I found this particular problem on AOPS : https://artofproblemsolving.com/community/c6h2129220p15547913
A real number $c > 0$ is given. Ana and Bob play the following game on the $x$, $y$-coordinate plane. First, Bob will choose a function $f \colon \mathbb{R} \to [0, 2020]$, then the set of nice points will be defined as $\{(x, y) \in \mathbb{R}^2 \mid f(x + f(y)) = y\}$. After this, Ana will choose to stand on one nice point and then Bob will choose to stand on another one. After this, they will take turns moving with Ana making the first move. On each of their turn, Bob can move to any nice point not more than $2020$ units from his current position and Ana can move to any nice point not more than $c$ units from her current position. Ana wins if and only if, after finitely many turns, she ends up at the same point as Bob. Find the least number $c$ such that Ana can win no matter how Bob choose the function $f$, or show that such number $c$ doesn’t exist.
I have had trouble with the proof provided. It is split into a proof of the lower and upper bounds, but only the latter is in any meaningful way expanded upon. I have had trouble with the lower bound proof, which states that $f(x)=2020\{\frac{x}{2020}$}. How exactly would this proof that $c=4040$ is the minimal solution?
The proof that Ana can reach any point from any other point is also written somewhat confusingly, I am not sure I understand how the points where chosen and what is the significance of $1234$.
Your help would we much appreciated!
A much more detailed solution can be found in this AoPS post. To prove the lower bound, the solution there uses the same function and explains the reasoning, so I think that the post will be very helpful.
As for the significance of $1234$, I think this can be explained with a more detailed reasoning of the claim. (The linked AoPS post uses a similar way to prove the upper bound. Therefore, the following explanation will also somewhat follow it.)
Claim 2. Let $(x_1, y_1)$ and $(x_2, y_2)$ be any two nice points. Then Ana can go from $(x_1, y_1)$ to $(x_2, y_2)$ via a series of jumps.
To prove this claim, we first prove some smaller claims first.
Claim 2a. For any nice point $(x, y)$, we have $0 \leq y \leq 2020$.
Proof: By the definition of nice points, $(x, y)$ is a nice point only if $y$ is in the range of $f$. $\square$
Claim 2b. Let $(x_s, y_s)$ and $(x_t, y_t)$ be any two nice points with $|x_s - x_t| \leq 1234$. Then Ana can go from $(x_s, y_s)$ to $(x_t, y_t)$ in one jump.
Proof: We have $$\sqrt{(x_s - x_t)^2 + (y_s - y_t)^2} \leq \sqrt{1234^2 + 2020^2} \leq 4040.$$ (We know $(y_s - y_t)^2 \leq 2020^2$ due to Claim 2a.) In other words, the distance between $(x_s, y_s)$ and $(x_t, y_t)$ is not more than $4040$, which implies Ana can go from $(x_s, y_s)$ to $(x_t, y_t)$ in one jump. $\square$
Let us now prove the original claim.
Proof of Claim 2: If $|x_1 - x_2| \leq 1234$, then we use Claim 2b and we are done. So, let us consider the case when $|x_1 - x_2| > 1234$.
Without loss of generality, let $x_1 < x_2$. (If Ana can go from $(x_1, y_1)$ to $(x_2, y_2)$, then Ana can clearly go from $(x_2, y_2)$ to $(x_1, y_1)$.)
Let $(x, y)$ be an arbitrary nice point. Consider the point $(x', y')$ where \begin{align*} x' &= x + 2021 - f(f(x + 2021)), \\ y' &= f(x + 2021). \end{align*} By substituting into the definition of nice points, we can easily verify that $(x', y')$ must also be a nice point. Moreover, we have $0 \leq y, y' \leq 2020$ due to Claim 2a, and $$x' - x = 2021 - f(f(x + 2021)) \in [1, 2021].$$ Therefore, the distance from $(x, y)$ to $(x', y')$ is $$\sqrt{(x - x')^2 + (y - y')^2} \leq \sqrt{2021^2 + 2020^2} < 4040.$$ So, Ana can jump from $(x, y)$ to $(x', y')$ and vice versa. Furthermore, since $1 \leq x' - x \leq 2021$, that means Ana can always increase her $x$-coordinate by some $k \in [1, 2021]$. (The value of $k$ depends on $(x, y)$.)
Observe that if Ana apply this jumping 'method', Ana will eventually reach some $(x_3, y_3)$ with $|x_2 - x_3| \leq 1234$. If not, then Ana must have reached some $(x_3^-, y_3^-)$ with $x_3^- + 1234 < x_2$, and then jump directly to some $(x_3^+, y_3^+)$ with $x_2 + 1234 < x_3^+$.
But this is impossible; the distance from $(x_3^-, y_3^-)$ to $(x_3^+, y_3^+)$ is at least $2 \cdot 1234 = 2468$, yet the jumping method guarantees the distance is at most $2021$. This is a contradiction.
So, Ana must be able to jump to some $(x_3, y_3)$ with $|x_2 - x_3| \leq 1234$. Using Claim 2b, we know Ana can do one more jump from $(x_3, y_3)$ to $(x_2, y_2)$. This completes the proof.
As the proof illustrates, $1234$ is not so special; any $M$ works, as long as $\sqrt{M^2 + 2020^2} \leq 4040$ (so that Claim 2b holds) and $2M > 2021$ (so that our jumping method works).