2020 USAMO Problems/Problem 4

Problem 4

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


Let's start off with just $(a_1, b_1), (a_2, b_2)$ and suppose that it satisfies the given condition. We could use $(1, 1), (1, 2)$ for example. We should maximize the number of conditions that the third pair satisfies. We find out that the third pair should equal $(a_1+a_2, b_1+b_2)$:

We know this must be true: \[|a_1b_2-a_2b_1| = 1\]

So \[a_1b_2-a_2b_1 = 1\]

We require the maximum conditions for $(a_3, b_3)$ \[|a_3b_2-a_2b_3| = 1\] \[|a_3b_1-a_1b_3| = 1\]

Then one case can be: \[a_3b_2-a_2b_3 = 1\] \[a_3b_1-a_1b_3 = -1\]

We try to do some stuff such as solving for $a_3$ with manipulations: \[a_3b_2a_1-a_2b_3a_1 = a_1\] \[a_3b_1a_2-a_1b_3a_2 = -a_2\] \[a_3(a_1b_2-a_2b_1) = a_1+a_2\] \[a_3 = a_1+a_2\] \[a_3b_2b_1-a_2b_3b_1 = b_1\] \[a_3b_1b_2-a_1b_3b_2 = -b_2\] \[b_3(a_1b_2-a_2b_1) = b_1+b_2\] \[b_3 = b_1+b_2\]

We showed that 3 pairs are a complete graph; however, 4 pairs are not a complete graph. We will now show that: \[a_4 = a_1+2a_2\] \[b_4 = b_1+2b_2\] \[|a_1b_1+2a_2b_1-a_1b_1-2a_1b_2| = 1\] \[2|a_2b_1-a_1b_2| = 1\]

This is clearly impossible because $1$ is not even and also $|a_2b_1-a_1b_2| = 1$. The answer is as follows: \[0+1+2+\ldots+2\] $a_1$ has $0$ subtractions that follow condition while $a_2$ has $1$ and then the rest has $2$. There are $n$ terms, so our answer be $2n-3$ and in case of $n=100$ that means \[\boxed{N=197}.\]~Lopkiloinm

Solution 2

We claim the answer is $197$.

Study the points $(0, 0), (a_i, b_i), (a_j, b_j)$. If we let these be the vertices of a triangle, applying shoelace theorem gives us an area of $\frac{1}{2}|0\times{b_i}+{a_i}\times{b_j}+{b_i}\times{0}-0\times{a_i}-{b_i}\times{a_j}-{b_j}\times{0} = \frac{1}{2}|a_ib_j - a_j b_i| = \frac{1}{2}$. Therefore, the triangle formed by the points $(0, 0), (a_i, b_i), (a_j, b_j)$ must have an area of $\frac{1}{2}$.

Two cases follow. Case 1: Both $(a_i, b_i), (a_j, b_j)$ have exactly one coordinate equal to $0$. Here, one point must be on the $x$ axis and the other on the $y$ axis in order for the triangle to have a positive area. For the area of the triangle to be $\frac{1}{2}$, it follows that the points must be $(1, 0), (0, 1)$ in some order.

Case 2: At least one of $(a_i, b_i), (a_j, b_j)$ does not have exactly one coordinate equal to $0$. Define $S[l]$ to be a list of lines such that each line in the list has some two lattice points that, with $(0, 0)$, form a triangle with area $\frac{1}{2}$. Note that for any such line that passes through such two lattice points, we may trivially generate infinite lattice points on the line that have nonnegative coordinates.

Note that lines $y=1$ and $x=1$ are included in $S[l]$, because the points $(1, 1), (2, 1)$ serve as examples for $y=1$ and $(1, 1), (1, 2)$ serve as examples for $x=1$. For the optimal construction, include the points $(1, 0)$ and all the points $(0, 1), (0, 2), (0, 3), ... , (0, 99)$, in that order. In this case, every adjacent pair of points would count ($98$), as well as picking $(0, 1)$ and a nonadjacent point ($99$), so this would be $98+99=197$.

To prove that this is the maximum, consider the case where some $n$ number of points were neither on $x=1$ nor on $y=1$. In this case, we would be removing $n$ adjacent pairs and $n$ options to choose from after choosing $(0, 0)$, resulting in a net loss of $2n$. By having $n$ points on some other combination of lines in $S[l]$, we would trivially have a maximum gain of $n-1$ pairs of points on the lines such that there are no lattice points between those pairs. Because these points are not on $x=1$ or $y=1$, the altitude from a given point to the line formed by $(0, 0)$ and $(0, 1)$ and $(1, 0)$ is not $1$, and so the area of the triangle cannot be $\frac{1}{2}$. Thus, by not having all points on lines $x=1$ and $y=1$, we cannot exceed the maximum of $197$. Thus, $\boxed{197}$ is our answer.


See also

2020 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6
All USAMO Problems and Solutions

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