Slovenija 2008

Let us call a non-rectangular isosceles trapezoid with the bases parallel to two of the edges of the board regular. We show that the second player can always win. After the first player's every move the second player should check to see if they can form a regular trapezoid by the placing of a single coin. If not, the second player should put the coin onto the board in such a way, that after their move the arrangement of the coins will have the vertical line across the middle of the board as its axis of symmetry (i.e. the second player's move mirrors the first player's move with regard to this line).

Assume that the first player wins by placing a coin onto the square which we denote by $a_0$. Let $a_1,a_2$ and $a_3$ denote the other three squares that together with $a_0$ form a regular trapezoid (i.e. all four are different and each contains a coin). Let $b_0,b_1,b_2$ and $b_3$ be the corresponding mirror images. Squares $b_1,b_2$ and $b_3$ contain coins and $b_0$ is empty. First, we show that none of the squares $a_0,a_1,a_2,a_3,b_0,b_1,b_2$ and $b_3$ coincide.

Since the length of the board is even, we have $a_i\ne b_i$ for all $i=0,1,2,3$. If $a_i=b_j$ for some $i\ne j$, then $b_i=a_j$, so the side $a_i{a}_j$ of the trapezoid $a_0{a}_1{a}_2{a}_3$ is horizontal. If this is the case, the side $a_k{a}_l$ for $\{k,l\}=\{0,1,2,3\}\setminus \{i,j\}$ is also horizontal, because a regular trapezoid has no right angles. Since this trapezoid is isosceles, we get $b_k=a_l$ and $a_l=b_k$ for $l\ne k$. This implies $b_0=a_m$ for some $m\ne 0$. A contradiction, the square $b_0$ is empty and squares $a_1,a_2,a_3$ contain coins.

Hence, all those squares are necessarily different and before the first player made his or her last move the triples $a_1,a_2,a_3$ and $b_1,b_2,b_3$ were covered with coins. This implies that at least one of the triples was covered with coins before the second player made his or her last move, so following the strategy the second player should already have won.