[PYTHON] import sympy as sp

Define sqrt

sqrt = sp.sqrt

Points

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

Verify distances

AB = (A - B).norm() BC = (B - C).norm() BD = (B - D).norm() AH = (A - H).norm() assert sp.simplify(AB - 4) == 0 assert sp.simplify(BC - 2sqrt(5)) == 0 assert sp.simplify(BD - 2sqrt(5)) == 0 assert sp.simplify(AH - 4) == 0

Centroid G of triangle ABH

G = (A + B + H) / 3

Plane ACD normal and distance from G to plane

n_ACD = (D - C).cross(A - C) dist_G_to_ACD = sp.Abs(n_ACD.dot(G)) / n_ACD.norm() r = dist_G_to_ACD

Plane T normal = G - A

n_T = G - A

Plane T constant: (G - A)·P = G·G - A·G - r**2

c_T = G.dot(G) - A.dot(G) - r**2

Distance from G to plane T

dist_G_to_T = sp.Abs(n_T.dot(G) - c_T) / n_T.norm()

Circle radius (intersection of sphere and plane)

circle_radius = sp.sqrt(r2 - dist_G_to_T2)

Plane ABC normal

AB_vec = B - A AC_vec = C - A n_ABC = AB_vec.cross(AC_vec)

Cosine of angle between normals

cos_phi = sp.Abs(n_T.dot(n_ABC)) / (n_T.norm() * n_ABC.norm())

Projected area

area_proj = sp.pi * circle_radius**2 * cos_phi

Simplify

area_proj_simplified = sp.simplify(area_proj) print(area_proj_simplified) [/PYTHON]

[THOUGHT] We placed CD on the x‑axis with C=(0,0,0), D=(4,0,0).
The foot H of the perpendicular from A to CD must be the midpoint, so H=(2,0,0).
Since BC=BD=2√5, B lies on the circle of radius 4 in the plane x=2; we chose B=(2,4,0).
AB=4 forces A to be 60° away on the same circle, giving A=(2,2,2√3).
All given lengths are satisfied.

The centroid G of triangle ABH is G=(2,2,2√3/3).
Plane ACD has normal n_ACD=(0,-8√3,8); its distance from G is r=2√3/3, so the sphere S has radius r.

The locus T is the intersection of S with the plane (G−A)·P = G·G−A·G−r², which is horizontal (z=√3).
Its radius is 1, so area(T)=π.

Plane ABC has normal n_ABC=(-8√3,4√3,4).
The normal of T is n_T=(0,0,-4√3/3).
Thus cos φ=|n_T·n_ABC|/(|n_T||n_ABC|)=1/4.

The orthogonal projection of T onto plane ABC scales area by cos φ, giving area = π·(1/4)=π/4.
Hence k=4. [/THOUGHT] \boxed{4}

[THOUGHT] The Python output from the executed code is "pi/4", which directly gives the area as π/4. According to the problem statement, if the area is expressed as π/k, then k must be 4. This result is consistent with the geometric setup and the code verification, so no further computation is needed. [/THOUGHT] \boxed{4}