Solution:

Let $r_i = A_1A_i$ for $2 \le i \le 11,$ and let $\theta$ be the common angle, with $\cos\theta = \frac{12}{13}$ and $\sin\theta = \frac{5}{13}.$ Each area condition says $\frac{1}{2} r_i r_{i+1} \cdot \frac{5}{13} = 1,$ so $r_i r_{i+1} = \frac{26}{5}$ for $i = 2, \ldots, 10.$ Consecutive products being equal forces the $r_i$ to alternate between two values $a = r_2 = r_4 = \cdots$ and $b = r_3 = r_5 = \cdots,$ with $ab = \frac{26}{5};$ in particular $r_{11} = b.$

By the law of cosines, every side $A_iA_{i+1}$ with $2 \le i \le 10$ has the same length $s,$ where $$s^2 = a^2 + b^2 - 2ab \cdot \tfrac{12}{13} = (a + b)^2 - 2ab - \tfrac{48}{5} = (a+b)^2 - 20.$$ Writing $u = a + b,$ the perimeter condition is $9\sqrt{u^2 - 20} + u = 20.$ Squaring $9\sqrt{u^2 - 20} = 20 - u$ gives $81u^2 - 1620 = 400 - 40u + u^2,$ which simplifies to $4u^2 + 2u - 101 = 0,$ so $u = \frac{-1 + 9\sqrt{5}}{4}$ (the positive root; then $20 - u \gt 0$ as required).

Thus $A_1A_2 + A_1A_{11} = a + b = \frac{9\sqrt{5} - 1}{4},$ with $5$ squarefree and no prime dividing all of $9,$ $1,$ $4.$ The answer is $9 + 5 + 1 + 4 = 19.$