Given a circle of radius 1, the locus of all P such that there are points A, B, C on the circle such that ∠APB = ∠BPC = ∠CPA = 90 is a surface of a volume that has a volume V. Find V.
This is a 3D generalization of the fact that the locus of all on a circle makes a right triangle with the diameter.
(In 3 dimensions) the locus of points, P, for which <APB = 90° is a sphere with diameter AB.
To satisfy all three angle relations, P must be on the intersection of three spheres with diameters AB, BC, CA. Note that the three spheres can intersect for propitiously chosen A, B, C.
The three spheres having diameters AB, BC, CA will intersect iff the smallest side (=diameter) is at least as large as twice the circumradius of triangle ABC:
Min(a, b, c) >= 2abc / sqrt( (a+b+c)(a+b-c)(a-b+c)(-a+b+c) )
where a=BC, b=CA, c=AB.
Edited on May 1, 2019, 4:54 am
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Posted by FrankM
on 2019-05-01 04:53:59 |
Solution, I think