Combinatorics – Game Coloring Points Rioplatense Olympiad 1999

Question

This problem is from the Rioplatense MO 1999/3 L3. The question is the following:

Two players $A$ and $B$ play the following game:
$A$ chooses a point, with integer coordinates, on the plane and colors it green, then $B$ chooses $10$ points of integer coordinates, not yet colored, and colors them yellow. The game always continues with the same rules; $A$ and $B$ choose one and ten uncolored points and color them green and yellow, respectively.

a) The goal of $A$ is to achieve $111^2$ green points that are the intersections of $111$ horizontal lines and $111$ vertical lines (parallel to the coordinate axes). $B$'s goal is to stop him. Determine which of the two players has a strategy that ensures they achieve their goal.

b) The goal of $A$ is to achieve $4$ green points that are the vertices of a square with sides parallel to the coordinate axes. $B$'s goal is to stop him. Determine which of the two players has a strategy that ensures they achieve their goal.

I'm having trouble with part b).
For part a), A has the winning strategy. The winning strategy is to place at least $111 \times 11 ^{110}$ green points in a horizontal line. The strategy is basically to form row by row (horizontally), and put a sufficient number of points so that it can form another horizontal row.
$111 \times 11 ^{110}$ is the minimum because if we need to place $f_i$ points in the $ith$ row, we need to place $10f_{i} + f_{i} = 11f_{i}$ points in the $i-1th$ row. Then it's easy too see that it's the minimum.

I am aware that it is not formally explained, if necessary I can write it formally and better explained, although I think this idea is quite intuitive.

For part b) I have no idea how to continue with the problem. I spent a long time with this part and I'm not even sure who has the winning strategy.
I found this problem in a Pigeonhole principle resource, so that may be useful.

Answer 1

Hint 1:

If $A$ can win, their penultimate move must setup $11$ different squares so $B$ can't block them all. What would this look like?

Hint 2:

The first part of the question is a hint for the second part.

Solution:

First $A$ creates the configuration from a). Imagine a box around all the points that have already been colored by either player. Choose a 45-degree line outside this box. This will be the diagonal of the square. $A$ colors $11$ of the intersections of the horizontal lines from a) with this diagonal. ($B$ can't block this by coloring only $100$ additional points before $A$'s eleventh move.) Then, consider the reflections of the $111^2$ intersections from a) across the diagonal. There are $111$ horizontal lines passing through these points. Because $B$ has only colored $110$ additional points, there must be one with no yellow point. $A$'s next move is to color the intersection of this line with the diagonal. Then, there are $11$ ways $A$ can make a square on the next move playing on the same line and $B$ can't block them all: these are the intersections of that line with the vertical lines passing through each of $A$'s first $11$ points on the diagonal. (Those vertical lines are the reflections of the horizontal lines that was used to construct those points.)