = \left(0, \dfrac 23\right)$ . Then the equation of the three parabolas are:

$\begin{cases} 6y = 9x^2 - 1 & ...(1) \ \rm bottom \\ 9 (x-y)^2+12 (x+y)-4 = 0 & ...(2) \ \rm right \\ 9 (x+y)^2-12 (x-y)-4 = 0 & ...(3) \ \rm left \end{cases}$

The region shaded blue, supposedly bounded by a red curve but I hate to redo the figure, is the part of the triangle which is nearer to the centroid $G$ than an edge. Since the perimeter is symmetrical about the y-axis, we can find the length of curve of the right half and then multiply it by $2$ .

From $(1)$ , we get $y = \dfrac 32 x^2 - \dfrac 16 = y_1$ and from $(2)$ , $y=x+\dfrac {2\sqrt{2-6x}}3 - \dfrac 23 =y_2$ . Let parabolas $y_1$ and $y_2$ meet at $P(a,b)$ . From $y_1 = y_2$ , $a = \dfrac {1-\sqrt 2 + 2\sqrt{\sqrt 2-1}}3$ . And the perimeter of the blue region:

$\begin{aligned} p & = 2 \int_0^a \left(\sqrt{1 + \left(\frac {dy_1}{dx}\right)^2} + \sqrt{1 + \left(\frac {dy_2}{dx}\right)^2} \right) dx \\ & = 2 \int_0^a \left(\sqrt{1 + (3x)^2} + \sqrt{1 + \left(1 - \frac 2{\sqrt{2-6x}} \right)^2} \right) dx \\ & \approx \boxed{1.54} \end{aligned}$