Answer: 20

Small Hint:

Internal tangency with the excircle opposite $A$ forces the radius of $\omega$ to be $r + 2r_A,$ where $r$ is the inradius and $r_A$ that exradius

Big Hint:

With $BC$ on the $x$-axis, the $B$-excircle has radius equal to the height $h$ from $A;$ the external tangency collapses to one linear relation between base and leg

Solution:

Let $BC = a$ and $AB = AC = b.$ Place $B = \left(-\frac{a}{2}, 0\right),$ $C = \left(\frac{a}{2}, 0\right),$ $A = (0, h)$ with $h = \sqrt{b^2 - \frac{a^2}{4}},$ so the semiperimeter is $s = b + \frac{a}{2}$ and the area is $K = \frac{ah}{2}.$ The inradius and exradii are $r = \frac{K}{s} = \frac{ah}{a + 2b},$ $r_A = \frac{K}{s - a} = \frac{ah}{2b - a},$ and $r_B = \frac{K}{s - b} = h.$ The incenter is $I = (0, r)$ and the $A$-excircle has center $(0, -r_A).$ The $B$-excircle touches line $BC$ at distance $s$ from $B,$ that is, at $x = b,$ so its center is $(b, h).$

Internal tangency with the $A$-excircle: the center distance is $r + r_A,$ so the radius $\rho$ of $\omega$ satisfies $\rho - r_A = r + r_A,$ i.e. $\rho = r + 2r_A.$ External tangency with the $B$-excircle requires $b^2 + (h - r)^2 = (\rho + h)^2$ $= (h + r + 2r_A)^2,$ which rearranges to $b^2 = 4(r + r_A)(h + r_A).$ Since $$r + r_A = \frac{4abh}{4b^2 - a^2}$$ and $$h + r_A = \frac{2bh}{2b - a},$$ and $h^2 = \frac{4b^2 - a^2}{4},$ the condition becomes $b^2 = \frac{8ab^2}{2b - a},$ that is, $2b - a = 8a,$ so $2b = 9a.$

For integer sides, $a = 2t$ and $b = 9t$ for a positive integer $t,$ giving perimeter $20t.$ The minimum is $20,$ achieved by the triangle with sides $9,$ $9,$ $2.$