Let $ABC$ be a triangle and let $\Omega$ be its circumcircle. The internal bisectors of angles $A, B$ and $C$ intersect $\Omega$ at $A_1, B_1$ and $C_1$, respectively, and the internal bisectors of angles $A_1, B_1$ and $C_1$ of the triangles $A_1 A_2 A_ 3$ intersect $\Omega$ at $A_2, B_2$ and $C_2$, respectively. If the smallest angle of the triangle $ABC$ is $40^{\circ}$, what is the magnitude of the smallest angle of the triangle $A_2 B_2 C_2$ in degrees?
1. Let the angles of triangle \(ABC\) be \(A\), \(B\), and \(C\). Given that the smallest angle of triangle \(ABC\) is \(40^\circ\), we can assume without loss of generality that \(A = 40^\circ\).

2. The internal angle bisectors of angles \(A\), \(B\), and \(C\) intersect the circumcircle \(\Omega\) at points \(A_1\), \(B_1\), and \(C_1\) respectively. The angles of triangle \(A_1B_1C_1\) are given by:
   \[
   \angle A_1 = \frac{B}{2} + \frac{C}{2}, \quad \angle B_1 = \frac{A}{2} + \frac{C}{2}, \quad \angle C_1 = \frac{A}{2} + \frac{B}{2}
   \]

3. Since \(A = 40^\circ\), we have:
   \[
   \angle B_1 = \frac{40^\circ}{2} + \frac{C}{2} = 20^\circ + \frac{C}{2}
   \]

4. The internal angle bisectors of angles \(A_1\), \(B_1\), and \(C_1\) of triangle \(A_1B_1C_1\) intersect the circumcircle \(\Omega\) at points \(A_2\), \(B_2\), and \(C_2\) respectively. The angles of triangle \(A_2B_2C_2\) are given by:
   \[
   \angle A_2 = \frac{\angle B_1 + \angle C_1}{2}, \quad \angle B_2 = \frac{\angle A_1 + \angle C_1}{2}, \quad \angle C_2 = \frac{\angle A_1 + \angle B_1}{2}
   \]

5. Substituting the expressions for \(\angle A_1\), \(\angle B_1\), and \(\angle C_1\):
   \[
   \angle A_2 = \frac{\left(\frac{A}{2} + \frac{C}{2}\right) + \left(\frac{A}{2} + \frac{B}{2}\right)}{2} = \frac{\frac{A}{2} + \frac{C}{2} + \frac{A}{2} + \frac{B}{2}}{2} = \frac{A + \frac{B}{2} + \frac{C}{2}}{2}
   \]

6. Since \(A = 40^\circ\), we have:
   \[
   \angle A_2 = \frac{40^\circ + \frac{B}{2} + \frac{C}{2}}{2} = \frac{40^\circ + 90^\circ}{2} = \frac{130^\circ}{2} = 65^\circ
   \]

7. Similarly, we can find the other angles:
   \[
   \angle B_2 = \frac{\left(\frac{B}{2} + \frac{C}{2}\right) + \left(\frac{A}{2} + \frac{C}{2}\right)}{2} = \frac{\frac{B}{2} + \frac{C}{2} + \frac{A}{2} + \frac{C}{2}}{2} = \frac{\frac{B}{2} + \frac{C}{2} + \frac{A}{2} + \frac{C}{2}}{2} = \frac{B + \frac{A}{2} + \frac{C}{2}}{2}
   \]
   \[
   \angle C_2 = \frac{\left(\frac{A}{2} + \frac{B}{2}\right) + \left(\frac{B}{2} + \frac{C}{2}\right)}{2} = \frac{\frac{A}{2} + \frac{B}{2} + \frac{B}{2} + \frac{C}{2}}{2} = \frac{\frac{A}{2} + B + \frac{C}{2}}{2}
   \]

8. Since \(A = 40^\circ\), the smallest angle of triangle \(A_2B_2C_2\) is:
   \[
   \angle A_2 = 65^\circ
   \]

The final answer is \(\boxed{65^\circ}\).