Problem 1551

Given an acute triangle $\triangle ABC$ with orthocenter $H$, we need to find the perimeter of the triangle given that the sum of the distances from the vertices $B$ and $C$ to the orthocenter $H$ is equal to $4r$, where $r$ is the radius of the inscribed circle, and $BC = a$.

1. **Express the distances $BH$ and $CH$ in terms of the circumradius $R$:**
$$BH = 2R \cos \beta \quad \text{and} \quad CH = 2R \cos \gamma$$
where $R$ is the circumradius of $\triangle ABC$.

2. **Sum the distances $BH$ and $CH$:**
$$BH + CH = 2R \cos \beta + 2R \cos \gamma = 2R (\cos \beta + \cos \gamma)$$

3. **Given that $BH + CH = 4r$, equate and simplify:**
$$2R (\cos \beta + \cos \gamma) = 4r$$
$$R (\cos \beta + \cos \gamma) = 2r$$

4. Use the identity for the sum of cosines in terms of the angles of the triangle:
$$\cos \beta + \cos \gamma = 2 \cos \left( \frac{\beta + \gamma}{2} \right) \cos \left( \frac{\beta - \gamma}{2} \right)$$
Since $\alpha + \beta + \gamma = 180^\circ$, we have $\beta + \gamma = 180^\circ - \alpha$, thus:
$$\cos \left( \frac{\beta + \gamma}{2} \right) = \cos \left( \frac{180^\circ - \alpha}{2} \right) = \sin \left( \frac{\alpha}{2} \right)$$

5. Substitute back into the equation:
$$R \left( 2 \sin \left( \frac{\alpha}{2} \right) \cos \left( \frac{\beta - \gamma}{2} \right) \right) = 2r$$
$$R \sin \left( \frac{\alpha}{2} \right) \cos \left( \frac{\beta - \gamma}{2} \right) = r$$

6. **Use the relationship between the inradius $r$ and the circumradius $R$:**
$$r = (p - a) \tan \left( \frac{\alpha}{2} \right)$$
where $p$ is the semi-perimeter of the triangle.

7. Equate and simplify:
$$R \sin \left( \frac{\alpha}{2} \right) \cos \left( \frac{\beta - \gamma}{2} \right) = (p - a) \tan \left( \frac{\alpha}{2} \right)$$
$$R \sin \left( \frac{\alpha}{2} \right) \cos \left( \frac{\beta - \gamma}{2} \right) = (p - a) \frac{\sin \left( \frac{\alpha}{2} \right)}{\cos \left( \frac{\alpha}{2} \right)}$$
$$R \cos \left( \frac{\beta - \gamma}{2} \right) = \frac{p - a}{\cos \left( \frac{\alpha}{2} \right)}$$

8. Simplify further:
$$R \cos \left( \frac{\alpha}{2} \right) \cos \left( \frac{\beta - \gamma}{2} \right) = p - a$$
$$R \sin \left( \frac{\beta + \gamma}{2} \right) \cos \left( \frac{\beta - \gamma}{2} \right) = b + c - a$$
$$R (\sin \beta + \sin \gamma) = b + c - a$$

9. **Using the identity $\sin \beta + \sin \gamma = 2 \sin \left( \frac{\beta + \gamma}{2} \right) \cos \left( \frac{\beta - \gamma}{2} \right)$:**
$$2R (\sin \beta + \sin \gamma) = 2(b + c - a)$$
$$b + c = 2(b + c) - 2a$$
$$b + c = 2a$$

10. **Find the perimeter of $\triangle ABC$:**
$$\text{Perimeter} = a + b + c = 3a$$

The final answer is $\boxed{ 3a }$.