for $n=3k+2$ I think a strategy that forces the player making the second move to increase the smallest distance between the 3 counters at every step works.

Recast the game as one about triples of numbers by taking the angles of the triangles in radians and multiplying them by $n/\pi$. This gives us three positive integers summing to $n$. We start with $1, 1, n-2$, and the game ends when we get $[n/3], [(n+1)/3]$, $[(n+2)/3]$ where $[n]$ denotes the floor of $n$. A move is choosing two numbers and moving them closer together while keeping them integers with the same sum.

(EDIT: now that I've taken the time to iron out the details, I think OHO's post is correct)

I have a different answer. Write $n=3+2^k m$ where $m$ is odd. Then the first person wins if and only if $k$ is odd. Simplify the game the same way as MellowMelon, and prove that $x\geq y\geq z$ is a losing position if and only if $\{x-y,y-z\}=\{0,2^k m\}$ for some $m$ odd and $k$ even. The proof is by induction on $x-z$. 1. If $\{x-y,y-z\}=\{0,2^k m\}$ for some $m$ odd and $k$ even then any move will bring it to $x'\geq y'\geq z'$ with $\{x'-y',y'-z'\}\not=\{0,2^{k'} m'\}$ for all $m'$ odd and $k'$ even. Done by induction since $x'-z'<x-z$. 2. If $\{x-y,y-z\}=\{0,2^k m\}$ for some $m$ odd and $k$ odd then (by replacing $x,z$ with $\frac{x+z}{2},\frac{x+z}{2}$) it can be moved to $x'\geq y'\geq z'$ with $\{x'-y',y'-z'\}=\{0,2^{k-1} m\}$. 3. If $0\not\in\{x-y,y-z\}$ then we move it to either $(y,y,x+z-y)$ or $(y,\frac{x+z}{2},\frac{x+z}{2})$. One of these has the form $x'\geq y'\geq z'$ with $\{x'-y',y'-z'\}=\{0,2^{k'} m'\}$ for some $m'$ odd and $k'$ even because their $\max-\min$ are $|x+z-2y|$ and $\frac{|x+z-2y|}{2}$.

@ above : What if $x+z$ is odd for your 3rd case?

If $x+z$ is odd then $(y,y,x+z-y)$ works because $\max-\min=|x+z-2y|$ is odd.

Is there a more constructive way to show the answer (i.e. one where the answer doesn't appear to fall from the sky)?

http://rmms.lbi.ro/rmm2015/index.php?id=solutions_math One may notice the answer by checking the small value of n(n<25 ),but that would take a lot of time

