[THOUGHT] Let me work through this problem step by step.

First, I need to set up coordinates for the tetrahedron ABCD with the given constraints:

• AB = CD = 4
• BC = BD = 2√5
• H is the foot of perpendicular from A to line CD
• AH = 4
• Planes ABH and BCD are perpendicular

Let me place the coordinate system strategically. Since H is on line CD and AH ⊥ CD, I'll work with coordinates.

Let me start by setting up the geometry and finding all the points. [/THOUGHT] [PYTHON] import numpy as np from scipy.optimize import fsolve import sympy as sp from sympy import sqrt, Rational, pi, symbols, solve, simplify, Matrix

Let's set up coordinates

Place H at origin, CD along x-axis

Since AH ⊥ CD and AH = 4, A is at distance 4 from H perpendicular to CD

Let H = (0, 0, 0)

CD is along x-axis, so C = (-c, 0, 0) and D = (d, 0, 0) for some c, d > 0

CD = 4 means c + d = 4

A is perpendicular to CD from H, so A is in the yz-plane

Let A = (0, a_y, a_z) with a_y^2 + a_z^2 = 16 (since AH = 4)

Plane ABH contains A, B, H

Plane BCD contains B, C, D

These planes are perpendicular

Let's use symbolic computation

c, d, a_y, a_z = symbols('c d a_y a_z', real=True, positive=True) b_x, b_y, b_z = symbols('b_x b_y b_z', real=True)

H at origin

H = Matrix([0, 0, 0])

C and D on x-axis

C = Matrix([-c, 0, 0]) D = Matrix([d, 0, 0])

A perpendicular to x-axis from H

A = Matrix([0, a_y, a_z])

B general position

B = Matrix([b_x, b_y, b_z])

Constraints:

1. CD = 4: c + d = 4

2. AH = 4: a_y^2 + a_z^2 = 16

3. AB = 4: b_x^2 + (b_y - a_y)^2 + (b_z - a_z)^2 = 16

4. BC = 2√5: (b_x + c)^2 + b_y^2 + b_z^2 = 20

5. BD = 2√5: (b_x - d)^2 + b_y^2 + b_z^2 = 20

From 4 and 5: (b_x + c)^2 = (b_x - d)^2

=> b_x^2 + 2b_xc + c^2 = b_x^2 - 2b_xd + d^2

=> 2b_x(c + d) = d^2 - c^2 = (d-c)(d+c)

=> 2b_x4 = (d-c)*4

=> b_x = (d-c)/2

Plane BCD: normal vector is (B-C) × (D-C)

BC_vec = B - C DC_vec = D - C

Normal to BCD

n_BCD = BC_vec.cross(DC_vec) print("Normal to BCD (symbolic):", n_BCD)

Plane ABH: contains A, B, H

Vectors in plane: A-H = A, B-H = B

n_ABH = A.cross(B) print("Normal to ABH (symbolic):", n_ABH) [/PYTHON]

[THOUGHT] Let me continue the calculation to find all the coordinates and then compute the final answer. [/THOUGHT] [PYTHON] import numpy as np from sympy import *

Let's set up with specific coordinates

H at origin, CD along x-axis

From constraint c + d = 4 and b_x = (d-c)/2

Planes ABH and BCD perpendicular means their normals are perpendicular

n_BCD = (0, b_z*(c+d), -b_y*(c+d)) = (0, 4b_z, -4b_y)

n_ABH = (a_yb_z - a_zb_y, a_zb_x, -a_yb_x)

n_BCD · n_ABH = 0

0 + 4b_za_zb_x + 4b_ya_yb_x = 0

4b_x(a_zb_z + a_yb_y) = 0

Either b_x = 0 or a_zb_z + a_yb_y = 0

Let's try b_x = 0 first (meaning d = c = 2)

Then C = (-2, 0, 0), D = (2, 0, 0)

With b_x = 0:

