# 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.

## Solution

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 $(a_1+a_2, b_1+b_2)$ or $(|a_1-a_2|, |b_1-b_2|)$

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: $$a_4 = a_1+2a_2$$ $$b_4 = b_1+2b_2$$ $$a_1b_1+2a_2b_1-a_1b_1-2a_1b_2 = \pm1$$ $$2(a_2b_1-a_1b_2) = \pm1$$

This is clearly impossible because RHS must be even. OK, so we done. Not yet. We not solve answer yet. The answer is clearly that: $$0+1+2+\ldots+2$$ $a_1$ 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

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