Brazilian Math Olympiad

a. The probability that the cone of the four vectors is proper is $\frac{7}{8}$ so the probability that the cone is all of ${\mathbb{R}}^3$ is $\frac{1}{8}$. Construct a vector $u_{12}$ that is normal to the plane spanned by $v_1$ and $v_2$ oriented so that $v_3\cdot u_{12}>0$. Then the half-space $\{w\mid w\cdot u_{12}\ge 0\}$ contains the cone generated by $\{v_1,v_2,v_3\}$. It follows that if $v_4\cdot u_{12}>0$, the cone generated by all four vectors will be contained in the same half-space. So to keep the cone from being proper, we must assume that $v_4\cdot u_{12}<0$.

Similarly, find $u_{13}$ orthogonal to $v_1$ and $v_3$ with $v_2\cdot u_{13}>0$ and $u_{23}$ orthogonal to $v_2$ and $v_3$ with $v_1\cdot u_{13}>0$. The cone is proper – contained in a half space – if and only if at least one of the three values $v_4\cdot u_{ij}>0$. I further claim that if all three of those dot products are negative, then the cone covers all of space. If $v_1,v_2$ and $v_3$ are fixed, the three signs of the dot products $v_4\cdot u_{ij}$ are not independent. But if we average over all choices of the vectors, then $-v_1$ occurs exactly as often as $+v_1$ and so on. We conclude that on average the three dot products in question are negative with probability $\frac{1}{8}$.

b. Given $v_1,v_2$ and $v_3$ then $v_4$ lies in the interior of the cone generated by those three if and only if $v_4\cdot u_{ij}>0$ for all three such dot products. So there is a $\frac{1}{8}$ chance that $v_4$ lies in $C(v_1,v_2,v_3)$. Similarly, there is a $\frac{1}{8}$ chance that $v_2$ lies in $C(v_1,v_3,v_4)$. These two events are disjoint: only one vector can be in the interior of a triangle of the other three. So the probability we seek is the union of four disjoint events, each of probability $\frac{1}{8}$, which gives a probability of $\frac{1}{2}$.