What are the conditions for zero to belong to a convex hull of a set of points What are the conditions on a set of data points such that zero belongs to the convex hull defined by these points. (A part from expressing zero as a positive combination of the data points with coefficients equal to one).
I am looking for some established results for the case of dimension 3 involving distances sign of the coordinates...
Thank in advance.
 A: An algorithm to determine if 0 belongs to the interior of the convex hull defined by $n$ points $v_1,\ldots, v_n$ in $\mathbb{R}^3$. Without loss of generality, assume no two of the vectors are linearly dependent and not all of the points lie on the same plane. 
For distinct $i<j$, define:
$$A_{ij}=\{v_k \cdot (v_i \times v_j): k\neq i,j \}.$$
Claim 1. If for all $i<j$, the set $A_{ij}$ contains both positive and negative numbers, then 0 belongs to the interior of the convex hull of $v_1,\ldots v_n$. The converse also holds. 
Proof. First suppose 0 belongs to the interior of the convex hull so that $0=\sum_{i=1}^n \lambda_i v_i$ and all $\lambda_i >0$ with $\sum_i \lambda_i=1$. For a fixed $i<j$, take the dot product of the two sides of the equation with $v_i \times v_j$ to get $0=\sum_{k\neq i,j} \lambda_k v_k \cdot (v_i \times v_j)$. Since $\lambda_k>0$ for all $k$ and the numbers $\lambda_k v_k \cdot (v_i \times v_j)$ add up to zero, at least one must be positive and at least one must be negative. Note that if they are all zero, then all of $v_k$s must lie on the plane spanned by $v_i$ and $v_j$ in which case the convex hull has empty interior.  
Conversely, suppose 0 is not in the convex hull of $v_1,\ldots, v_n$. Let $P$ be a plane passing through the origin so that the entire convex hull lies on one side of the plane. Rotate the plane around any axis until it touches one of the points say $v_1$. All other points still lie on one side of the plane. Now rotate the plane around the line passing through the origin and $v_1$ until it touches another point say $v_2$. We have found a plane through the origin and $v_1$, $v_2$ such that one side of the plane contains no points from $v_1,\ldots, v_n$. Clearly all of the $v_i$s either make an acute or right angle with $v_1\times v_2$ or an obtuse or right angle with $v_1 \times v_2$. Therefore $A_{ij}$ cannot contain both positive and negative numbers.
Claim 2. If for all $i<j$, the set $A_{ij}$ is not a set of all positive or all negative numbers, then 0 belongs to the closed convex hull of the points. 
Proof is similar to the proof of Claim 1. If 0 is not in the interior but in the closed convex hull, then clearly it is on the boundary.
