[PYTHON] import numpy as np

def invert_point(p): # Center A(-2, 0), Power 9 # Q = A + 9 * (P - A) / |P - A|^2 x, y = p dx = x - (-2) dy = y - 0 dist_sq = dx2 + dy2 scale = 9.0 / dist_sq qx = -2 + scale * dx qy = scale * dy return np.array([qx, qy])

Define points

R1 = np.array([-0.5, np.sqrt(3)/2]) R2 = np.array([-0.5, -np.sqrt(3)/2]) O_prime = np.array([-1, 0]) # On C1 O = np.array([0, 0]) # On C2

Map points

R1_prime = invert_point(R1) R2_prime = invert_point(R2) O_prime_mapped = invert_point(O_prime) # Image of O' O_mapped = invert_point(O) # Image of O

print(f"R1': {R1_prime}") print(f"R2': {R2_prime}") print(f"O' mapped: {O_prime_mapped}") print(f"O mapped: {O_mapped}")

Function to find circle from 3 points

def circle_from_3_points(p1, p2, p3): x1, y1 = p1 x2, y2 = p2 x3, y3 = p3

D = 2 * (x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2))
center_x = ((x1**2 + y1**2) * (y2 - y3) + (x2**2 + y2**2) * (y3 - y1) + (x3**2 + y3**2) * (y1 - y2)) / D
center_y = ((x1**2 + y1**2) * (x3 - x2) + (x2**2 + y2**2) * (x1 - x3) + (x3**2 + y3**2) * (x2 - x1)) / D

radius = np.sqrt((center_x - x1)**2 + (center_y - y1)**2)
return (center_x, center_y), radius

Image Circle 1 (from R1, R2, O')

Note: R1, R2 are on both. O' is on C1.

center1, radius1 = circle_from_3_points(R1_prime, R2_prime, O_prime_mapped) print(f"Image Circle 1: Center {center1}, Radius {radius1}")

Image Circle 2 (from R1, R2, O)

center2, radius2 = circle_from_3_points(R1_prime, R2_prime, O_mapped) print(f"Image Circle 2: Center {center2}, Radius {radius2}")

Function to calculate arc length

def get_arc_length(center, radius, p_start, p_mid, p_end): cx, cy = center

def get_angle(p):
    return np.arctan2(p[1] - cy, p[0] - cx)

ang_start = get_angle(p_start)
ang_mid = get_angle(p_mid)
ang_end = get_angle(p_end)

# Normalize angles to [0, 2pi) for easier logic, or handle diffs
# We need the arc from start to end passing through mid.
# Calculate angular distance.

# Let's use vectors and cross products/dot products to be robust, or just handle the wrap around.
# Simple way: check the two possible arcs (clockwise vs counter-clockwise) and see which contains mid.

# Total angle is either abs(ang_end - ang_start) or 2pi - ...
# But we need the specific one.

# Let's shift angles so ang_start is 0.
a_mid = (ang_mid - ang_start) % (2 * np.pi)
a_end = (ang_end - ang_start) % (2 * np.pi)

# If a_mid < a_end, then we go start -> mid -> end in counter-clockwise direction. Length is a_end * r.
# If a_mid > a_end, then we go start -> mid -> end in clockwise direction (which is the long way around in this frame).
# Wait.
# Case 1: 0 < a_mid < a_end. Arc is CCW, angle is a_end.
# Case 2: a_end < a_mid. Arc is the other way?
# Actually, if we move CCW from start:
# If we hit mid before end, then the CCW arc is the one. Angle = a_end.
# If we hit end before mid, then the CCW arc is NOT the one. We must go the other way (CW). Angle = 2pi - a_end.

if a_mid < a_end:
    angle = a_end
else:
    angle = 2 * np.pi - a_end
    
return radius * angle

len1 = get_arc_length(center1, radius1, R1_prime, O_prime_mapped, R2_prime) len2 = get_arc_length(center2, radius2, R1_prime, O_mapped, R2_prime)

print(f"Length 1: {len1}") print(f"Length 2: {len2}") print(f"Total Length: {len1 + len2}") [/PYTHON]

[THOUGHT] The problem asks for the length of the locus of point $Q$ as point $P$ moves along a region $\mathcal{R}$. The relationship between $P$ and $Q$ is given by $\overrightarrow{AP} \parallel \overrightarrow{AQ}$ and $\overrightarrow{AP} \cdot \overrightarrow{AQ} = 9$. Since the dot product is positive, $P$ and $Q$ lie on the same ray from $A$, and $|AP| \cdot |AQ| = 9$. This is an inversion centered at $A(-2,0)$ with power $9$.

The region $\mathcal{R}$ consists of two arcs:

1. Arc of $C_1$: $C_1$ is $x^2 + y^2 = 1$. The arc connects $R_1$ and $R_2$ and contains $O'(-1,0)$.

   • $R_1, R_2$ are the intersection points of $C_1$ and $C_2$.
   • $C_1: x^2 + y^2 = 1$.
   • $C_2: (x+1)^2 + y^2 = 1 \implies x^2 + 2x + 1 + y^2 = 1 \implies 1 + 2x + 1 = 1 \implies 2x = -1 \implies x = -0.5$.
   • $y^2 = 1 - (-0.5)^2 = 0.75 \implies y = \pm \sqrt{3}/2$.
   • $R_1(-0.5, \sqrt{3}/2)$, $R_2(-0.5, -\sqrt{3}/2)$.
   • $O'(-1,0)$ is on $C_1$.
   • The image of this arc under inversion is an arc of a circle (since $A$ is not on $C_1$).
   • From the Python code, the image circle has radius $r=3$ and the arc angle is $240^\circ$ ($4\pi/3$ radians).
   • Length $L_1 = 3 \times \frac{4\pi}{3} = 4\pi$.

2. Arc of $C_2$: $C_2$ is $(x+1)^2 + y^2 = 1$. The arc connects $R_1$ and $R_2$ and contains $O(0,0)$.

   • $A(-2,0)$ lies on $C_2$. Thus, the image of $C_2$ is a straight line.
   • The image of the arc containing $O(0,0)$ (which does not contain $A$) is the finite line segment connecting the images of $R_1$ and $R_2$.
   • Image of $R_1$: $R_1' = (2.5, \frac{3\sqrt{3}}{2})$.
   • Image of $R_2$: $R_2' = (2.5, -\frac{3\sqrt{3}}{2})$.
   • Length $L_2$ is the distance between $R_1'$ and $R_2'$, which is $3\sqrt{3}$.

Total length $k = 4\pi + 3\sqrt{3}$. Using approximations: $4\pi \approx 12.56637$ $3\sqrt{3} \approx 5.19615$ $k \approx 17.7625$

The value of $\lfloor k \rfloor$ is 17. [/THOUGHT]

\boxed{17}