Let $A,B,C$ be points on the unit circle in the $xy$ plane, and $P$ be a point in space such that $PABC$ is a trirectangular tetrahedron with $P$ at the vertex. Find the locus of $P$ as $A,B,C$ vary on the circle.
From this stackexchange post, I know that $P$ is uniquely defined for any $A,B,C$, but how would I go about finding the locus of all points $P$?
Comment: One way for constructing a trirectangular tetrahedron is as per comment by user achille hui:
find the orthocenter of triangle (H).
draw a perpendicular at H on plane the triangle ABC locates, like in figure xy plane.
Find the mid point of one side, say AC.
Draw a sphere center at midpoint and crossing point A and C.
Use intersect command to find the intersection of the perpendicular and the sphere, this point is P which is the vertex of trirectangular tetrahedron .
Now if the triangle is equilateral, then the vertex will be I which is lower than point T(intersection of perpendicular from G and sphere based on circumcircle. If the triangle is right angled, then the orthocenter is on the circucircle , and the sphere on the circumcircle is also the one constructed using the mid point of the hypotenuse, which is the center of the circumccircl(also the sphere), that is the vertex is on circumccircle. For other types of triangle the vertex P is somewhere between I and plane xy. Hence the locus of P seems to be an ellipsoid inscribed in the sphere on circumcircle. The formula is:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$
Here $b=r$, $a=r$ and $c=OI$, since $OI<r\rightarrow c<a$ the ellipsoid is oblate.