AUT_ABooklet_2023

a) In the following, let $A_n$ denote the end-point of the segment that Alice drew in her $n$-th turn (assuming the game has not ended by then), and let $B_n$ denote the end-point of Bob's $n$-th segment. Furthermore, let $B_0$ denote the starting point of Bob's first segment. If Alice can force an end to the game, so can Bob by applying the same strategy and ignoring Alice's first move. It is therefore sufficient to prove that Alice can force an end. Bob must always choose the $n$-th end-point $B_n$ on the circle with radius $1$ and center in $A_n$. We name this circle $k_n$. Furthermore, let $l_n$ denote the line perpendicular to $B_{n-1}{A}_n$ through $A_n$. We now note that if Bob chooses his end-point in such a way that his segment forms an acute angle with the preceding segment (such that $B_n$ lies on the same side of $l_n$ as the segment $\overline{B_{n-1}{A}_n}$), Alice can end the game with her next move. Now let $h_n$ denote the part of $k_n$ on the opposite side of $l_n$ from $\overline{B_{n-1}{A}_n}$ (including the intersection points of $l_n$ and $k_n$). In the following, we only need to consider the case in which Bob chooses the point $B_n$ on the semi-circle $h_n$. Let $B$ denote the set of all points, whose distance from the first drawn segment is less than $1$. The set $B$ consists of a $1\times 2$ rectangle and the interior of two semi-circles. Bob chooses $B_1$ on $h_1$. In the next move, Alice can choose $A_2$ as close as she wishes to $A_1$. Let $r$ denote the distance between $A_2$ and $A_1$. We now consider two cases.

Case 1: $B_0,A_1,B_1$ do not lie on a common line.

If Alice chooses $A_2=A_1$ (which she is not allowed to do, according to the rules), $h_2$ will overlap with the semicircular edge of $B$ at one end. The other end of $h_2$ must therefore lie in the interior of the rectangular section of $B$, which means that this end must have a positive distance from the edge of $B$. Since Alice can choose an arbitrarily small value of $r$, she can (for reasons of continuity) move the point $A_2$ away slightly from $A_1$ towards the rectangular section of $B$ such that $h_2$ comes to lie completely in the interior of $B$. This means that $B_2$ will lie completely in the interior of $B$, and all its points thus have a distance less than $1$ from the first segment. Alice can therefore certainly choose her next segment in such a way that it intersects the first segment.





Case 2: $B_0,A_1,B_1$ lie on a common line. In this case, Alice cannot choose $A_2$ in such a way that $h_2$ lies completely in the interior of $B$. If Bob chooses $B_2$ in the interior of $B$, Alice can choose her next segment in such a way that it intersects the first segment, ending the game. We can therefore assume that Bob chooses $B_2$ on $h_2$ outside of $B$. In this case, Alice can choose her next point $A_3$ in such a way that its distance from $A_2$ is at most $r$. By the triangle inequality, the distance from $A_3$ to $A_1$ is then at most $2r$. If $r=0$ (which is not allowed by the rules), we would have $A_3=A_1$. In this case, analogously to the previous case, $h_3$ would overlap with the semicircular edge of $B$, and the other end would lie in the interior of the rectangular part of $B$ with a positive distance from the edge. Since Alice can choose $2r$ arbitrarily small, she can (again by reasons of continuity) move $A_3$ slightly away from $A_1$ toward the part of $h_3$ in the interior of the rectangular section of $B$, such that $h_3$ comes to lie completely in the interior of $B$. Then $B_3$ lies in the interior of $B$, and Alice can choose her next segment in such a way that it intersects the first segment.

b) We will show that each of the players can always make a move with which they do not lose. This is trivially the case for the first two moves, so we assume without loss of generality that at least two segments have already been drawn. Let $s$ denote the last segment drawn and $t$ the one drawn immediately before that. Furthermore, let $S$ denote the union of all segments that were drawn before $s$ and $t$. Let $r$ denote the smallest distance between any of the points of $s$ and $S$. Since $s$ and $S$ are assumed to not have any common points, we certainly have $r>0$.

Now let $B$ denote the set of all points $x$, whose distance from $s$ is less than $r/2$. $B$ certainly does not contain any point from $S$. The only segments among those that have been drawn to this point that contain any of the points in $B$ are thus $s$ and $t$. Extending $s$ to a line, we divide the Euclidean plane into two half-planes, one of which certainly does not include any of the points of $t$. We choose this half-plane and determine its intersection with $B$. We can certainly find a segment of length $1$ in this part of $B$, with one end in the end of $s$, that does not intersect either $t$ or any of the other segments.