USAMO 2011 Question 2 - Solitaire on a pentagon 
An integer is assigned to each vertex of a regular pentagon so that
  the sum of the five integers is 2011. A turn of a solitaire game
  consists of subtracting an integer $m$ from each of the integers at
  two neighbouring vertices and adding 2m to the opposite vertex, which
  is not adjacent to either of the first two vertices. (The amount $m$
  and the vertices chosen can vary from turn to turn.) The game is won
  at a certain vertex if, after some number of turns, that vertex has
  the number 2011 and the other four vertices have the number 0. Prove
  that for any choice of the initial integers, there is exactly one
  vertex at which the game can be won.

I have realised that the order of the turns does not matter, however I cannot seem to get any further. 
 A: I'm going to deal with the general case for any odd number of vertices and any starting sum (with noted exceptions), because I like giving general solutions :)
Let there be $n$ vertices (where $n$ is odd), labelled $0$ through $n-1$, with vertex $r$ having an associated integer $x_{r}$, and with $\Sigma x = \sum_{r=0}^{n-1}x_{r}$. In your case, $n=5$ and $\Sigma x=2011$.
I also presume that $x_{r}$ can be negative, since the question merely states "integer", and because otherwise there are some states that are unwinnable - for example, in your scenario, any state with one vertex having $2010$ and any other having $1$ is unwinnable if $x_{r}\ge0$ is enforced.
Method:


*

*Find some function $I$ of $\{x_{r}\}$ that is invariant under the allowed transformations of $\{x_{r}\}$. These transformations take the form:
$$\begin{align}x_{r} &\rightarrow x_{r}-m\\x_{r+1} &\rightarrow x_{r+1}-m\\x_{r+k+1} &\rightarrow x_{r+k+1}+2m\end{align}$$
where $n=2k+1$ and indices are considered modulo $n$. Try building a quantity that is invariant modulo $n$ - you don't need to get too complicated, a linear combination of $\{x_{r}\}$ will suffice. This invariant will never change throughout the game, so it's a useful quantity to know.

*Consider the $n$ possible winning states (one for each vertex), and the corresponding values of your invariant - $I_{r}$ being the value of the invariant for the win state where only $x_{r}$ is non-zero. If there are no repeated members in $\{I_{r}\}$, then any initial state can have at most one winning state, since there is at most one winning state you can reach. If $I \notin \{I_{r}\}$, then you cannot win from this state, since you can only reach states that have invariant $I$, and no winning state has invariant $I$. To answer the question, we need $\{I_{r}\}$ to comprise distinct elements, and to contain every possible value of $I$ - hence every state leads to one and only winning state.

*Now plug in values to solve for the specific case (admittedly, I solved for the specific case first then generalised, you might find it easier to do the process with specific values of $n$ and $\Sigma x$ in mind).
Answer in spoilers below:
Part 1 (picking an invariant):

 Choose the function $I=(\sum_{r=0}^{n-1}rx_{r})\ (\mathrm{mod}\ n)$, which transforms to $I - rm - (r+1)m + (r+k+1)2m\ (\mathrm{mod}\ n) = I + (2r+2k+2-2r-1)m\ (\mathrm{mod}\ n) = I + (2k+1)m\ (\mathrm{mod}\ n) = I$and is hence invariant since it didn't matter what $r$ or $m$ was. It doesn't matter ultimately which vertex is vertex $0$ - while $I$ will depend on the labelling of vertices, it will still indicate the same winning vertex - proof after part 2.

Part 2 (using invariant to deduce possible winning states):

 The function's codomain is the set $\mathbb{Z}_{n}$, the integers modulo $n$, and we have $I_{r} = r\Sigma x\ (\mathrm{mod}\ n)$. For any $\Sigma x$ coprime with $n$, $\{I_{r}\}$ will be a set equivalent to $\mathbb{Z}_{n}$; the winning state is then the $I_{r}$ that equals $I$ - i.e. the winning vertex is $r = I\cdot(\Sigma x)^{-1} \ (\mathrm{mod}\ n)$, where $(\Sigma x)^{-1}$ is the multiplicative inverse of $\Sigma x$ modulo $n$ - hence why we require $\Sigma x$ coprime to $n$ as otherwise $(\Sigma x)^{-1}$ does not exist.

Proof why labelling doesn't matter (uses results from parts 1 and 2):

 To prove why the choice of which vertex is vertex $0$ is irrelevant, increment all indices by some amount $a$ (and then reduce modulo $n$). $I$ will then increase by $a\Sigma x\ (\mathrm{mod}\ n)$, thus the winning vertex's index will increase by $a\Sigma x(\Sigma x)^{-1}\ (\mathrm{mod}\ n) = a$, so the winning vertex is the same. As long as you label the vertices consecutively, it doesn't matter which vertex you start from.

Part 3 (solution for specific case):

 Now we have the general solution, in your specific case, all we need show is that $2011$ is coprime with $5$; it clearly is, and in fact $2011 = 1\ (\mathrm{mod}\ 5)$, which makes life very easy as its multiplicative inverse is just $1$. Now just compute $I = x_{1}+2x_{2}+3x_{3}+4x_{4}$, and then $I\ (\mathrm{mod}\ 5)$ gives you the winning vertex.

