Problem Analysis #29

[THOUGHT] To solve this problem, we need to understand the movement of point $Q$ relative to point $P$ as $P$ moves along the region $\mathcal{R}$.

Step 1: Condition Analysis

• $\overrightarrow{AP}$ and $\overrightarrow{AQ}$ are parallel, so $\overrightarrow{AQ} = \lambda \overrightarrow{AP}$ for some scalar $\lambda$.
• Given $\overrightarrow{AP} \cdot \overrightarrow{AQ} = 9$, we have $\overrightarrow{AP} \cdot (\lambda \overrightarrow{AP}) = \lambda \|\overrightarrow{AP}\|^2 = 9$. Therefore, $\lambda = \frac{9}{\|\overrightarrow{AP}\|^2}$.

Step 2: Region $\mathcal{R}$ Analysis

• $\mathcal{R}$ consists of two arcs:
  • $R_1O'R_2$ on $C_1$ (circle centered at origin with radius 1)
  • $R_1OR_2$ on $C_2$ (circle centered at $(-1,0)$ with radius 1)
• These circles intersect at two points, which we need to find.

Step 3: Finding Intersection Points

• $C_1$: $x^2 + y^2 = 1$
• $C_2$: $(x+1)^2 + y^2 = 1$
• Expanding $C_2$ gives $x^2 + 2x + 1 + y^2 = 1$, which simplifies to $x^2 + y^2 + 2x = 0$.
• Using $x^2 + y^2 = 1$ from $C_1$, we substitute to get $1 + 2x = 0$ or $x = -\frac{1}{2}$.
• Substituting $x = -\frac{1}{2}$ into $C_1$, we get $\left(-\frac{1}{2}\right)^2 + y^2 = 1$, or $\frac{1}{4} + y^2 = 1$, hence $y^2 = \frac{3}{4}$, so $y = \pm \frac{\sqrt{3}}{2}$.
• Thus, the intersection points are $R_1 = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $R_2 = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$.

Step 4: Calculate Length of Arcs

• The arc $R_1O'R_2$ on $C_1$ is a 120° arc (since the angle $R_1OR_2$ is 120°, given symmetry and equal radii). The length is $\frac{120}{360} \times 2\pi \times 1 = \frac{1}{3} \times 2\pi = \frac{2\pi}{3}$.
• Similarly, the arc $R_1OR_2$ on $C_2$ is also a 120° arc. The length is $\frac{120}{360} \times 2\pi \times 1 = \frac{2\pi}{3}$.
• Total arc length for $P$ along $\mathcal{R}$ is $\frac{2\pi}{3} + \frac{2\pi}{3} = \frac{4\pi}{3}$.

Step 5: Locus of $Q$

• As $P$ moves on an arc, $\overrightarrow{AQ} = \frac{9}{\|\overrightarrow{AP}\|^2} \overrightarrow{AP}$.
• Since $P$ moves on a circular arc around $A$, $Q$ will also move on a circular arc around $A$, but with a varying radius determined by $\|\overrightarrow{AP}\|$. However, the total angle covered by $Q$ will be the same as $P$ because the direction of $\overrightarrow{AP}$ and $\overrightarrow{AQ}$ is the same.
• Thus, $Q$ traces the same 120° arc twice on two different circles, each proportional to the distance from $A$.

Step 6: Calculate the Length of Locus $k$ for $Q$

• The length of each arc traced by $Q$ is proportional to the length of the arc traced by $P$. Since $Q$ is effectively moving at the same angular speed as $P$, the total length $k$ of the locus traced by $Q$ is the same as the total length of the locus traced by $P$, which is $\frac{4\pi}{3}$.

Step 7: Final Calculation

• The value of $k$ is $\frac{4\pi}{3} \approx 4.1888$.
• $\lfloor k \rfloor = \lfloor 4.1888 \rfloor = 4$.

[/THOUGHT] $\boxed{4}$