The answer is that the first player wins iff $v_2(n-3)$ is odd. The motivation for this will be made clear in the following solution: First we rephrase the question as follows: Given a triple of positive integers $(a,b,c)$ with $a+b+c=n$, a legal move consists of replacing two of $a$, $b$, and $c$ with two other numbers that add to the same value but have a smaller difference. Initially the triple $(n-2,1,1)$ is given, both players alternate making a legal move, and whoever cannot make a legal move loses. This is equivalent to the problem because $(a,b,c)$ corresponds to the triangle that, when traversing counterclockwise along the circumcircle of the $n$-gon, has $\frac{2a\pi}{n}$, $\frac{2b\pi}{n}$, and $\frac{2c\pi}{n}$ as arcs between consecutive vertices. Now, say that two triples of integers $(a,b,c)$ and $(x,y,z)$ are equivalent if every legal move on $(a,b,c)$ corresponds to a legal move on $(x,y,z)$ and vice versa. Denote by $(a,b,c)\sim(x,y,z)$ the equivalence of $(a,b,c)$ and $(x,y,z)$. Clearly, $(a,b,c)\sim(a_0,b_0,c_0)$, where $a_0,b_0,c_0$ is any permutation of $a,b,c$. Lemma: If $k$ is any integer, then $(a,b,c)\sim(a+k,b+k,c+k)$ and $(a,b,c)\sim(-a,-b,-c)$. Proof: The legal move $(a,b,c)\rightarrow(a+i,b-i,c)$ corresponds to $(a+k,b+k,c+k)\rightarrow(a+k+i,b+k-i,c+k)$ and $(-a,-b,-c)\rightarrow(-a-i,-b+i,c)$ respectively. As a corollary, $(a,a,0)\sim(-a,-a,0)\sim(0,0,a)\sim(a,0,0)$. Now, note that $(n-2,1,1)\sim(n-3,0,0)$ by the lemma. Let $m=n-3$. By examining small cases, we notice that the first player wins the game of $(m,0,0)$ iff $v_2(m)$ is odd, and we will prove this by strong induction. Throughout the induction, we say that a player wins if they make a legal move that results in a triple that is equivalent to $(0,0,0)$ or $(1,0,0)$. This is clearly equivalent to the winning condition of the problem. For the base cases, note that for $(2,0,0)$, the first player makes the move $(2,0,0)\rightarrow(1,1,0)\sim(1,0,0)$ and wins. For $(3,0,0)$, the first player is forced to make the move $(3,0,0)\rightarrow(2,1,0)$, and the second player wins by making the move $(2,1,0)\rightarrow(1,1,1)\sim(0,0,0)$. For the inductive step, we will show that if $v_2(m)$ is odd, then the first player can give the second player the triple $(m',0,0)$ where $v_2(m')$ is even and $m'<m$. This is now equivalent to the second player going first with a triple that results in a loss for the first player by the inductive hypothesis. We will show that the opposite is true if $v_2(m)$ is even. Suppose that $v_2(m)$ is odd. Then, $2\mid m$, so the first player can make the move $(m,0,0)\rightarrow\left(\frac{m}{2},\frac{m}{2},0\right)\sim\left(\frac{m}{2},0,0\right)$. Because $v_2\left(\frac{m}{2}\right)$ is even, we are done. Suppose that $v_2(m)$ is even. Then, the first player must make the move $(m,0,0)\rightarrow(m-k,k,0)$ for some positive integer $k<m$. Without loss of generality, suppose that $m-k\geq k$. If $m-k$ is odd, then the second player makes the move $(m-k,k,0)\rightarrow(m-2k,k,k)\sim(m-3k,0,0)\sim(|m-3k|,0,0)$. Since $|m-3k|$ is also odd, we are done. If $m-k$ is even, then the second player can either make the move $(m-k,k,0)\rightarrow(m-2k,k,k)\sim(m-3k,0,0)\sim(|m-3k|,0,0)$ or the move $(m-k,k,0)\rightarrow\left(\frac{m-k}{2},k,\frac{m-k}{2}\right)\sim\left(k,\frac{m-k}{2},\frac{m-k}{2}\right)\sim\left(\frac{3k-m}{2},0,0\right)\sim\left(\frac{|m-3k|}{2},0,0\right)$. Because one of $v_2(|m-3k|)$ and $v_2\left(\frac{|m-3k|}{2}\right)$ is even, the second player can always force the first player into a losing position by the inductive hypothesis, so we are done.

Let $a,b,c$ be the arc lengths between the three counters, so in particular, $a,b,c$ are all positive integers that sum to $n$. The information of the state of the counters is fully captured by the multiset $\{a,b,c\}$. A move consists of replacing two of the elements $x$ and $y$ with $x'$ and $y'$ such that $(x')^2+(y')^2<x^2+y^2$ (or whatever). Due to the monovariant $a^2+b^2+c^2$, we see that the game must eventually stop. We have the following claim that classifies all winning positions. Claim: The losing positions are exactly \[\{k,k,n-2k\}\]where $1\le k<n/2$ and $v_2(n-3k)$ is even or infinite. Proof: Call a position losing if it satisfies the above criteria, and winning otherwise. Since the game terminates after a finite amount of moves, it suffices to verify three things. Any position that has no further moves available is losing. Any move from a losing position leads to a winning position. For any winning position, there is some move that leads to a losing position. We begin by proving the first item. Indeed, a position which has no further moves available is one of \[\{\ell,\ell,\ell\},\{\ell,\ell,\ell+1\},\{\ell,\ell+1,\ell+1\},\]all of which can be checked to be losing. We now verify the second item. Suppose we are in the losing position $\{k,k,n-2k\}$ where $v_2(n-3k)$ is even. Indeed, only states which have two equal numbers have the potential to be losing, so the only possible states that are one move from $\{k,k,n-2k\}$ and are not winning are \[\left\{k,\frac{n-k}{2},\frac{n-k}{2}\right\},\{4k-n,n-2k,n-2k\}.\]However note that \[v_2\left(n-3\frac{n-k}{2}\right)=v_2\left(\frac{3k-n}{2}\right)=v_2(n-3k)-1\]and \[v_2\left(n-3(n-2k)\right)=v_2\left(6k-2n\right)=v_2(n-3k)+1,\]both of which are odd. Thus, all positions one move away from $\{k,k,n-2k\}$ are winning, as desired. We now verify the third item. There are two (similar) cases. Case 1: Suppose we are in the winning position $\{a<b<c\}$. Replace $\{a,c\}$ with $\{b,n-2b\}$, which is a valid move since $a<b<c$. If $\{b,b,2n-b\}$ is losing, then we are done. Thus, we may suppose WLOG that it is winning, so \[v_2(n-3b)=v_2(a+c-2b)\]is odd. In particular, $a+c\equiv 2b\pmod{2}$, so $a+c\equiv 0\pmod{2}$. We now claim that the position \[\left\{b,\frac{a+c}{2},\frac{a+c}{2}\right\}\]is losing, which finishes. Indeed, \[v_2\left(n-3\frac{a+c}{2}\right)=v_2\left(\frac{2b-a-c}{2}\right)=v_2(a+c-2b)-1,\]which is even, as desired. Case 2: Suppose we are in the winning position $\{k,k,n-2k\}$ (so $k\ne n-2k$). Thus, we have $v_2(n-3k)$ odd, so in particular, $n\equiv k\pmod{2}$. So we may apply a move to go the position \[\left\{k,\frac{n-k}{2},\frac{n-k}{2}\right\}.\]This is losing as \[v_2\left(n-3\frac{n-k}{2}\right)=v_2\left(\frac{3k-n}{2}\right)=v_2(n-3k)-1,\]which is even, as desired. This proves the third item, which completes the proof of the claim. $\blacksquare$ To finish, we see that the first player wins if and only if $\{1,1,n-2\}$ is winning, or that $v_2(n-3)$ is odd. Remark: The hardest part of the problem is to find the claim. I could only do this by bashing out all the positions and finding which ones are winning and losing for $5\le n\le 16$. Once I found the claim though, proving it was straightforward.

