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2022 USAJMO Problems/Problem 2

Revision as of 21:28, 9 June 2022 by Mryang6859 (talk | contribs) (→‎Solution 1: Pigeonhole)

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: Induction

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