# 2022 USAJMO Problems/Problem 2

## Problem

Let $a$ and $b$ be positive integers. The cells of an $(a + b + 1)\times (a + b + 1)$ grid are colored amber and bronze such that there are at least $a^2+ab-b$ amber cells and at least $b^2+ab-a$ bronze cells. Prove that it is possible to choose $a$ amber cells and $b$ bronze cells such that no two of the $a+b$ chosen cells lie in the same row or column.

## Solution 1: Pigeonhole

Let $n=a+b+1$. There are $n!$ ways to select $n$ cells such that no two are in the same row or column. Each such selection can be specified by $p_i=(p_{i1},p_{i2},\cdots,p_{in})$, a permutation of $1,2,3,\cdots,n$, such that in the $k^\text{th}$ row, the cell in column $p_{ik}$ is selected. Let $A(p_i)$ be the number of amber cells in the selection $p_i$. We just need to prove there exists a $p_i$ such that $$A(p_i)=a,\text{ or } A(p_i)=a+1.$$ Using contradiction, assuming no such $p_i$ exists.

In calculating $\sum A(p_i)$, each amber cell is counted $(n-1)!$ times. We have $$\sum A(p_i) \ge (n-1)!(a^2+ab-b).$$ $$\Rightarrow\max(A(p_i))\ge\left\lceil\frac{(n-1)!(a^2+ab-b)}{n!}\right\rceil=\left\lceil\frac{n(a-1)+1}{n}\right\rceil = a.$$ WLOG, let $A(p_1)=\max(A(p_i))$. With the contradiction assumption $$A(p_1)\ge a+2.$$ Similarly, let $A(p_2)=\min(A(p_i))$. $$\sum A(p_i) \le (n-1)![n^2-(b^2+ab-a)] \Rightarrow \min(A(p_i))\le a+1 \Rightarrow A(p_2)\le a-1.$$

One can always move from $p_1$ to $p_2$ via a path in $p$ space $$p_1\rightarrow q_1 \rightarrow q_2 \rightarrow \cdots \rightarrow q_k \rightarrow p_2$$ such that in each step two numbers in the starting permutation are swapped to get the next permutation. For example if $$p_1=(5,4,3,1,2), p_2=(1,2,3,4,5)$$ one such path is $$(5,4,3,1,2)\rightarrow (1,4,3,5,2) \rightarrow (1,2,3,5,4) \rightarrow (1,2,3,4,5).$$ Then the value of $A$ along the path changes by at most 2 in each step. On the other hand since $A(p_i)\neq a, a+1$, there is a gap of at least width 3 in the range of $A$. A path starting above the gap and ending below the gap with max step size two should not exist. Contradiction.

Therefore there must be an $i$ such that either $A(p_i)=a$ or $A(p_i)=a+1$. (Y)

## Solution 2

We claim that it is possible to choose $a$ amber cells such that no two lie in the same row or column.

Number the rows and columns from $1$ to $a+b+1.$ In cell $(x,y),$ write the number $x+y\pmod{a+b+1}.$ Then two same-numbered cells will be on different rows and columns.

There are $a+b+1$ different numbers and $\geq a^2+ab-b = (a-1)(a+b+1)+1$ amber cells. Using Pigeonhole, we have $a+b+1$ holes, and cells in the same hole are in different rows and columns, and $\geq(a-1)(a+b+1)+1$ pigeons, so there exists $a$ amber cells such that no two lie in the same row or column. Similarly, there exists b bronze cells such that no two lie in the same row or column.

Choose those $a$ amber cells and $b$ bronze cells. If no two chosen cells lie in the same row or column then we are done. Otherwise, since there are $a+b+1$ rows and columns, there exists a row and a column with no chosen cell. If their intersection is an amber cell, choose that cells and unchoose an amber cell that is in the same row or column as a bronze cell. If the intersection is a bronze cell, choose that cell and unchoose a chosen bronze cell that is in the same row or column as an amber cell. Repeating this process, we can obtain $a$ amber cells and $b$ bronze cells such that no two lie in the same row or column.

## Solution 3: Graph Theoretic

We can prove this statement by using the Hall's marriage theorem. Let us define a bipartite graph $G = (A \cup B, E)$, where $A$ is the set of rows of the grid, $B$ is the set of columns of the grid, and $E$ is the set of edges between $A$ and $B$. We say that a row $a \in A$ is connected to a column $b \in B$ if and only if the cell in row $a$ and column $b$ is colored amber.

Now, we need to show that the conditions of Hall's marriage theorem are satisfied. That is, for any subset $S \subseteq A$, the number of columns connected to $S$ is at least $|S|$. To prove this, let $S$ be any subset of $A$, and let $T$ be the set of columns connected to $S$. Then, the number of amber cells in the rows of $S$ is at least $a|S|$, and the number of amber cells in the rows not in $S$ is at least $b(a+b+1)-a|S|$. Therefore, the total number of amber cells is at least $a|S| + b(a+b+1)-a|S| = ab + b(a+b)$, which is at least $a^2+ab-b$ by assumption. Hence, the number of columns connected to $S$ is at least $b$.

Similarly, we can define a bipartite graph $G' = (A \cup B', E')$, where $B'$ is the set of rows of the grid, and an edge between a row $a$ and a column $b' \in B'$ exists if and only if the cell in row $a$ and column $b'$ is colored bronze. We can show that the conditions of Hall's marriage theorem are also satisfied for $G'$.

Therefore, by Hall's marriage theorem, there exist matchings $M$ and $M'$ in $G$ and $G'$, respectively, such that each row and column in the grid is incident to at most one edge in $M$ or $M'$. Let $X$ be the set of cells in the grid that are covered by $M$, and $Y$ be the set of cells in the grid that are covered by $M'$. Then, $|X| = |Y| = a+b$, since each matching covers a total of $a+b$ cells. Moreover, no two cells in $X$ or $Y$ lie in the same row or column, since each row and column is incident to at most one edge in $M$ or $M'$.

Finally, we can choose $a$ amber cells and $b$ bronze cells from $X$ and $Y$, respectively, such that no two of the chosen cells lie in the same row or column. This is because $|X| = |Y| = a+b$, and there are at least $a^2+ab-b$ amber cells in the grid and at least $b^2+ab-a$ bronze cells in the grid. Therefore, there are at least $a^2+ab-b$ cells colored amber in $X$, and at least $b^2+ab-a$ cells colored bronze in $Y$. This implies that there are at least $a$ amber cells in $X$ and at least $b$ bronze cells in $Y$, so we can choose a subset of $a$ amber cells from $X$ and a subset of $b$ bronze cells from $Y$, such that no two of the chosen cells lie in the same row or column.