2011 USAMO Problem 3, Hexagons. 
In hexagon $ABCDEF$, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy $\angle A = 3\angle D$, $\angle C = 3 \angle F$, and $\angle E = 3\angle B$. Furthermore $AB = DE$, $BC = EF$, and $CD = FA$. Prove that diagonals $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ are concurrent.

There are solutions here at AoPS, but I was wondering if anyone had any alternate takes on this problem/has a copy of the official solutions.
 A: Somehow I have a copy of the official solutions, see here.
I have not actually read the official solutions yet myself past a quick glance, so I would like to talk a bit only about the solutions on the AoPS Wiki you linked (mostly their "Solution 2"). While Solutions 1 and 3 there are genuinely semi-bashes, Solution 2 is really synthetic in nature. I am fairly sure that you can replace the use of complex numbers by something else and get an even shorter solution.
Let me explain further. Given the points $A$, $C$, and $E$, it is possible to construct $B$, $D$, and $F$ such that we have a valid hexagon $ABCDEF$. Now it takes $3$ parameters to give us a triangle $ACE$: its $3$ side lengths. So if we are given $3$ numbers, we can construct a valid hexagon $ABCDEF$.
On the other hand, if we look at the conditions they give us on $ABCDEF$, we are told that it is possible that we can get any valid hexagon via this construction: the "space of valid hexagons" is $3$-dimensional, and we have a construction of points in this space depending on $3$ parameters. If we can prove that all valid hexagons can be gotten this way, then we will immediately be done.
So what is the main issue with doing this? The main issue is that we do not know that the space of valid hexagons is connected. The space of valid hexagons could be, say, the union of $2$ $3$-dimensional balls, and our construction could give only one of these two balls. How do we get around this?
For simplicity, consider a $1$-dimensional analogue: the hyperbola $xy = 1$ has $2$ connected components (here, we mean connected in the sense that both sides are connected). But this is actually sort of an illusion: if we look at complex solutions to $xy = 1$ instead of the real solutions, we will get something connected. And so it is here: transferring to complex numbers allows us to prove that every valid hexagon can be gotten in this way without too much difficulty.
I should mention this is a relatively complicated example of this heuristic; this is why the proof the "phantom hexagon lemma," used in Solution 2 on the AoPS Wiki, requires a certain amount of explicit work. In cases where we have only algebraic equations, i.e. $xy = 1$, as opposed to equations with magnitudes and arguments, i.e. $|x| = 1$, this connectedness heuristic works perfectly (though there is something to check).
