Fill the table with T,M,O

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Fill the table with T,M,O

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Thailand online MO

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Source: 2021 Thailand Online MO P10 (Mock TMO contest)

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#1 h Apr 6, 2021, 9:15 AM • 1 Y

Y by FaThEr-SqUiRrEl

Each cell of the board with $2021$ rows and $2022$ columns contains exactly one of the three letters $T$ , $M$ , and $O$ in a way that satisfies each of the following conditions:

In total, each letter appears exactly $2021\times 674$ of times on the board.
There are no two squares that share a common side and contain the same letter.
Any $2\times 2$ square contains all three letters $T$ , $M$ , and $O$ .

Prove that each letter $T$ , $M$ , and $O$ appears exactly $674$ times on every row.

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kankan

#2 h Apr 7, 2021, 9:32 PM • 1 Y

Y by FaThEr-SqUiRrEl

incomplete sol
I'm not entirely sure how to finish.

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#3 h Apr 8, 2021, 2:05 AM • 1 Y

Y by FaThEr-SqUiRrEl

Hint for @above

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#4 h Apr 8, 2021, 7:56 AM • 1 Y

Y by FaThEr-SqUiRrEl

Slightly troll. Let f(T)=O, f(O)=M, f(M)=T.

Let’s drop the condition each letter shows up on the board the same number of times for now.

The motivation for this observation is that there are limited colorings for this grid, as the two rules on coloring are very strong conditions. The main observation is that if one row is colored such that $a_i$ denotes the $i$ th letter on that row, and $b_i$ denote the $i$ th letter on the next row then one of the two are true: $f(a_i)=b_i \forall i$ or $f(b_i)=a_i$ .

Notice there are two choices for the first entry, and if three cells of a $2\cdot 2$ squares are colored, we can easily determine the color of the last. Therefore, there are exactly two colorings of the next row.

It remains to show both coloring work. Note $f$ is a bijection, so clearly no two squares sharing a side contain the same letter. Also, easy casework reveals that any 2 by 2 square contains all colors.

Let $t_i$ be the number of $t$ ’s on row $i$ and define $m_i, o_i$ similarly. By the third condition, $\sum\limits_{i=1}^{2021} (t_i+m_i\omega_3+o_i\omega_3^2)=0$ . On the other hand, we know $t_i+m_i\omega_3+o_i\omega_3^2=\omega_3^{k_i} (t_1+m_1\omega_3+o_1\omega_3^2)$ for some $0\le k_i <3$

Let $a_i$ be the number of solutions to $k_j=i$ . Then, our sum is equal to $(t_1+m_1\omega_3+o_1\omega_3^2)(a_0+a_1\omega_3+a_2\omega_3^2)=0$ note $k_0, k_1, k_2\in\mathbb{Z}, a_0+a_1+a_2=2021$ , so $a_0+a_1\omega_3+a_2\omega_3^2\ne 0,$ so $t_1=m_1=o_1,$ as desired.

This post has been edited 2 times. Last edited by CANBANKAN, Apr 15, 2021, 3:48 PM

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#5 h Apr 12, 2021, 4:20 PM

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Replace $T,M,O$ with $X,Y,Z$ . Consider any subsquare $2\times 2$ . Observe that, it has to be in two diagonal form. $X-(\star),b -X(\bullet), -XX-$ with $X$ . $-$ denotes the arbitrary letter in $\{X,Y,Z\}$ . Call $\cal A$ is the first row. Consider the second row, combine with the first row to form a well-formed array of the validate squares of $2\times 2$ .There will be $B,C$ is dependent what we consider with the first row, a diagonal $2\times 2$ square $(\star),\, \text{and},\,\,(\bullet)$ $\textbf{Wlog},$ assume that row is $B$ . Continue with the third row. There are $2$ choices first row, or row $C$ . We have all rows, which has to be of type $A,B,C$ . Denote their numbers with $p,q,r$ . In three rows $\{\mathcal{A},B,C\}$ the total number of $|X|=\text{number of }\,\,|Y|$ . I will write down rest prove bit later, sorry.

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