Problem 1786

1. Define the coordinates and semicircles:

Given points $A(0,0)$, $B(2b,0)$, and $C(2c,0)$, the midpoints $F_1$ and $F_2$ are:
$$F_1(b,0) \quad \text{and} \quad F_2(b+c,0)$$

2. Equations of the semicircles:

The semicircle above segment $AB$ with center $F_1$ and radius $b$ is:
$$(x-b)^2 + y^2 = b^2$$

The semicircle above segment $BC$ with center $F_2$ and radius $c-b$ is:
$$(x-b-c)^2 + y^2 = (c-b)^2$$

3. **Coordinates of points $P$ and $Q$:**

Points $P$ and $Q$ are the points of tangency of the line segment $PQ$ with the semicircles. Using the properties of tangents and symmetry, we find:
$$P\left(\frac{2b^2}{c}, \frac{2b\sqrt{b(c-b)}}{c}\right) \quad \text{and} \quad Q\left(\frac{2b(2c-b)}{c}, \frac{2(c-b)\sqrt{b(c-b)}}{c}\right)$$

4. **Slope and distance of line $PQ$:**

The slope $m_{PQ}$ of the line $PQ$ is calculated as:
$$m_{PQ} = \frac{\frac{2(c-b)\sqrt{b(c-b)}}{c} - \frac{2b\sqrt{b(c-b)}}{c}}{\frac{2b(2c-b)}{c} - \frac{2b^2}{c}} = \frac{2(c-b) - 2b}{2b(2c-b) - 2b^2} = \frac{c-2b}{2\sqrt{b(c-b)}}$$

The distance $PQ$ is:
$$PQ = 2\sqrt{b(c-b)}$$

5. **Equations of the semicircles above $AF_2$ and $F_1C$:**

The semicircle above segment $AF_2$ with center $\left(\frac{b+c}{2}, 0\right)$ and radius $\frac{b+c}{2}$ is:
$$\left(x - \frac{b+c}{2}\right)^2 + y^2 = \left(\frac{b+c}{2}\right)^2$$

The semicircle above segment $F_1C$ with center $\left(\frac{b+2c}{2}, 0\right)$ and radius $\frac{2c-b}{2}$ is:
$$\left(x - \frac{b+2c}{2}\right)^2 + y^2 = \left(\frac{2c-b}{2}\right)^2$$

6. **Coordinates of points $R$ and $S$:**

Points $R$ and $S$ are the points of tangency of the line segment $RS$ with the semicircles. Using the properties of tangents and symmetry, we find:
$$R\left(\frac{b(b+c)}{c}, \frac{(b+c)\sqrt{b(c-b)}}{c}\right) \quad \text{and} \quad S\left(\frac{b(3c-b)}{c}, \frac{(2c-b)\sqrt{b(c-b)}}{c}\right)$$

7. **Slope and distance of line $RS$:**

The slope $m_{RS}$ of the line $RS$ is calculated as:
$$m_{RS} = \frac{\frac{(2c-b)\sqrt{b(c-b)}}{c} - \frac{(b+c)\sqrt{b(c-b)}}{c}}{\frac{b(3c-b)}{c} - \frac{b(b+c)}{c}} = \frac{(2c-b) - (b+c)}{b(3c-b) - b(b+c)} = \frac{c-2b}{2\sqrt{b(c-b)}}$$

The distance $RS$ is:
$$RS = \sqrt{b(c-b)}$$

8. Conclusion:

Since the slopes of $PQ$ and $RS$ are equal, $PQ \parallel RS$. Additionally, the distance $PQ$ is twice the distance $RS$:
$$PQ = 2 \cdot RS$$

$\blacksquare$

The final answer is $\boxed{ PQ \parallel RS }$ and $|PQ| = 2 \cdot |RS|$