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.

  • $\begingroup$ 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. $\endgroup$
    – Jean Marie
    Jul 4, 2021 at 18:37
  • 1
    $\begingroup$ @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. $\endgroup$
    – Zerox
    Jul 4, 2021 at 18:43
  • $\begingroup$ @Narasimham Non-rotational hyperboloid still has central symmetry. $\endgroup$
    – Zerox
    Jul 8, 2021 at 20:17
  • $\begingroup$ Ok, just asked what you meant. $\endgroup$
    – Narasimham
    Jul 8, 2021 at 22:09

1 Answer 1


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 $O$.


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