An Unusual Equable Triangle

So we can use $s$ for the triangle's semiperimeter, let's define the shortest side to be $a$ (which matches the diagram in the question).

All equable triangles (indeed, all equable tangential polygons) have inradius $2$ ; just equate the area $rs$ and the perimeter $2s$ .

Now it's time to look up formulas for the distances between centres. In any triangle, the distance between the incentre $I$ and the centre of the nine-point circle $N$ is $IN = \frac{R-2r}{2}$

The distance between the incentre and the centroid $G$ is $IG =\frac13 \sqrt{5r^2-16Rr+s^2}$

In this case, we want $r=IG=IN=2$ ; solving the above we get $R=8,\;\;\;s=4\sqrt{17}$

By definition of $s$ , we have $a+b+c=8\sqrt{17}$ . Now, using $2Rr=\frac{abc}{a+b+c}$ we get $abc=256\sqrt{17}$

Equating area formulas, $\begin{aligned} rs&=\sqrt{s(s-a)(s-b)(s-c)} \\ r^2 s &= (s-a)(s-b)(s-c) \\ &= s^3-(a+b+c)s^2+(bc+ca+ab)s-abc \\ &= -s^3+(bc+ca+ab)s-abc \end{aligned}$

Rearranging, this gives $bc+ca+ab=\frac{r^2 s+s^3+abc}{s} = 340$

So the sidelengths $a,b,c$ are the roots of the cubic $x^3-8\sqrt{17}x^2+340x-256\sqrt{17}=0$

I don't think this factorises nicely; numerically, the smallest root is $5.76622\ldots$ giving the answer $\boxed{576622}$ .