[THOUGHT] Set coordinates so plane BCD is z=0 with C(-2,0,0), D(2,0,0), and B determined by BC=BD=2√5, giving B(0,4,0). Use the perpendicular-plane condition to deduce H is midpoint of CD, then use AB=4 and AH=4 to get A. Compute centroid G of triangle ABH. Find radius r1 of sphere S as distance from G to plane ACD. Locus T is intersection of S with the sphere having diameter AG (right angle condition), which is a circle in a plane normal to (A−G). Compute its radius ρ via sphere-sphere intersection, then project its area onto plane ABC using area scaling factor |cos(angle between normals)|. Simplify to form π/k. [/THOUGHT] [PYTHON] import sympy as sp

sqrt = sp.sqrt

Coordinates

C = sp.Matrix([-2,0,0]) D = sp.Matrix([2,0,0]) H = sp.Matrix([0,0,0]) B = sp.Matrix([0,4,0]) A = sp.Matrix([0,2,2*sqrt(3)])

Centroid of triangle ABH

G = (A + B + H)/3

A, B, C, D, H, G print("G =", G)

plane ACD normal

n_acd = (C-A).cross(D-A) print("n_acd =", sp.simplify(n_acd))

distance from point G to plane ACD (r1)

r1 = sp.Abs(n_acd.dot(G-A)) / sp.sqrt(n_acd.dot(n_acd)) r1_simplified = sp.simplify(r1) print("r1 =", r1_simplified)

Sphere S2: diameter AG

AG = A-G d = sp.sqrt((AG.dot(AG)))/2 # distance between centers G and M equals |AG|/2 print("d =", sp.simplify(d))

Intersection circle radius rho

x = r1_simplified2 / (2*d) # since r2=d rho2 = sp.simplify(r1_simplified2 - x**2) print("rho^2 =", rho2)

plane normals: circle plane normal n1 parallel to AG, projection plane ABC normal

n1 = AG # normal to circle plane n_abc = (B-A).cross(C-A) cos_factor = sp.simplify(sp.Abs(n1.dot(n_abc)) / (sp.sqrt(n1.dot(n1))*sp.sqrt(n_abc.dot(n_abc)))) print("cos_factor =", cos_factor)

projected area

area_proj = sp.simplify(sp.pi * rho2 * cos_factor) print("area_proj =", area_proj)

express as pi/k

k = sp.simplify(sp.pi / area_proj) print("k =", k) print("k rational?", sp.nsimplify(k)) [/PYTHON][THOUGHT] Set coordinates so plane BCD is z=0 with C(-2,0,0), D(2,0,0), and B determined by BC=BD=2√5, giving B(0,4,0). Use the perpendicular-plane condition to deduce H is midpoint of CD, then use AB=4 and AH=4 to get A. Compute centroid G of triangle ABH. Find radius r1 of sphere S as distance from G to plane ACD. Locus T is intersection of S with the sphere having diameter AG (right angle condition), which is a circle in a plane normal to (A−G). Compute its radius ρ via sphere-sphere intersection, then project its area onto plane ABC using area scaling factor |cos(angle between normals)|. Simplify to form π/k. [/THOUGHT] [PYTHON] import sympy as sp

sqrt = sp.sqrt

Coordinates

C = sp.Matrix([-2,0,0]) D = sp.Matrix([2,0,0]) H = sp.Matrix([0,0,0]) B = sp.Matrix([0,4,0]) A = sp.Matrix([0,2,2*sqrt(3)])

Centroid of triangle ABH

G = (A + B + H)/3 print("G =", G)

Plane ACD normal

n_acd = (C-A).cross(D-A) print("n_acd =", sp.simplify(n_acd))

distance from point G to plane ACD (r1)

r1 = sp.Abs(n_acd.dot(G-A)) / sp.sqrt(n_acd.dot(n_acd)) r1 = sp.simplify(r1) print("r1 =", r1)

Sphere S2: diameter AG

AG = A-G d = sp.sqrt(AG.dot(AG))/2 # distance between centers G and M equals |AG|/2 print("d =", sp.simplify(d))

Intersection circle radius rho

x = r12 / (2*d) # since r2=d rho2 = sp.simplify(r12 - x**2) print("rho^2 =", rho2)

plane normals: circle plane normal n1 parallel to AG, projection plane ABC normal

n1 = AG n_abc = (B-A).cross(C-A) cos_factor = sp.simplify(sp.Abs(n1.dot(n_abc)) / (sp.sqrt(n1.dot(n1))*sp.sqrt(n_abc.dot(n_abc)))) print("cos_factor =", cos_factor)

projected area

area_proj = sp.simplify(sp.pi * rho2 * cos_factor) print("area_proj =", area_proj)

express as pi/k

k = sp.simplify(sp.pi / area_proj) print("k =", k) print("k simplified numeric:", sp.N(k)) print("k rational guess:", sp.nsimplify(k)) PYTHON

[THOUGHT] The Python computation gives projected area (=\pi/4), so (\pi/k=\pi/4\Rightarrow k=4). \boxed{4}