BC = 2√5: c^2 + b_y^2 + b_z^2 = 20 => 4 + b_y^2 + b_z^2 = 20 => b_y^2 + b_z^2 = 16

AB = 4: (b_y - a_y)^2 + (b_z - a_z)^2 = 16

AH = 4: a_y^2 + a_z^2 = 16

Let's parameterize: a_y = 4cos(θ), a_z = 4sin(θ)

b_y = 4cos(φ), b_z = 4sin(φ)

AB^2 = (4cos(φ) - 4cos(θ))^2 + (4sin(φ) - 4sin(θ))^2 = 16

= 16*(cos(φ) - cos(θ))^2 + 16*(sin(φ) - sin(θ))^2 = 16

= 16*(cos²φ - 2*cos(φ)cos(θ) + cos²θ + sin²φ - 2sin(φ)*sin(θ) + sin²θ) = 16

= 16*(2 - 2*cos(φ-θ)) = 16

=> 2 - 2*cos(φ-θ) = 1

=> cos(φ-θ) = 1/2

=> φ - θ = ±π/3

Let's take θ = 0 (A in y-direction): A = (0, 4, 0)

Then φ = π/3: B = (0, 4cos(π/3), 4sin(π/3)) = (0, 2, 2√3)

Let me verify all constraints

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

print("AB =", np.linalg.norm(A - B)) print("CD =", np.linalg.norm(C - D)) print("BC =", np.linalg.norm(B - C)) print("BD =", np.linalg.norm(B - D)) print("AH =", np.linalg.norm(A - H))

Check H is foot of perpendicular from A to CD

CD is along x-axis, A = (0, 4, 0), so perpendicular foot is (0, 0, 0) = H ✓

Check planes ABH and BCD are perpendicular

Normal to BCD: BC × BD

BC = C - B BD = D - B n_BCD = np.cross(BC, BD) print("n_BCD =", n_BCD)

Normal to ABH: HA × HB

HA = A - H HB = B - H n_ABH = np.cross(HA, HB) print("n_ABH =", n_ABH)

Check perpendicularity

dot_product = np.dot(n_BCD, n_ABH) print("n_BCD · n_ABH =", dot_product) [/PYTHON]

[THOUGHT] The coordinates are verified. Now I need to continue with finding G (centroid of triangle ABH), the sphere S, the locus T, and finally the projection onto plane ABC. [/THOUGHT] [PYTHON] import numpy as np from numpy.linalg import norm

Verified coordinates

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

G = centroid of triangle ABH

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

Sphere S has center G and is tangent to plane ACD

Find equation of plane ACD

Vectors in plane: AC = C - A, AD = D - A

AC = C - A AD = D - A n_ACD = np.cross(AC, AD) print("Normal to ACD:", n_ACD) n_ACD_unit = n_ACD / norm(n_ACD) print("Unit normal to ACD:", n_ACD_unit)

Distance from G to plane ACD

Plane through A with normal n: n · (P - A) = 0

Distance = |n · (G - A)| / |n|

dist_G_to_ACD = abs(np.dot(n_ACD, G - A)) / norm(n_ACD) print("Distance from G to plane ACD (radius of S):", dist_G_to_ACD)

r_S = dist_G_to_ACD

T is locus of points P on S such that angle APG = π/2

This means AP ⊥ PG, so P lies on sphere with diameter AG

T = intersection of sphere S (center G, radius r_S) with sphere with diameter AG

Sphere with diameter AG has center at midpoint of AG and radius |AG|/2

M = (A + G) / 2 r_AG = norm(A - G) / 2 print("Center of sphere with diameter AG:", M) print("Radius of sphere with diameter AG:", r_AG)

T is the intersection of two spheres

Sphere 1: center G, radius r_S

Sphere 2: center M, radius r_AG

Distance between centers

d_GM = norm(G - M) print("Distance between G and M:", d_GM)

The intersection is a circle. Let's find its radius.

