# 2020 USOJMO Problems/Problem 5

## Problem

Suppose that $(a_1,b_1),$ $(a_2,b_2),$ $\dots,$ $(a_{100},b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i,j)$ satisfying $1\leq i and $|a_ib_j-a_jb_i|=1$. Determine the largest possible value of $N$ over all possible choices of the $100$ ordered pairs.

## Solution

Call the pair $(i, j)$ good if $1\leq i < j \leq 100$ and $|a_ib_j-a_jb_i|=1$. Note that we can reorder the pairs $(a_1, b_1), (a_2, b_2), \ldots, (a_{100}, b_{100})$ without changing the number of good pairs. Thus, we can reorder them so that $a_1\leq a_2\leq\ldots\leq a_{100}$. Furthermore, reorder them so that if $a_i=a_j$ for some $i, then $b_i.

Now I claim the maximum value of $N$ is $\boxed{197}$. First, we will show $N\leq 197$.

The key claim is that for any fixed integer $m$, $1\leq m \leq 100$, then there are at most 2 values of $i$ with $1\leq i with the pair $(i, m)$ good. We will prove this with contradiction.

Assume there are 3 integers $i$, $j$, and $k$ that are less than $m$, such that $(i, m)$, $(j, m)$, and $(k, m)$ are all good. Note that $gcd(a_m, b_m)$ divides $a_ib_m-a_mb_i=\pm 1$, so $gcd(a_m, b_m)=1$. From $a_ib_m-a_mb_i=\pm 1$, we have $$a_ib_m\equiv \pm 1 \mod a_m$$ Similarly, $$a_jb_m\equiv \pm 1 \mod a_m$$ $$a_kb_m\equiv \pm 1 \mod a_m$$

From $gcd(a_m, b_m)=1$ we have $$a_i\equiv \pm a_j \equiv \pm a_k \mod a_m$$

But since $i$, $j$, and $k$ are less than $m$, from our reordering we have $0\leq a_i, $0\leq a_j, and $0\leq a_k. Thus our congruence gives us that $a_j=a_i$ or $a_j=a_m-a_i$, and similar for $a_k$. Thus at least two of $a_i, a_j, a_k$ are equal. WLOG let them be $a_i$ and $a_j$. Now we have 4 cases.

Case 1: $a_m>2$

We have $a_ib_m-a_mb_i=\pm 1$ and $a_jb_m-a_mb_j=\pm 1$, so since $a_i=a_j$ we can subtract the two equations to get $a_mb_i-a_mb_j$ equals $\pm 2, 0$. But $a_m>2$, so $|a_m(b_i-b_j)|\leq 2$ implies that $b_i=b_j$. But all pairs are distinct and $a_i=a_j$, so this is a contradiction.

Case 2: $a_m=2$

We have 2 subcases.

Subcase 2a: $a_i\equiv a_j\equiv a_k\equiv 1 \mod a_m$

Clearly we must have $a_i=a_j=a_k=1$ in this case, as all 3 variables are less than $a_m=2$. Then $1\cdot b_m-a_mb_i=\pm 1$, $1\cdot b_m-a_mb_j=\pm 1$, and $1\cdot b_m-a_mb_k=\pm 1$. So at least 2 of $b_i$, $b_j$, and $b_k$ must be equal, but this contradicts the fact that all pairs $(a_x, b_x)$ are distinct.

Subcase 2b: $a_i\equiv a_j\equiv a_k\equiv 0 \mod a_m$

$a_i$ must be even in this case, but since $a_m$ is also even then $a_ib_m-a_mb_i$ must be even, which contradicts $|a_ib_m-a_mb_i|=1$.

Case 3: $a_m=1$

We must have $a_i=0,1$, and similarly for $a_j$ and $a_k$. Thus at least 2 of the 3 must be equal. Now we have 2 subcases again.

Subcase 3a: 2 of the 3 are equal to 0

WLOG let $a_i=a_j=0$. Then $a_ib_m-a_mb_i=\pm 1$ and $a_m=1$ so $b_i=\pm 1$. Since $b_i$ is nonnegative, $b_i=1$. Similarly $b_j=1$, but all pairs are distinct so we have a contradiction.

Subcase 3b: 2 of the 3 are equal to 1

WLOG let $a_i=a_j=0$. Then $a_ib_m-a_mb_i=\pm 1$, so $b_m-b_i=\pm 1$. But $i and $a_i=a_m=1$, so by our ordering of the pairs we have $b_i so $b_m-b_i=1$. But similarly $b_m-b_j=1$, so $b_i=b_j$, which is a contradiction to all pairs being distinct.

Case 4: $a_m=0$

We must have $a_i=0$, since $0\leq a_i\leq a_m=0$. Thus $a_ib_m-a_mb_i=0$, but $|a_ib_m-a_mb_i|=1$, contradiction.$\blacksquare$

Thus our claim is proved. Now for $m=1$, there is clearly no $i$ with $1\leq i < m \leq 100$ and $(i, m)$ good. For $m=2$, there is at most one value of i with $1\leq i < m \leq 100$ and $(i, m)$ good (namely, this is when $i=1$). By our claim, for all $2 there are at most 2 values of $i$ with $1\leq i < m \leq 100$ and $(i, m)$ good, so adding these up gives at most $2\cdot 98+1=197$ good pairs. Now we will show that 197 good pairs is achievable.

The sequence $(1,1), (1,2), (2,3),\ldots ,(99,100)$ has 197 good pairs - namely, $(1, n)$ is good for all $1, and $(i, i+1)$ is good for all $1. This adds up for a total of 197 good pairs, so we are done. $\blacksquare$

~MortemEtInteritum

## Solution 2

We claim the answer is $2n-3$ (so $197$), where $100$ is to be replaced with $n$.

The problem essentially counts unordered pairs $(P,Q)$ of lattice points so that $POQ$ has area $\frac{1}{2}$ due to Shoelace ($O$ is the origin).

For a construction, take the points $(0,1),(1,1),(1,2), \dots (1,n-1)$.

We now show by induction that $2n-3$ is the maximum, where $n=2$ obviously holds as a base case.

To reduce to the inductive hypothesis of $n-1$ points from $n$ points, we are allowed to consider the point with the greatest sum of coordinates $R$.

The key claim is that $R$ is part of at most $2$ pairs (triangles) with area $\frac{1}{2}$.

Before we proceed, note that $R$ must have relatively prime coordinates $(m,n)$, or else any triangles found can be reduced down by a greatest common factor along $OR$, impossible since $\frac{1}{2}$ is the minimum positive area.

Call the hypothetical point(s) $X$. With triangle $XOR$ with area $\frac{1}{2}$, we know the length of $OR$, so indeed the locus of $X$ is two lines evenly spaced next to $OR$ each parallel to $OR$ (before we consider the lattice condition).

When we add in the lattice condition, we see that each of these two lines contain at most one valid $X$. This is because if two lattice points with non-negative integers lied on one of these lines, one would have to be a multiple of $m$ more than the other on the $x$-coordinates and a multiple of $n$ more on the $y$-coordinates. This contradicts the coordinate sum maximality of $R$.

Thus, by deleting $R$, we lose at most two pairs $(X,R)$, completing the induction and the problem.