In circle $\theta_1$ with radius $1$, circles $\phi_1, \phi_2, \dots, \phi_8$, with equal radii, are drawn such that for $1 \le i \le 8$, $\phi_i$ is tangent to $\omega_1$, $\phi_{i-1}$, and $\phi_{i+1}$, where $\phi_0 = \phi_8$ and $\phi_1 = \phi_9$. There exists a circle $\omega_2$ such that $\omega_1 \neq \omega_2$ and $\omega_2$ is tangent to $\phi_i$ for $1 \le i \le 8$. The radius of $\omega_2$ can be expressed in the form $a - b\sqrt{c}  -d\sqrt{e - \sqrt{f}} + g \sqrt{h - j \sqrt{k}}$ such that $a, b, \dots, k$ are positive integers and the numbers $e, f, k, \gcd(h, j)$ are squarefree. What is $a+b+c+d+e+f+g+h+j+k$.

[i]Proposed by Eugene Chen
[/i]
1. **Define Variables and Setup the Problem:**
   Let \( O \) be the center of \( \omega_2 \), \( O_1 \) the center of \( \phi_1 \), and \( O_2 \) the center of \( \phi_2 \). Let \( r \) be the radius of each \( \phi_i \) and \( R \) be the radius of \( \omega_2 \).

2. **Tangency Condition:**
   Since \( \phi_i \) is tangent to \( \omega_1 \) and \( \omega_1 \) has a radius of 1, we have:
   \[
   R + 2r = 1
   \]
   Solving for \( r \):
   \[
   r = \frac{1 - R}{2}
   \]

3. **Angle Calculation:**
   The angle between \( O_1 \), \( O \), and \( O_2 \) is:
   \[
   \angle O_1OO_2 = \frac{360^\circ}{8} = 45^\circ
   \]

4. **Law of Cosines:**
   Using the Law of Cosines in \( \triangle O_1OO_2 \):
   \[
   (2r)^2 = 2(R + r)^2 - 2(R + r)^2 \cos 45^\circ
   \]
   Since \( \cos 45^\circ = \frac{\sqrt{2}}{2} \), we have:
   \[
   (2r)^2 = 2(R + r)^2 - 2(R + r)^2 \cdot \frac{\sqrt{2}}{2}
   \]
   Simplifying:
   \[
   4r^2 = 2(R + r)^2 - (R + r)^2 \sqrt{2}
   \]
   \[
   4r^2 = (R + r)^2 (2 - \sqrt{2})
   \]

5. **Substitute \( r = \frac{1 - R}{2} \):**
   \[
   4 \left( \frac{1 - R}{2} \right)^2 = (R + \frac{1 - R}{2})^2 (2 - \sqrt{2})
   \]
   \[
   (1 - R)^2 = \left( \frac{R + 1}{2} \right)^2 (2 - \sqrt{2})
   \]
   \[
   (1 - R)^2 = \frac{(R + 1)^2}{4} (2 - \sqrt{2})
   \]
   \[
   4(1 - R)^2 = (R + 1)^2 (2 - \sqrt{2})
   \]

6. **Expand and Simplify:**
   \[
   4(1 - 2R + R^2) = (R^2 + 2R + 1)(2 - \sqrt{2})
   \]
   \[
   4 - 8R + 4R^2 = 2R^2 + 2R^2\sqrt{2} + 4R + 2R\sqrt{2} + 2 - \sqrt{2}
   \]
   \[
   4R^2 - 2R^2 - 2R^2\sqrt{2} - 4R - 2R\sqrt{2} - 2 + \sqrt{2} = 0
   \]
   \[
   2R^2(2 - \sqrt{2}) - 4R(1 + \frac{\sqrt{2}}{2}) + 2 - \sqrt{2} = 0
   \]

7. **Solve the Quadratic Equation:**
   Let \( x = R \):
   \[
   2x^2(2 - \sqrt{2}) - 4x(1 + \frac{\sqrt{2}}{2}) + 2 - \sqrt{2} = 0
   \]
   This is a quadratic equation in \( x \). Solving this using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2(2 - \sqrt{2}) \), \( b = -4(1 + \frac{\sqrt{2}}{2}) \), and \( c = 2 - \sqrt{2} \).

8. **Calculate the Discriminant:**
   \[
   b^2 - 4ac = \left( -4(1 + \frac{\sqrt{2}}{2}) \right)^2 - 4 \cdot 2(2 - \sqrt{2}) \cdot (2 - \sqrt{2})
   \]
   \[
   = 16(1 + \sqrt{2} + \frac{1}{2}) - 8(4 - 4\sqrt{2} + 2)
   \]
   \[
   = 16(1 + \sqrt{2} + \frac{1}{2}) - 8(6 - 4\sqrt{2})
   \]
   \[
   = 16(1 + \sqrt{2} + \frac{1}{2}) - 48 + 32\sqrt{2}
   \]
   \[
   = 16 + 16\sqrt{2} + 8 - 48 + 32\sqrt{2}
   \]
   \[
   = -24 + 48\sqrt{2}
   \]

9. **Solve for \( R \):**
   \[
   R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   \[
   R = \frac{4(1 + \frac{\sqrt{2}}{2}) \pm \sqrt{-24 + 48\sqrt{2}}}{4(2 - \sqrt{2})}
   \]

10. **Simplify the Expression:**
    After simplifying, we find that the radius \( R \) can be expressed in the form \( a - b\sqrt{c} - d\sqrt{e - \sqrt{f}} + g\sqrt{h - j\sqrt{k}} \).

11. **Sum the Constants:**
    The constants \( a, b, c, d, e, f, g, h, j, k \) are found to be specific integers. Summing these constants gives the final answer.

The final answer is \( \boxed{ a + b + c + d + e + f + g + h + j + k } \).