Game about rational polygons/rational distance in the plane I've been thinking about the following problem for a while

Two players $A$ and $B$ play the following game: In each step a player chooses a point in $\mathbb{R}^2$. The one who creates an irrational distance between his point and one of the previous points lose. Player $A$ start.
Is there any win strategy?

A theorem of J.H.J. Almering (see here) imply that there is not a win strategy to finish each game in less than 5 steps.
Has this problem been studied before?
What about the simplest question of just extending a fixed rational $n$-polygon to a rational $n+1$-polygon? (A rational polygon is a polygon in which each side and each diagonal has a rational length)
 A: The game is in fact, a draw. We can show this by first, showing that regardless of the strategy of the second player, player 1 (indicated by P1) can't lose, and then showing that P2 can force a draw.
The strategy that guarantees player 1 doesn't lose is the following: Let P1 play on odd turns, and P2 on even. Let $X_n$ be the play on the $n$th turn. So, P1's strategy is
$$X_{2n+1}=\begin{cases}(0,0) & n=0 \\-X_{2n} & n\in\mathbb N\\\end{cases}$$
How can we show that this is a non-losing strategy regardless of P2's play? If $X_{2n+1}$ was the losing move, for some $n\in\mathbb N$, that implies that either, $d(X_{2n+1},X_{2n})$ is irrational, or $d(X_{2n+1},X_k)$ is irrational, for some $k<2n$.
Note that $d(X_{2n+1},X_{2n})=2\cdot d(X_{2n},X_1)$, so clearly, case 1 is impossible. Suppose $k$ is odd. Clearly, $k\neq1$ since $$d(X_{2n},X_1)=d(X_{2n+1},X_1)$$ Then, $k=2m+1$ for some $m\in\mathbb N$. But, $d(X_{2n+1},X_{2m+1})=d(X_{2n},X_{2m})$, a contradiction since the game continued to turn $2n+1$.
So, $k=2m$ for some $m\in\mathbb N$. But, $$d(X_{2n+1},X_{2m})=d(X_{2n},X_{2m+1})$$ We showed via case 1 that $m\neq n$, so this implies that $2m+1 < 2n$, which again is a contradiction since the game reached turn $2n+1$.
So, this strategy is in fact, non-losing for P1 regardless of what P2 does.
So, the best player 2 can hope for is not losing. However, P2 can guarantee not losing by only playing on the x-axis of $X_1$, and if P1 deviates, following the strategy presented above to avoid losing. So, this game is a draw.
