In $\triangle ABC$, the sides have integers lengths and $AB=AC$. Circle $\omega$ has its center at the incenter of $\triangle ABC$. An [i]excircle[/i] of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$, and the other two excircles are both externally tangent to $\omega$. Find the minimum possible value of the perimeter of $\triangle ABC$.
1. **Given Information and Assumptions:**
   - We have an isosceles triangle \( \triangle ABC \) with \( AB = AC \).
   - The incenter of \( \triangle ABC \) is the center of circle \( \omega \).
   - The excircle tangent to \( \overline{BC} \) is internally tangent to \( \omega \).
   - The other two excircles are externally tangent to \( \omega \).

2. **Define Key Points and Circles:**
   - Let \( I \) be the incenter of \( \triangle ABC \).
   - Let \( I_A, I_B, I_C \) be the excenters opposite vertices \( A, B, C \) respectively.
   - Let \( \omega_A, \omega_B, \omega_C \) be the excircles opposite vertices \( A, B, C \) respectively.
   - Let \( \Omega \) be the circle internally tangent to \( \omega_A \) and externally tangent to \( \omega_B \) and \( \omega_C \).

3. **Geometric Relationships:**
   - Since \( \omega_B \) and \( \Omega \) are externally tangent, the points \( X, T_B, K \) are collinear, where \( X \) is the tangency point of \( \omega_B \) with \( BC \), \( T_B \) is the tangency point of \( \Omega \) with \( \omega_B \), and \( K \) is the antipode of \( T_A \) on \( \Omega \).
   - The line \( BC \) is the Monge D'Alembert Line, passing through the insimilicenters and exsimilicenters of the excircles.

4. **Radical Center and Harmonic Division:**
   - The radical center of \( \omega_A, \omega_B, \omega_C \) lies on the line \( T_BX \) and also on the line \( AM \) by symmetry.
   - The point \( K \) lies on both lines \( XT_B \) and \( AM \), making \( K \) the radical center of \( \omega_A, \omega_B, \omega_C \).

5. **Calculation of Perimeter:**
   - Let \( S \) be the midpoint of segment \( YZ \), where \( Y \) and \( Z \) are tangency points of \( \omega_B \) and \( \omega_A \) with \( AC \) respectively.
   - The power of point \( S \) with respect to \( \omega_A \) and \( \omega_B \) is equal, making \( KS \) the radical axis of the two circles.
   - Using the properties of similar triangles and harmonic division, we find the ratio \( \frac{AI}{AI_A} = \frac{4}{5} \).

6. **Trigonometric Relationships:**
   - Using the harmonic pencil property, we find \( \frac{IM}{MI_A} = \frac{4}{5} \).
   - This implies \( \left( \tan \frac{B}{2} \right)^2 = \frac{4}{5} \), leading to \( \tan \frac{B}{2} = \frac{2}{\sqrt{5}} \).
   - Solving for \( \cos B \), we get \( \cos B = \frac{1}{9} \).

7. **Final Calculation:**
   - Given \( \cos B = \frac{1}{9} \), we find \( BC = \frac{2}{9} \).
   - The minimum possible perimeter occurs when \( BC = 2 \) and \( AB = AC = 9 \), giving a perimeter of \( 2 + 9 + 9 = 20 \).

The final answer is \( \boxed{20} \).