Using formula for intersection of two spheres

If spheres have centers C1, C2, radii r1, r2, distance d between centers

The intersection circle has radius r = sqrt(r1^2 - h^2) where h is distance from C1 to plane of circle

h = (d^2 + r1^2 - r2^2) / (2d)

d = d_GM r1 = r_S r2 = r_AG

h = (d2 + r12 - r2**2) / (2*d) print("h (distance from G to plane of T):", h)

r_T = np.sqrt(r12 - h2) print("Radius of circle T:", r_T)

Now find the plane of T

The plane of T is perpendicular to line GM and passes through point G + h*(M-G)/|M-G|

GM_unit = (M - G) / norm(M - G) center_T = G + h * GM_unit print("Center of circle T:", center_T)

Normal to plane of T is GM_unit

normal_T = GM_unit print("Normal to plane of T:", normal_T) [/PYTHON]

[THOUGHT] Now I need to find the orthogonal projection of circle T onto plane ABC, and compute its area. [/THOUGHT] [PYTHON] import numpy as np from numpy.linalg import norm

Circle T has:

- Center: (0, 2.5, √3/2)

- Radius: 1

- Normal: (0, √3/2, -1/2)

Plane ABC: find its equation

A = np.array([0, 4, 0]) B = np.array([0, 2, 2*np.sqrt(3)]) C = np.array([-2, 0, 0])

AB = B - A AC = C - A n_ABC = np.cross(AB, AC) print("Normal to ABC:", n_ABC) n_ABC_unit = n_ABC / norm(n_ABC) print("Unit normal to ABC:", n_ABC_unit)

Normal to plane of T

normal_T = np.array([0, np.sqrt(3)/2, -0.5]) normal_T_unit = normal_T / norm(normal_T) print("Unit normal to T:", normal_T_unit)

When projecting a circle onto a plane, the projection is an ellipse

The area of projection = area of circle * |cos(θ)|

where θ is the angle between the plane of the circle and the target plane

cos(θ) = |n_T · n_ABC| (dot product of unit normals)

cos_theta = abs(np.dot(normal_T_unit, n_ABC_unit)) print("cos(theta):", cos_theta)

Area of circle T

area_T = np.pi * 1**2 print("Area of circle T:", area_T)

Area of projection

area_projection = area_T * cos_theta print("Area of projection:", area_projection)

Express as π/k

area_projection = π * cos_theta = π/k

So k = 1/cos_theta

k = 1 / cos_theta print("k =", k)

Let's compute this more precisely with sympy

import sympy as sp from sympy import sqrt, pi, Rational, simplify, cos, acos

Exact values

A_s = sp.Matrix([0, 4, 0]) B_s = sp.Matrix([0, 2, 2*sqrt(3)]) C_s = sp.Matrix([-2, 0, 0])

AB_s = B_s - A_s AC_s = C_s - A_s n_ABC_s = AB_s.cross(AC_s) print("Exact normal to ABC:", n_ABC_s)

Normal to T: (0, √3/2, -1/2) or equivalently (0, √3, -1)

n_T_s = sp.Matrix([0, sqrt(3), -1])

Compute dot product and magnitudes

dot = n_ABC_s.dot(n_T_s) mag_ABC = sp.sqrt(n_ABC_s.dot(n_ABC_s)) mag_T = sp.sqrt(n_T_s.dot(n_T_s))

cos_theta_exact = sp.Abs(dot) / (mag_ABC * mag_T) cos_theta_exact = simplify(cos_theta_exact) print("Exact cos(theta):", cos_theta_exact)

Area = π * r^2 * cos(theta) = π * 1 * cos(theta)

area_exact = pi * cos_theta_exact print("Exact area:", simplify(area_exact))

k = 1/cos(theta)

k_exact = 1 / cos_theta_exact print("Exact k:", simplify(k_exact)) [/PYTHON]

[THOUGHT] The calculation shows that the area of the orthogonal projection of T onto plane ABC is π/4. Therefore, k = 4. [/THOUGHT]

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