In the triangle $ABC$, the angle    $\alpha = \angle  BAC$ and the side $a = BC$ are given. Assume that $a = \sqrt{rR}$, where $r$ is the inradius and $R$ the circumradius. Compute all possible lengths of sides $AB$ and $AC.$
1. **Given Information and Definitions:**
   - In triangle \( \triangle ABC \), let \(\alpha = \angle BAC\).
   - The side \(a = BC\) is given.
   - It is given that \(a = \sqrt{rR}\), where \(r\) is the inradius and \(R\) is the circumradius.
   - Let \(\Delta\) denote the area of \(\triangle ABC\).

2. **Area Formulas:**
   - The area \(\Delta\) can be expressed in two ways:
     \[
     \Delta = \frac{abc}{4R} \quad \text{and} \quad \Delta = rs
     \]
     where \(s\) is the semi-perimeter of the triangle, \(s = \frac{a+b+c}{2}\).

3. **Manipulating the Area Formulas:**
   - Equating the two expressions for \(\Delta\):
     \[
     \frac{abc}{4R} = rs
     \]
   - Given \(a = \sqrt{rR}\), substitute \(a\) into the equation:
     \[
     \frac{\sqrt{rR} \cdot b \cdot c}{4R} = r \cdot \frac{\sqrt{rR} + b + c}{2}
     \]
   - Simplify and solve for \(b + c\):
     \[
     \frac{bc \cdot \sqrt{rR}}{4R} = r \cdot \frac{\sqrt{rR} + b + c}{2}
     \]
     \[
     \frac{bc \cdot \sqrt{rR}}{4R} = \frac{r \sqrt{rR}}{2} + \frac{r(b + c)}{2}
     \]
     \[
     \frac{bc \cdot \sqrt{rR}}{4R} - \frac{r \sqrt{rR}}{2} = \frac{r(b + c)}{2}
     \]
     \[
     \frac{bc \cdot \sqrt{rR} - 2rR \sqrt{rR}}{4R} = \frac{r(b + c)}{2}
     \]
     \[
     \frac{bc - 2rR}{2\sqrt{rR}} = b + c
     \]
     \[
     b + c = \frac{bc - 2rR}{2\sqrt{rR}} \quad (\diamond)
     \]

4. **Using the Cosine Rule:**
   - By the Cosine Rule:
     \[
     b^2 + c^2 - 2bc \cos \alpha = a^2
     \]
   - Substitute \(a = \sqrt{rR}\):
     \[
     b^2 + c^2 - 2bc \cos \alpha = rR
     \]
   - Express \(b^2 + c^2\) in terms of \(b + c\) and \(bc\):
     \[
     (b + c)^2 - 2bc(1 + \cos \alpha) = rR
     \]
   - Substitute \(b + c = \frac{bc - 2rR}{2\sqrt{rR}}\) and let \(t = bc\):
     \[
     \left(\frac{t - 2rR}{2\sqrt{rR}}\right)^2 - 2t(1 + \cos \alpha) = rR
     \]
     \[
     \frac{(t - 2rR)^2}{4rR} - 2t(1 + \cos \alpha) = rR
     \]
     \[
     \frac{t^2 - 4t rR + 4r^2 R^2}{4rR} - 2t(1 + \cos \alpha) = rR
     \]
     \[
     \frac{t^2}{4rR} - t + rR - 2t(1 + \cos \alpha) = rR
     \]
     \[
     \frac{t^2}{4rR} - t - 2t(1 + \cos \alpha) = 0
     \]
     \[
     \frac{t^2}{4rR} - t(1 + 2(1 + \cos \alpha)) = 0
     \]
     \[
     \frac{t^2}{4rR} - t(3 + 2\cos \alpha) = 0
     \]
     \[
     t^2 - 4rR(3 + 2\cos \alpha)t = 0
     \]
   - Since \(t = bc \neq 0\), divide by \(t\):
     \[
     t = 4rR(3 + 2\cos \alpha)
     \]

5. **Solving for \(b\) and \(c\):**
   - Substitute \(t = bc\) back into the equation:
     \[
     bc = 4rR(3 + 2\cos \alpha)
     \]
   - From \((\diamond)\):
     \[
     b + c = \frac{4rR(3 + 2\cos \alpha) - 2rR}{2\sqrt{rR}}
     \]
     \[
     b + c = \frac{4rR(3 + 2\cos \alpha) - 2rR}{2\sqrt{rR}}
     \]
     \[
     b + c = \frac{2rR(6 + 4\cos \alpha - 1)}{2\sqrt{rR}}
     \]
     \[
     b + c = \frac{2rR(5 + 4\cos \alpha)}{2\sqrt{rR}}
     \]
     \[
     b + c = \sqrt{rR}(5 + 4\cos \alpha)
     \]
   - We now have a quadratic in \(b\):
     \[
     x^2 - (\sqrt{rR}(5 + 4\cos \alpha))x + 4rR(3 + 2\cos \alpha) = 0
     \]
   - Solve this quadratic equation for \(b\) and \(c\).

The final answer is \( \boxed{ b } \) and \(c\) are the roots of the quadratic equation: