Choosing lines and points in D^2 I recently heard of a game between two players "Line" and "Point" and wanted to look for more information on it. However, without knowing the name of it (if it has one) finding more information is hard, has anyone heard of it? Is there a winning strategy for one of the players?
The game is as follows, it is played on the unit disk $D^2$ in $\mathbb{R}^2$ with the point $p_0 = (0,0)$ marked to begin with. Play alternates between L and P (starting with L) and on turn $n$ they do the following:
L chooses a new line $l_n$ through point $p_{n-1}$ and then P chooses a new point $p_n$ on line $l_n$ inside $D^2$.
This forms a sequence of points $(p_n)_{n = 1}^\infty$ in $D^2$. L wins if this sequence converges to a point in $D^2$, P wins if it does not.
As far as I can tell P has a winning strategy, but I my formal proof for this is a sketch at best.
 A: Line actually has a winning strategy: it can force a convergent sequence. The problem was posed and solved in the following paper:
J. Maly and M. Zeleny (2006), A note on Buczolich's solution of the Weil gradient problem: a construction based on an infinite game, Acta Mathematica Hungarica, Vol. 113, pp. 145-158.
A: Edit: The following has a serious gap, but I'll leave it up, for now, in case it gives someone an idea for a correct proof.  The error is that, because P always chooses the more distant (from $p_{n-1}$) of the two options for $p_n$, $L$ can cause him to go back and forth, rather than "farther and farther".  
I think P can win with the following strategy.  Fix a series $\sum_1^\infty t_n$ of positive terms, with sum 1, such that $\sum_1^\infty \sqrt{t_n}$ diverges.  Let $r_n=\sum_1^n t_k$.  Let P choose his $n$-th point $p_n$ so that its distance from the origin is $r_n$.  This is always possible, because $r_n>r_{n-1}$ and $l_n$ extends from $p_{n-1}$ all the way to the boundary of the disk.  In fact, P always has two options at the desired distance from the origin; let $p_n$ be the one farther from $p_{n-1}$ (or either one in case of a tie).  This completes the description of the strategy.  Now why does it win?  Easy estimates show that, once $r_n$ is close to 1 (i.e., for sufficiently large $n$), the angle between the radii from the origin to $p_{n-1}$ and from the origin to $p_n$ is at least of order $\sqrt{t_n}$.  (The smallest angle occurs when $l_n$ is perpendicular to the radius through $p_{n-1}$, and this smallest angle is close to $\sqrt{2t_n}$ if I've done the arithmetic correctly.)  Since $\sum_1^\infty \sqrt{t_n}$ diverges, it follows that the radii keep rotating farther and farther, not approaching a limit.  Therefore, the sequence $(p_n)$ fails to converge, and P wins.
A: P can always choose $p_n$ to be at distance $2^{-n}$ from $p_{n-1}$. 
EDIT: Never mind, I misread the problem, thought P would win if the sequence converged. 
