Let $\omega$ be a circle with radius $1$. Equilateral triangle $\vartriangle ABC$ is tangent to $\omega$  at the midpoint of side $BC$ and $\omega$  lies outside  $\vartriangle ABC$. If line $AB$ is tangent to $\omega$ , compute the side length of  $\vartriangle  ABC$.
1. Let \( s \) be the side length of \(\triangle ABC\). Since \(\omega\) is tangent to \(\triangle ABC\) at the midpoint of side \(BC\) and lies outside \(\triangle ABC\), \(\omega\) is the \(A\)-excircle of \(\triangle ABC\).

2. The radius of the \(A\)-excircle is given as \(1\). The \(A\)-excircle is tangent to \(BC\) at its midpoint, which means the distance from the \(A\)-excenter to the midpoint of \(BC\) is the radius of the excircle, which is \(1\).

3. In an equilateral triangle, the exradius \(r_a\) (radius of the \(A\)-excircle) is given by:
   \[
   r_a = \frac{s \sqrt{3}}{2}
   \]
   where \(s\) is the side length of the equilateral triangle.

4. Given that the radius of the excircle is \(1\), we set up the equation:
   \[
   \frac{s \sqrt{3}}{2} = 1
   \]

5. Solving for \(s\):
   \[
   s \sqrt{3} = 2
   \]
   \[
   s = \frac{2}{\sqrt{3}}
   \]
   \[
   s = \frac{2 \sqrt{3}}{3}
   \]

Conclusion:
The side length of \(\triangle ABC\) is \(\boxed{\frac{2 \sqrt{3}}{3}}\).