ignore this

All my limp rooks in one basket. The answer is $2022^2 + 1$. More generally, we claim that the answer is $4k^2 + 1$ for a $2k+1 \times 2k+1$ board. For minimality take $k^2$ cycles of size $4$ and force $4$ beetles to go along those cycles. Now take a board and beetle config with no intersections ever. Put it as a lattice in $\mathbb{Z}^2$ such that it has corners $(1, 1)$ and $(2k+1,2k+1)$. Split the elements of the lattice into sets $F_{i,j}$ which consist of the beetles $(x, y) \equiv (i, j) \pmod{2}$. Then $|F_{1,1}| \le (k + 1)^2, |F_{1,0}|, |F_{0,1}| \le k(k+1), |F_{0,0}| \le k^2$. Note that every two moves, the beetles in $F_{i,j}$ and $F_{i+1,j+1}$ swap due to the turning right condition. Then since $F_{0,0}$ and $F_{1,1}$ swap, we get $|F_{1,1}| \le k^2$. Then since every move, the unions $F_{0,0} \sqcup F_{1,1}$ and $F_{0,1} \sqcup F_{1,0}$ swap, we get that $|F_{0,1}| + |F{1,0}| \le k^2$. As such, the total number of beetles is at most $|F_{0,0}|+|F_{1,0}|+|F_{0,1}|+|F_{1,1}| \le 4k^2$.

Tintarn wrote: On a $2023 \times 2023$ board, there are beetles on some of the cells, with at most one beetle per cell. After one minute, each beetle moves a cell to the right or to the left or to the top or to the bottom. After each further minute, the beetles continue to move to adjacent fields, but they always make a $90^\circ$ turn, i.e. when a beetle just moved to the right or to the left, it now moves to the top or to the bottom, and vice versa. What is the minimal number of beetles on the board such that no matter where they start and how they move (according to the rules), at some point two beetles will end up in the cell? Can the bettles go outside of the board? Or they need to move so that they always stay inside the board?

lian_the_noob12 wrote: Tintarn wrote: On a $2023 \times 2023$ board, there are beetles on some of the cells, with at most one beetle per cell. After one minute, each beetle moves a cell to the right or to the left or to the top or to the bottom. After each further minute, the beetles continue to move to adjacent fields, but they always make a $90^\circ$ turn, i.e. when a beetle just moved to the right or to the left, it now moves to the top or to the bottom, and vice versa. What is the minimal number of beetles on the board such that no matter where they start and how they move (according to the rules), at some point two beetles will end up in the cell? Can the bettles go outside of the board? Or they need to move so that they always stay inside the board? I don't think so... They should stay in the board

CHKMO 2023/3 Only yesterday I got told some countries use some problems from Hong Kong selection tests...

For an $m\times n$ board, the answer is $4\left \lfloor \frac{m}{2}\right \rfloor \left \lfloor \frac{n}{2}\right \rfloor +1$. Colour the boxes that belong to an even row and an even column (starting from the upper left box). There are $\left \lfloor \frac{m}{2}\right \rfloor \left \lfloor \frac{n}{2}\right \rfloor$ such boxes. An example with $4\left \lfloor \frac{m}{2}\right \rfloor \left \lfloor \frac{n}{2}\right \rfloor$ beetles is obtained by placing the beetles on any $\left (2\left \lfloor \frac{m}{2}\right \rfloor \right )\times \left (2\left \lfloor \frac{n}{2}\right \rfloor \right )$ rectangle, partitioning it into $2\times 2$ squares (there ar $\left \lfloor \frac{m}{2}\right \rfloor \left \lfloor \frac{n}{2}\right \rfloor$ such squares) and letting the movement on each $2\times 2$ square be$$\begin{array}{|c|c|}\hline \downarrow & \leftarrow \\\hline \rightarrow & \uparrow \\\hline \end{array}$$repeatedly. Observe that after the first $4$ movements each beetle was exactly one time on a coloured box, so according to the Pigeonhole Principle we get that for $4\left \lfloor \frac{m}{2}\right \rfloor \left \lfloor \frac{n}{2}\right \rfloor +1$ beetles there was a moment at which at least $\left \lfloor \frac{m}{2}\right \rfloor \left \lfloor \frac{n}{2}\right \rfloor +1$ were on coloured boxes, and using the Pigeonhole Principle again this means that there were two beetles on the same box. LLL2019 wrote: CHKMO 2023/3 Only yesterday I got told some countries use some problems from Hong Kong selection tests... The same problem with smaller numbers was also given on an intermediate stage of the Argentina Mathematical Olympiad in $2018$.