Remark: Totally worth doing. Note that each position can be uniquely defined by an ordered triple $(i, j, k)$, where $i, j, k$ are the number of arcs between counters, pairwise, where the numbers always satisfy $i + j + k = n$. The initial position is described by $(n-2, 1, 1)$. Furthermore, valid move consists of choosing two of $i, j, k$ from $(i, j, k)$, WLOG $i < j$, keeping their sum constant, but making their difference $j - i$ smaller. Furthermore, note that the state of the position $(a, b, c)$ (winning or losing) is the same as any position of the form $(a + \ell, b + \ell, c + \ell)$ where $\ell$ can be any integer, or of the form $(-a, -b, -c)$, since only differences between the numbers matter. Thus, we want to find when $(n-2, 1, 1) = (n-3, 0, 0)$ is a winning position. I claim that in general, the position $(x, 0, 0)$ is winning if and only if $v_2(x)$ is odd. We will do this with induction. The base case clearly holds. Now, suppose this claim holds true for all $x < m$ for some $m$. If $v_2(m)$ is odd, we will show player $1$ can force player $2$ into to a losing position. Player $1$ can move to\[(\tfrac{m}{2}, \tfrac{m}{2}, 0) = (0, 0, -\tfrac{m}{2}) = (\tfrac{m}{2}, 0, 0)\]and since $v_2(\tfrac{m}{2}) = v_2(m) - 1$ is clearly even, this next positition for player $2$ is losing, so indeed $(m, 0, 0)$ is winning. If $v_2(m)$ is even, we will show player $2$ can force player $1$ into a losing position. Player $1$'s must move to a position of the form $(m-i, i, 0)$ where $i < \tfrac{m}{2}$. If $m-i$ is odd, then player $2$ can move to $(m-2i, i, i) = (m-3i, 0, 0)$ and since $m-i$ is odd, so is $m-3i$, and thus $v_2(m-3i) = 0$ is even. Hence, player $1$ is forced into a losing position. If $m-i$ is even, then player $2$ either copy the previous case and move to $(m-2i, i, i) = (m-3i, 0, 0)$, or he can move to\[(\tfrac{m-i}{2}, i, \tfrac{m-i}{2}) = (0, \tfrac{3i-m}{2}, 0) = (\tfrac{m - 3i}{2}, 0, 0).\]Note that $v_2(m-3i)$ and $v_2(\tfrac{m-3i}{2})$ are consecutive so one of them is even and thus one of the moves leads to a losing position. So player $2$ chooses that move and force player $1$ into a losing position. We have exhausted all cases in the necessary inductive step, so we have proven our claim. So indeed, for the original problem, player $1$ wins when $v_2(n - 3)$ is odd. $\blacksquare$

FWIW, I think it's worth mentioning that the "characterization of all winning / losing positions" may be much simplified by noting that we can take a geometrical approach, by expressing all the triples in terms of the corresponding coordinates in the triangle for Dumbass Notation (the one in inequalities). This gives us a way to induct up, as we can continue just adding "rings" (a.k.a. adding 3 to the overall side length of the triangle) to the triangle, and this triangle of the $n+3$ case will still have the same "states of the game" as the triangle of the $n$ case inscribed within it. This does need some casework though, as the $3$ cases (separated $\bmod 3$) should (?) be done independently.

