Consider the tetrahedron with and . Let be the foot of the perpendicular from to line . Suppose that the planes and are perpendicular to each other and that . Let be the centroid of triangle , and let be the sphere with center tangent to the plane . Let be the locus of all points on such that . Find the area of the orthogonal projection of onto the plane . [4 points]\subsection*{Numerical answer}If Area = , compute .
[THOUGHT] The coordinates are A(2,2,2sqrt(3)), B(2,4,0), C(0,0,0), D(4,0,0), H(2,0,0), G(2,2,2sqrt(3)/3). Sphere S: center G, radius 2sqrt(3)/3. Sphere with diameter AG: center M(2,2,4sqrt(3)/3), radius 2sqrt(3)/3. T is circle in plane z=sqrt(3), center (2,2,sqrt(3)), radius 1. Plane ABC normal n=(-2sqrt(3), sqrt(3), 1), |n|=4. Plane T normal (0,0,1). Cosine factor |1|/4 = 1/4. Area = pi * 1^2 * (1/4) = pi/4. Thus k=4. Verify normal magnitude and dot with code. [/THOUGHT] [PYTHON] import numpy as np n = np.array([-2*np.sqrt(3), np.sqrt(3), 1]) mag = np.linalg.norm(n) nt = np.array([0,0,1]) dot = np.dot(nt, n) factor = abs(dot) / (1 * mag) print(factor) print(mag) </PYTHON]
[THOUGHT] Previous numpy code had syntax error, likely tool issue with np.array or linalg. Compute with math module for verification. Expected mag=4, factor=0.25 confirming projection factor 1/4. [/THOUGHT] [PYTHON] import math a = -2 * math.sqrt(3) b = math.sqrt(3) c = 1 mag = math.sqrt(a2 + b2 + c**2) dot = c factor = abs(dot) / mag print(factor) print(mag) [/PYTHON]
[THOUGHT] Python confirms factor=0.25=1/4, mag=4. Projection area = π * 1² * (1/4) = π/4. Thus k=4. [/THOUGHT] \boxed{4}
Explanation
The reasoning derives from the geometric setup of the tetrahedron and subsequent constructions. Coordinates are established as A(2, 2, 2√3), B(2, 4, 0), C(0, 0, 0), D(4, 0, 0), H(2, 0, 0), and G(2, 2, (2√3)/3), the centroid of triangle ABH. Sphere S has center G and radius (2√3)/3. Locus T, defined by points P on S where ∠APG = π/2, forms a circle in the plane z = √3, with center (2, 2, √3) and radius 1.
The orthogonal projection of T onto plane ABC requires the formula for the projected area of a circle: π r² |cos θ|, where r is the circle's radius and θ is the angle between the normals of the planes. Plane T has normal (0, 0, 1). Plane ABC, spanned by points A, B, C, has normal n = (-2√3, √3, 1), with magnitude |n| = 4, verified by computation.
The cosine of θ is the absolute value of the dot product of the unit normals: |n_T · n_ABC| / (|n_T| |n_ABC|) = | (0,0,1) · (-2√3, √3, 1) | / (1 * 4) = |1| / 4 = 1/4. Thus, the projected area is π (1)² (1/4) = π/4.
Python verification computes the dot product and magnitudes: dot = 1, |n| ≈ 4, factor ≈ 0.25, confirming |cos θ| = 1/4. Therefore, the area is π/4, so k = 4.