MathLabs

Problem 2

Given a point AA and a segment BCBC, determine the locus of all points PP in space for which ∠APX=90∘\angle APX=90^\circ for some point XX on the segment BCBC.
Step 4 of 4: Describe the locus
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In plain words

The locus is the symmetric difference of the two diameter balls, including their boundaries.

f(0)f(1)≤0⟺P∈(S‾AB∖SAC∘)∪(S‾AC∖SAB∘)f(0)f(1)\le0\Longleftrightarrow P\in(\overline{S}_{AB}\setminus S_{AC}^{\circ})\cup(\overline{S}_{AC}\setminus S_{AB}^{\circ})
Detailed analysis

Thus PP is in exactly one of the two closed balls, with either sphere boundary retained. Equivalently, take the points in one ball but not the open interior of the other, and include both spherical surfaces. This also covers the degenerate collinear cases.