[THOUGHT] Model the condition “AP ∥ AQ and AP·AQ=9” as an inversion about the circle centered at A with power 9: if P is at distance r from A, then Q lies on ray AP with distance 9/r from A. Thus Q is the inversion image of P with respect to the circle centered at A of radius 3. The path of P is two circular arcs (on C1 and C2) between the intersection points R1,R2. Under inversion: C1 (not through A) maps to a circle, and C2 (through A) maps to a line. Compute images of key points to identify which arc/segment is traced, then compute total length k and take floor. Verify numerically with Python. [/THOUGHT] [PYTHON] import sympy as sp sqrt = sp.sqrt pi = sp.pi

Define points

A = sp.Point(-2, 0) R1 = sp.Point(-sp.Rational(1,2), sqrt(3)/2) R2 = sp.Point(-sp.Rational(1,2), -sqrt(3)/2) B = sp.Point(-1, 0) # point on C1 arc O = sp.Point(0, 0)

def invert(P): # inversion about circle centered at A with power 9: Q = A + 9/|AP|^2 * (P-A) v = sp.Matrix([P.x - A.x, P.y - A.y]) r2 = (v[0]**2 + v[1]**2) t = sp.Rational(9,1) / r2 Q = sp.Point(A.x + tv[0], A.y + tv[1]) return sp.simplify(Q)

R1i, R2i, Bi, Oi = map(invert, [R1, R2, B, O]) print("R1i =", R1i) print("R2i =", R2i) print("Bi =", Bi) print("Oi =", Oi)

Image of C1 should be circle with center (4,0) radius 3

center1 = sp.Point(4,0); rad1 = sp.Rational(3,1) print("Check R1i on circle:", sp.simplify(R1i.distance(center1) - rad1)) print("Check Bi on circle:", sp.simplify(Bi.distance(center1) - rad1))

Determine arc length on image circle between R1i and R2i that contains Bi

Compute angles of R1i, R2i, Bi about center1

def angle(P, C): return sp.atan2(P.y-C.y, P.x-C.x)

a1 = angle(R1i, center1) a2 = angle(R2i, center1) ab = angle(Bi, center1) print("angles (rad):", a1, a2, ab)

In this symmetric case: endpoints at 2pi/3 and -2pi/3 (or 4pi/3), Bi at 0.

So arc through Bi is the major arc with central angle 4pi/3.

arc_len = rad1 * (4*pi/3) seg_len = R1i.distance(R2i) # segment on line image of C2 k = sp.simplify(arc_len + seg_len) print("arc_len =", sp.simplify(arc_len)) print("seg_len =", sp.simplify(seg_len)) print("k exact =", k) print("k numeric =", float(k.evalf())) print("floor(k) =", int(sp.floor(k)))

Numerical verification by sampling points along original arcs and computing length in Q-space

import numpy as np, math

def invert_num(x,y): vx, vy = x - (-2.0), y - 0.0 r2 = vxvx + vyvy t = 9.0 / r2 return (-2.0 + tvx, 0.0 + tvy)

Parametrize original arcs

C1 arc from angle 2pi/3 to 4pi/3 (through pi)

def sample_arc_C1(n=200000): th = np.linspace(2math.pi/3, 4math.pi/3, n) x = np.cos(th); y = np.sin(th) return x,y

C2 center (-1,0): arc through O corresponds to angles -pi/3 to pi/3

def sample_arc_C2(n=200000): th = np.linspace(-math.pi/3, math.pi/3, n) x = -1 + np.cos(th); y = np.sin(th) return x,y

def length_of_image(x,y): # compute polyline length of inverted points qx = np.empty_like(x); qy = np.empty_like(y) for i,(xx,yy) in enumerate(zip(x,y)): qx[i], qy[i] = invert_num(xx,yy) dx = np.diff(qx); dy = np.diff(qy) return np.sum(np.hypot(dx,dy))

x1,y1 = sample_arc_C1(20000) x2,y2 = sample_arc_C2(20000) len_num = length_of_image(x1,y1) + length_of_image(x2,y2) print("numerical length approx:", len_num) print("difference numeric-exact:", len_num - float(k.evalf())) [/PYTHON]

[THOUGHT] From the verified computation, the locus length is (k=3\sqrt{3}+4\pi\approx 17.7625), hence (\lfloor k\rfloor=17). \boxed{17}