Let A be the player that moves first and B be the player that moves second. We will prove that A wins if and only if $v_2(n-3)$ is odd. As in previous solutions, rephrase the problem in terms of integer triples; but for convenience, we further reduce all three entries by 1 (so any legal position $(a,b,c)$ has sum $a+b+c=n-3$). We first prove the following magical claim: Claim. Suppose that some player (say A) is faced with a position $(a,b,c)$ at some point, where $a\le b\le c$. If $a<b<c$ (call such situations bad), then A wins. Proof: Because $a<b<c$, the move $(a,b,c) \to (b,b,c+a-b)$ is legal. Now, suppose that A performs this move. If A wins from here on, then we are done. If A loses and B can win by responding $(b,b,c+a-b) \to (b,d,c+a-d)$, then A instead moves $(a,b,c) \to (b,d,c+a-d)$. This is clearly a legal move since it is the composition of two legal moves, and A wins in this case as well. $\square$ Now, we return to the problem. As an example, consider the case where $n$ is even. Then A's first move must result in a bad position, so B wins. For the general situation $n = 3 + 2^k p$ where $p$ is odd, based on our previous analysis, the optimal moves for both players are \[(0,0,2^kp) \to (0, 2^{k-1}p, 2^{k-1}p) \to (2^{k-2}p, 2^{k-2}p, 2^{k-1}p)\to (2^{k-2}p, 3\cdot 2^{k-3}p, 3\cdot 2^{k-3}p) \to \dots\]It is easy to see by induction that at each step, the three entries will include one odd multiple of $2^i$ and two equal odd multiples of $2^{i-1}$ for some $i$. Eventually one of the players would not be able to continue this, and the first player who depletes the powers of 2 wins. In other words, A wins if and only if $k$ is odd.

Despite this problem being somewhat easy, it is easily a masterpiece. The answer is all $n \ge 5$ such that $\nu_2(n-3)$ is odd. Part 1: Encoding the problem Understand each configuration as a game ``starting" with a triple $(a, b, c)$, where $a$, $b$, and $c$ are integers that denote the ratio of arc angles between the token positions such that $a+b+c=n$. We say that $(a, b, c)=(d, e, f)$ if and only if both configurations lead to Ash having a winning strategy or both lead to Ash not having a winning strategy. The area condition encodes as follows: a move is legal if and only if two of $a$, $b$, $c$ can be modified to have a smaller positive difference while keeping the same sum. Part 2: Triple equivalence properties What we can do is take an extension of the set of triples $(a, b, c)$ where we permit negatives; this way, we can prove equivalences more easily while keeping all calculations valid. We have the following properties: 1. $(a, b, c)=(b, c, a)=(c, a, b)$; 2. $(a, b, c)=(a+\delta, b+\delta, c+\delta)$ for any integer $\delta$; 3. $(a, b, c)=(-a, -b, -c)$. These are enough for us to prove that the set of desirable $n$ is the claimed one. Part 3: Triple chasing For each $n \ge 5$, write $m=(n-3)/2^{\nu_2(n-3)}$. Then triple chase as follows: \begin{align*} (n-2, 1, 1) = (n-3, 0, 0) = (2^{\nu_2(n-3)} \cdot m, 0, 0) = (2^{\nu_2(n-3)-1} \cdot m, 2^{\nu_2(n-3)-1} \cdot m, 0) \\ = (2^{\nu_2(n-3)-1} \cdot m, 2^{\nu_2(n-3)-2} \cdot m, 2^{\nu_2(n-3)-2} \cdot m) = (2^{\nu_2(n-3)-1} \cdot m, 2^{\nu_2(n-3)-2} \cdot m, 2^{\nu_2(n-3)-2} \cdot m) \\ = (2^{\nu_2(n-3)-2} \cdot m, 0, 0). \end{align*}Repeating the above iteratively, we find that \[ (n-2, 1, 1) = (2^{\nu_2(n-3) \bmod{2}}, 0, 0) = (2^{\nu_2(n-3) \bmod{2}}+1, 1, 1), \]so working out the RHS by hand, we find that $(n-2, 1, 1)$ has a winning strategy if and only if $\nu_2(n-3)$ is odd, as desired.