Number of points to define cone I am looking for the minimum number of surface points in three-dimensions required to identify an arbitrary cone. i.e. a cone with unknown origin, orientation, and angle. Note that the points must be arbitrarily chosen in the sense that they are not identifiable as being the origin for example.  
Hans Musgrave has come up with a better description of what I want in his answer For an arbitrary fixed cone C, determine the minimal integer n satisfying "if S is a set of surface points on that cone with size at least n then C is the only cone with S as a subset of its surface points" 
 A: Consider an arbitrary line in 3D as the axis of the cone.  The equation of the line is:
$$\frac{x-x_0}{l}= \frac{y-y_0}{m} = \frac{z-z_0}{n}$$
where  $x_0$,$y_0$,$z_0$ are the coordinates of a given point on the line, and $(l, m, n)$ are the direction vector of the line.
We can arbitrarily scale the direction vector so that $l=1$, leaving five arbitrary constants in the equation of the line.
We can pick a point on the line to be the apex of the cone by specifying values of $x$, and using the equation of the line to calculate and $y$ and $z$.  So the choice of apex introduces one more arbitrary constant.
Now we can calculate the distance from a general point to the axis, and the distance from a general point to the apex.
The ratio of these two numbers, line distance over apex distance, for points on the cone, must be a constant, the sine of the apex angle. Yet another arbitrary value.
So if we want to eliminate the seven constants introduced, we would need seven points on the cone substituted into this formula for the cosine.
A: If these points shouldn't be on the surface of the cone, we could use this as a "rule of thumb":


*

*origin has 3 "degrees of freedom" (i.e. you can determine it with 3 reals)

*orientation has 2

*angle has 1


This sums to 6 reals. This can be given by 2 points. For example:


*

*a point to determine its origin

*another point to determine


*

*its orientation (by saying that its rotation symmetry axe must be on it)

*and also its angle (by the distance from the origin)




However, if you want to determine it by the points of its surface, then the task is harder, because we have a limit on the allowed points. But the solution above serves as a lower limit.
If we would know the origin of the cone, then 3 surface points would be enough. This can be shown by the fact that we only need to find the plane, to which projecting the 3 points, we get a circle. Then this circle (and the origin) would determine the cone.
The construction algorithm for this:
We have the origin (call $O$), and the 3 points (call $P_1, P_2, P_3$). Then project these 3 points to the unit sphere with the center of the origin (getting $P_1', P_2', P_3'$). The circle determined by $P_{1-3}'$ will be fully on the surface of the cone.

What means, 3 surface points and the origin, determines a cone (in the general case).
Not knowing the origin, only a fourth surface point, we could do this algorithm with still infinite number of possible origins: as we could see above, if we have 3 surface points, we can fix the origin to anywhere to construct the cone.
If we have also a fourth point on the surface, making the same algorithm is not always successful: it will succeed only for the points, where the projected $P_{1-4}'$ points will be on a circle. Thus, having also a fourth point, significantly reduces the possible origins, from 3 unknowns to 1. It is because $P_4'$ can be anywhere on the circle of $P_{1-3}'$, thus it solves only 2 unknowns and not 3.
Having also a $\underline{\underline{\text{fifth}}}$ point will be enough in the general case.
A: There are a few plausible interpretations of your question. Since you mentioned orientation, I'm assuming 1-sided cones whose conic sections are ellipses, parabolas, or half-hyperbolas.


*

*One interpretation is to consider a fixed arbitrary cone and come up with a minimal set of points which are surface points of that cone and none other. In this case, $4$ points do not suffice, $5$ might, and $6$ do. In the case that $5$ points suffice, it is at least required that no $4$ such points are coplanar and no $3$ such points are colinear. To show that $6$ points suffice, consider the set $$\{(-2, 2, 0), (-1, 1, 0), (0, 0, 0), (1, 1, 0), (2, 2, 0), (0, 2, 2)\}.$$ This set determines a unique cone with a vertex at $(0,0,0)$, and since every cone is the image of that cone under a bijective transformation of $\mathbb R^3$, the same property holds for all cones. Showing $4$ points do not suffice is routine.

*Otherwise, one plausible way to think about the problem is to fix a cone $C$ and request a minimal integer $n$ satisfying "if $S$ is a set of surface points on that cone with size at least $n$ then $C$ is the only cone with $S$ as a subset of its surface points". This is a well-defined question for circles, and the answer is known to be $3$. In the case of cones, no number suffices. Any $n$ colinear points are on the surfaces of infinitely many cones.
