Looking at three wires, can I make it look like they meet? I was walking down the street and there were three wires (used by the tram) above my head. I looked up and by hazard it looked to me like they intersected all three in the same point. I asked myself if this is always the case for three wires.
Mathematically speaking, let there be three straight lines $a,b,c$ in the three dimensional space. What are the conditions that there exists a fourth line $d$ that actually intersects $a$, $b$ and $c$?
I found some questions about related things but there were things like quadrics and projective spaces and it went a bit over my head. But I could not come up with a simple, elegant proof, that can be understood by, say, a first semester student.
 A: Let $a$, $b$ and $c$ be three given lines: we must distinguish various cases to find when a line $d$ intersecting all three of them exists. First of all, if two among the given lines (for instance $a$ and $b$) meet at a certain point $P$, then we can take as $d$ the line passing through $P$ and through any point of $c$. In the following I will then assume that $a$, $b$, $c$ are pairwise disjoint.
If two among the given lines (for instance $a$ and $b$) are parallel, then they lie on a plane $\gamma$. If $c$ intersects $\gamma$ at a certain point $P$ then we can take as $d$ any line in $\gamma$ passing through $P$ and not parallel to $a$ and $b$. If, on the other hand, $c$ is parallel to $\gamma$ (and not lying on it), then no line $d$ can exist: any line intersecting $a$ and $b$ lies on $\gamma$ and cannot intersect $c$. 
It remains to examine the case when no two of the given lines are parallel, that is when $a$, $b$ and $c$ are pairwise skew. Given two skew lines, there exists a unique plane such that the first line lies on the plane and the second one is parallel to the plane. Let then $\alpha$ and $\beta$ be two planes passing through $c$ and such that $\alpha\parallel a$, $\beta\parallel b$. Any other plane $\gamma$ passing through $c$ and different from $\alpha$ and $\beta$ will intersect $a$ and $b$ at points $A$ and $B$ respectively. Line $AB$ is coplanar with $c$ but cannot be parallel to $c$ for every possible choice of $\gamma$: if it were so, we would then be able to construct other two points $A'\in a$ and $B'\in b$ such that $AB\parallel c$ and $A'B'\parallel c$ but in that case we would have $AB\parallel A'B'$ so that these four points would be coplanar, and so $a$ and $c$ would be coplanar, which is against our assumptions. We have then proved that in this case line $d$ always exists.
To summarize this discussion, the only case $d$ does not exist is when two of the given lines are parallel and their plane is parallel to the third line.
