[THOUGHT] Let me analyze this problem step by step.

First, I need to understand the setup:

• Point A is at (-2, 0)
• Points P and Q satisfy:
  • $\overrightarrow{AP}$ and $\overrightarrow{AQ}$ are parallel
  • $\overrightarrow{AP} \cdot \overrightarrow{AQ} = 9$

Since $\overrightarrow{AP}$ and $\overrightarrow{AQ}$ are parallel, Q lies on the ray from A through P (or the opposite ray).

Let $\overrightarrow{AP} = \vec{v}$, then $\overrightarrow{AQ} = t\vec{v}$ for some scalar $t$.

The dot product condition gives: $\vec{v} \cdot (t\vec{v}) = t|\vec{v}|^2 = 9$

So $t = \frac{9}{|\vec{v}|^2} = \frac{9}{|AP|^2}$

This means $|AQ| = |t| \cdot |AP| = \frac{9}{|AP|^2} \cdot |AP| = \frac{9}{|AP|}$

So $|AP| \cdot |AQ| = 9$, which is the definition of inversion with center A and radius 3.

Now I need to find the region $\mathcal{R}$:

• Circle $C_1$: $x^2 + y^2 = 1$ (center O at origin, radius 1)
• Circle $C_2$: $(x+1)^2 + y^2 = 1$ (center at (-1, 0), radius 1)

Let me find the intersection points $R_1$ and $R_2$.

Then I need to find the locus of Q when P moves along region $\mathcal{R}$, which is the inversion of $\mathcal{R}$ with center A(-2, 0) and radius 3.

Let me compute this with Python. [/THOUGHT] [PYTHON] import numpy as np from scipy import integrate

Find intersection points of C1 and C2

C1: x^2 + y^2 = 1

C2: (x+1)^2 + y^2 = 1

Subtracting: x^2 - (x+1)^2 = 0

x^2 - x^2 - 2x - 1 = 0

-2x - 1 = 0

x = -1/2

x_intersect = -1/2 y_intersect = np.sqrt(1 - x_intersect**2) print(f"Intersection points: R1 = ({x_intersect}, {y_intersect}), R2 = ({x_intersect}, {-y_intersect})")

Region R consists of:

1. Arc R1 O' R2 on C1 containing O'(-1, 0) - this is the left arc of C1

2. Arc R1 O R2 on C2 containing O(0, 0) - this is the right arc of C2

For inversion with center A(-2, 0) and radius r=3:

If P = (x, y), then Q = A + 9/|AP|^2 * (P - A)

Q = (-2, 0) + 9/((x+2)^2 + y^2) * (x+2, y)

A = np.array([-2, 0])

def invert_point(P): """Invert point P with respect to center A and radius 3""" AP = P - A dist_sq = np.dot(AP, AP) if dist_sq == 0: return None return A + 9 / dist_sq * AP

Let's parametrize the arcs and compute the inverted curves

Arc 1: Left arc of C1 (center at origin, radius 1)

This arc goes from R1 to R2 through O'(-1, 0)

R1 = (-1/2, sqrt(3)/2), R2 = (-1/2, -sqrt(3)/2)

Angle of R1 from center (0,0): arctan2(sqrt(3)/2, -1/2) = 2π/3

Angle of R2 from center (0,0): arctan2(-sqrt(3)/2, -1/2) = -2π/3 = 4π/3

angle_R1_C1 = np.arctan2(y_intersect, x_intersect) # 2π/3 angle_R2_C1 = np.arctan2(-y_intersect, x_intersect) # -2π/3

print(f"C1: angle_R1 = {angle_R1_C1}, angle_R2 = {angle_R2_C1}") print(f"C1: angle_R1 = {angle_R1_C1 * 180/np.pi} deg, angle_R2 = {angle_R2_C1 * 180/np.pi} deg")

The left arc goes from angle 2π/3 to 4π/3 (or -2π/3) through π

So we parametrize from 2π/3 to 4π/3

Arc 2: Right arc of C2 (center at (-1, 0), radius 1)

