2020 USOMO 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 $a_1b_2-a_2b_1 = \pm1$

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: $a_3b_2a_1-a_2b_3a_1 = a_1$ $a_3b_1a_2-a_1b_3a_2 = -a_2$ $a_3 = a_1+a_2$ $b_3 = b_1+b_2$

We showed that 3 pairs are a complete graph, but 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

-in collaboration with WONGBOB

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2020 USOMO Problems/Problem 4

Problem 4

Solution