Answer: 245

Small Hint:

With $AB = BC = a$ and $AC = b,$ the incircle touches $AC$ at its midpoint, so $AI^2 = r^2 + \frac{b^2}{4};$ Heron’s formula simplifies this to $AI^2 = \frac{ab^2}{2a + b}.$

Big Hint:

Set $\sqrt{\frac{a}{2a + b}} = \frac{p}{q}$ in lowest terms; the perimeter ratio becomes $125(q^2 - 2p^2)(q + 2p) = 6q^3,$ and divisibility pins down $q.$

Solution:

Let $a = AB = BC$ and $b = AC,$ so $s = a + \frac{b}{2}.$ The incircle touches $AC$ at its midpoint (tangent length from $A$ is $s - a = \frac{b}{2}$), so $AI^2 = CI^2 = r^2 + \frac{b^2}{4}.$ By Heron’s formula, $r^2 = \frac{(s-a)^2(s-b)}{s} = \frac{b^2}{4} \cdot \frac{2a - b}{2a + b},$ and therefore $$\begin{aligned} AI^2 &= \frac{b^2}{4}\left(\frac{2a - b}{2a + b} + 1\right) \\ &= \frac{ab^2}{2a + b}. \end{aligned}$$ The perimeter condition is $125\,(2\,AI + b) = 6\,(2a + b).$

Since $AI$ is rational, write $\sqrt{\frac{a}{2a + b}} = \frac{p}{q}$ in lowest terms. Then $aq^2 = p^2(2a + b)$ forces $a$ to be divisible by $p^2;$ writing $a = mp^2$ gives $b = m(q^2 - 2p^2),$ $2a + b = mq^2,$ and $AI = \frac{mp(q^2 - 2p^2)}{q}.$ The perimeter condition then loses $m$ entirely: $$125\,(q^2 - 2p^2)(2p + q) = 6q^3.$$ Since $\gcd(125, 6) = 1$ we get that $q$ is divisible by $5;$ and $q$ must be even, since for odd $q$ both factors on the left are odd while the right side is even. Writing $q = 10w$ and simplifying, $(50w^2 - p^2)(p + 5w) = 12w^3.$ Both factors on the left are coprime to $w$ (as $\gcd(p, q) = 1$). If a prime divided $w,$ the right side would make it divide their product, a contradiction; hence $w = 1.$ Now $(50 - p^2)(p + 5) = 12$ has the unique positive solution $p = 7.$

So $a = 49m,$ $b = 2m,$ and $AI = \frac{7m}{5},$ which is an integer exactly when $m$ is divisible by $5.$ Taking $m = 5$ gives $\triangle ABC$ with sides $245, 245, 10$ and $\triangle AIC$ with sides $7, 7, 10,$ whose perimeters $500$ and $24$ are indeed in ratio $125 : 6.$ The minimum possible $AB$ is $245.$