R1 = (-1/2, sqrt(3)/2), R2 = (-1/2, -sqrt(3)/2)

Relative to center (-1, 0):

R1 - (-1, 0) = (1/2, sqrt(3)/2), angle = π/3

R2 - (-1, 0) = (1/2, -sqrt(3)/2), angle = -π/3

angle_R1_C2 = np.arctan2(y_intersect, x_intersect + 1) # π/3 angle_R2_C2 = np.arctan2(-y_intersect, x_intersect + 1) # -π/3

print(f"C2: angle_R1 = {angle_R1_C2}, angle_R2 = {angle_R2_C2}") print(f"C2: angle_R1 = {angle_R1_C2 * 180/np.pi} deg, angle_R2 = {angle_R2_C2 * 180/np.pi} deg")

The right arc of C2 goes from angle π/3 to -π/3 through 0

[/PYTHON]

[THOUGHT] Let me continue the calculation to find the length of the locus traced by Q when P moves along region R. [/THOUGHT] [PYTHON] import numpy as np from scipy import integrate

A = np.array([-2, 0])

def invert_point(P): """Invert point P with respect to center A and radius 3""" AP = P - A dist_sq = np.dot(AP, AP) if dist_sq == 0: return None return A + 9 / dist_sq * AP

Arc 1: Left arc of C1 (center at origin, radius 1)

Goes from angle 2π/3 to 4π/3 (through π)

def arc1_point(theta): """Point on left arc of C1""" return np.array([np.cos(theta), np.sin(theta)])

def arc1_derivative(theta): """Derivative of point on C1""" return np.array([-np.sin(theta), np.cos(theta)])

Arc 2: Right arc of C2 (center at (-1, 0), radius 1)

Goes from angle π/3 to -π/3 (through 0)

def arc2_point(theta): """Point on right arc of C2""" return np.array([-1 + np.cos(theta), np.sin(theta)])

def arc2_derivative(theta): """Derivative of point on C2""" return np.array([-np.sin(theta), np.cos(theta)])

For inversion, if P(t) is a curve, then Q(t) = A + 9/|P-A|^2 * (P-A)

dQ/dt = d/dt[9(P-A)/|P-A|^2]

= 9 * [dP/dt * |P-A|^2 - (P-A) * 2(P-A)·dP/dt] / |P-A|^4

= 9/|P-A|^4 * [dP/dt * |P-A|^2 - 2(P-A)((P-A)·dP/dt)]

def inverted_derivative(P, dP): """Derivative of inverted point""" AP = P - A dist_sq = np.dot(AP, AP) dot_prod = np.dot(AP, dP) return 9 / dist_sq**2 * (dP * dist_sq - 2 * AP * dot_prod)

def arc_length_integrand_arc1(theta): P = arc1_point(theta) dP = arc1_derivative(theta) dQ = inverted_derivative(P, dP) return np.sqrt(np.dot(dQ, dQ))

def arc_length_integrand_arc2(theta): P = arc2_point(theta) dP = arc2_derivative(theta) dQ = inverted_derivative(P, dP) return np.sqrt(np.dot(dQ, dQ))

Arc 1: from 2π/3 to 4π/3

length1, _ = integrate.quad(arc_length_integrand_arc1, 2np.pi/3, 4np.pi/3) print(f"Length of inverted arc 1: {length1}")

Arc 2: from -π/3 to π/3

length2, _ = integrate.quad(arc_length_integrand_arc2, -np.pi/3, np.pi/3) print(f"Length of inverted arc 2: {length2}")

total_length = length1 + length2 print(f"Total length k: {total_length}") print(f"Floor of k: {int(np.floor(total_length))}") [/PYTHON]

[THOUGHT] The calculation shows that when point P moves along region R (consisting of the left arc of C1 and the right arc of C2), the corresponding point Q traces a locus with total length k ≈ 17.7625.

The floor of k is 17. [/THOUGHT]

\boxed{17}