IMTC

Let $f(x) = 12\,r_1 x^3 + 72\,r_2 x^2 + 432\,r_3 x$. Clearly $f(0) = 0$, so one of $r_1, r_2, r_3$ is equal to zero. If $r_1 = 0$, $f$ is quadratic and cannot have three distinct roots. If $r_3 = 0$, then $x^2$ divides $f(x)$ and $f$ must have a root $0$ with multiplicity two — forbidden by the distinctness condition. Thus $r_2 = 0$.

Now $f(x) = 12x(r_1 x^2 + 36\,r_3)$. The factor of $x$ gives the root $0$, so $r_1, r_3$ must be roots of $r_1 x^2 + 36\,r_3 = 0$. By Vieta's $r_1 + r_3 = 0$, so let $r_3 = -r_1$. Substituting gives $r_1(x^2 - 36) = 0$, and plugging $x = r_1$ yields $r_1 = \pm 6$, so $r_3 = \mp 6$.

The answer is $0^2 + (\pm 6)^2 + (\mp 6)^2 = \boxed{72}$.

We work backwards. For any $a_n$, either

Case 1: $a_n > 1$. The first equation gives a negative $a_{n-1}$, which is impossible for $n-1 \geq 1$ since every later term is an absolute value. The second gives a positive result less than $1$.

Case 2: $a_n < 1$. The first equation gives a positive solution greater than $1$, the second a positive solution less than $1$.

Case 3: $a_n = 1$. Always $a_{n-1} = \tfrac{1}{2}$. And it is impossible for any $a_n$ other than $a_{11}$ to equal $1$, because that would force $a_{n+1} = 0$ too early.

So for $2 \leq n \leq 11$: if $a_n > 1$ there is exactly one valid $a_{n-1}$ (with $a_{n-1} < 1$); if $0 < a_n < 1$ there are two, one satisfying $a_{n-1} > 1$ and one satisfying $0 < a_{n-1} < 1$.

We know $a_{12} = 0$, so $a_{11} = 1$ and $a_{10} = \tfrac{1}{2}$. Let $g_n$ and $l_n$ be the number of valid $a_n$ greater than and less than $1$ respectively, for $1 \leq n \leq 11$. Then $g_{n-1} = l_n$ and $l_{n-1} = l_n + g_n$, so $l_{n-1} = l_n + l_{n+1}$ — the Fibonacci recurrence in reverse.

With $l_{11} = 0$ and $l_{10} = 1$, we get $l_1 = 55$ (10th Fibonacci) and $g_1 = 34$ (9th Fibonacci). So $a_1$ has $55 + 34 = 89$ total solutions. Recalling that $a_0$ can be positive or negative, we double: $89 \cdot 2 = \boxed{178}$.

Let $M$ be the midpoint of $AB$. Since tangents $BA$ and $BC$ have equal length, the power of $B$ with respect to $\omega_1$ and $\omega_2$ is equal, so $B$ lies on the radical axis of $\omega_1, \omega_2$. Call the center of $\omega_1$ the point $J$.

Let $\omega_3$ be the degenerate circle of radius $0$ centered at $B$, and let $r$ be the radius of $\omega_1$. Since $B$ lies on the radical axis of $\omega_1, \omega_2$, we have $BJ^2 - r^2 = CB^2$. But $BJ = CJ$, so $CJ^2 - r^2 = CB^2$. The power of $C$ w.r.t. $\omega_3$ is $CB^2$, and w.r.t. $\omega_1$ is $CJ^2 - r^2$. These being equal, $C$ lies on the radical axis of $\omega_1, \omega_3$. Since $M$ is the midpoint of $AB$ its powers w.r.t. $\omega_1$ and $\omega_3$ also match, so $CM$ is the radical axis of $\omega_1, \omega_3$.

Construct $TB$ and let it meet $MC$ at $R$. Since $TB$ is the radical axis of $\omega_1, \omega_2$, $R$ is the radical center of $\omega_1, \omega_2, \omega_3$. So $BR = RT$, making $MR$ a midsegment of $\triangle BAT$ with $MR = \tfrac{AT}{2} = 26$.

Let the altitude from $B$ to $AT$ meet $MR$ at $F$ and $AT$ at $G$. Since $\triangle ABT$ is isosceles (equal tangents $AB, TB$),

Then $FC = \sqrt{BC^2 - BF^2} = 62$, $FR = \tfrac{MR}{2} = 13$, and so $RC = 49$. Power of a point at $R$ w.r.t. $\omega_2$ gives $RT^2 = RE \cdot RC$, i.e. $RE = \tfrac{35^2}{49} = 25$. Finally,