Consider a sphere and a plane π. For a variable point M∈π, exterior to the sphere, one considers the circular cone with vertex in M and tangent to the sphere. Find the locus of the centers of all circles which appear as tangent points between the sphere and the cone. Octavian Stanasila
Problem
Source: Romanian IMO Team Selection Test TST 1988, problem 1
Tags: geometry, 3D geometry, sphere, geometric transformation, rotation, geometry unsolved
01.10.2005 21:19
Let S be the center of the sphere σ and let N be the foot of a normal to the plane π from the point S. Denote r the radius of the sphere σ and h = NS the height of the center S above the plane π. First, assume that h≥r. The tangency circle (L) of the cone with the vertes M∈π is the intersection of the sphere σ with a sphere τ centered at the midpoint T of the segment SM and with radius TS=TM=MS2. The plane λ of the tangency circle (L) intersects the line m≡SM at a point L, the center of the circle (L). Since the problem is symmetrical with respect to rotation around the axis n≡NS, we need to consider only a point M moving on a fixed line p in the plane π passing through the point N, find the locus of these tangency circles (L) and rotate the locus around the axis n≡NS. Let μ be the plane passing through the points M, N, S. The spheres σ,τ project to the plane μ as circles (S), (T) intersecting at points A,B, the plane λ of the tangency circle (L) as a line l≡AB, l⊥m. The center L≡l∩m of the tangency circle (L) is identical with the midpoint of the segment AB. Since MS is a diameter of the circle (T) and the angle ∠MNS=90∘ is right, the circle (T) always passes through the point N. From the right angle triangle △MAS with the right angle ∠MAS=90∘ and the altitude AL to the hypotenuse MS, we have AL2=LS⋅LM MS⋅LS=(LM+LS)⋅LS=LM⋅LS+LS2=AL2+LS2=SA2 where r = SA is a constant radius of the sphere σ and its projection, the circle (S). Let the line l≡AB intersect the line n≡NS at a point Q. The right angle triangles △MNS∼△LQS are similar, having the angle ∠MSM≡∠QSL common. Consequently, QSLS=MSNS, QS=MS⋅LSNS=AL2NS=r2h Thus we see that the length of the segment QS does not depend on the position of the point M on the line p. Since the angle ∠QLS=90∘ is right, the locus of the points L is a circle with diameter QS. Rotating this circle around the axis n≡NS, the locus of the centers of all possible tangency circles (L) is a sphere ω with diameter QS=r2h≤r. If we assume h < r instead, then QS=r2h>r, the locus sphere ω intersects the given sphere σ in a circle (N) lying in the given plane π and the locus of all possible tangency circles (L) is only the portion of the locus sphere ω inside the given sphere σ.
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02.10.2005 21:04
Uhm ... isn't this construction simply the inversion of the plane in the sphere??