# Given three general lines in $\Bbb{R}^3$, can we construct the center and median plane of the one-sheet hyperboloid they determine?

It is well-known that $$3$$ general lines in $$\Bbb{R}^3$$ (that is, they are pairwise skew and their direction vectors are not coplanar) defines a unique hyperboloid of 1-sheet, i.e. the collection of lines intersecting all of them.

Suppose we know explicitly the position of the $$3$$ given lines but not the hyperboloid. Is it possible to construct the center and the median plane of the hyperboloid using only a straightedge (drawing lines and planes) and a compass (drawing spheres)? If not, neusis construction is also welcome.

Edit1: If we assume the $$3$$ lines to be $$\begin{cases} x=x_i+tl_i\\ y=y_i+tm_i\\ z=z_i+tn_i \end{cases} (t \in \Bbb{R}, 1 \le i \le 3)$$ and the unknown quadric to be $$Q(x,y,z)=A_1x^2+A_2y^2+A_3z^2+2(B_1yz+B_2zx+B_3xy)+2(C_1x+C_2y+C_3z)+D=0$$ then "$$3$$ lines are on the surface" is transformed into a set of linear equations in $$A_i,B_i,C_i,D$$ with coefficient polynomial about $$x_i,y_i,z_i,l_i,m_i,n_i$$ and the center is the only stationary point of $$Q$$, whose coordinates are linear in $$A_i,B_i,C_i$$. Thus the center is necessarily straightedge-and-compass constructible and the median plane should be constructed similarly as the axes of a hyperboloid are. However, the process above is too complicated, so I am looking forward to a much simpler method.

Edit2:I have successfully constructed the center using the central symmetry of the hyperboloid. Only the median plane is left.

• 1) What is the "neusis" method ? 2) What is the medial plane of a hyperboloid with one sheet ? 3) First time I encounter a straightedge and compass 3D problem : interesting. Jul 4, 2021 at 18:37
• @JeanMarie The definition of neusis construction can be found here. The 'median plane' (which is not a formal name) of a hyperboloid of 1-sheet is the only symmetric plane intersecting the surface by an ellipse. Jul 4, 2021 at 18:43
• @Narasimham Non-rotational hyperboloid still has central symmetry. Jul 8, 2021 at 20:17
• Ok, just asked what you meant. Jul 8, 2021 at 22:09

## 1 Answer

Construction of the center: Assume the $$3$$ lines to be $$l,m,n$$. Construct a plane $$p$$ through $$m$$ and parallel to $$l$$, then it will intersect $$n$$ at some point $$N$$. Construct a line $$l'$$ through $$N$$ and parallel to $$l$$. $$l'$$ intersects $$m$$ and $$n$$ both and is parallel to $$l$$, so it lies on the hyperboloid and is the symmetric image of $$l$$ with respect to the center $$O$$. Thus $$O$$ lies on the medial line of $$l$$ and $$l'$$. The same process can be applied to $$m$$ and $$n$$ and the $$3$$ medial lines concur at $$O$$.