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Difference between revisions of "2020 USAMO Problems/Problem 4"

(Created page with "==Problem 4== Suppose that <math>(a_1, b_1), (a_2, b_2), \ldots , (a_{100}, b_{100})</math> are distinct ordered pairs of nonnegative integers. Let <math>N</math> denote the n...")
 
(→‎Solution)
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So
 
So
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<cmath>a_1b_2-a_2b_1 = \pm1</cmath>
+
<cmath>a_1b_2-a_2b_1 = 1</cmath>
  
 
We require the maximum conditions for <math>(a_3, b_3)</math>
 
We require the maximum conditions for <math>(a_3, b_3)</math>
Line 22: Line 22:
 
<cmath>a_3b_2a_1-a_2b_3a_1 = a_1</cmath>
 
<cmath>a_3b_2a_1-a_2b_3a_1 = a_1</cmath>
 
<cmath>a_3b_1a_2-a_1b_3a_2 = -a_2</cmath>
 
<cmath>a_3b_1a_2-a_1b_3a_2 = -a_2</cmath>
 +
<cmath>a_3(a_1b_2-a_2b_1) = a_1+a_2</cmath>
 
<cmath>a_3 = a_1+a_2</cmath>
 
<cmath>a_3 = a_1+a_2</cmath>
−
<cmath>b_3 = b_1+b_2</cmath>
+
<cmath>a_3b_2b_1-a_2b_3b_1 = b_1</cmath>
 +
<cmath>a_3b_1b_2-a_1b_3b_2 = -b_2</cmath>
 +
<cmath>b_3(a_1b_2-a_2b_1) = b_1+b_2</cmath>
  
−
We showed that 3 pairs are a complete graph, but 4 pairs are not a complete graph. We will now show that
+
We showed that 3 pairs are a complete graph; however, 4 pairs are not a complete graph. We will now show that:
 
<cmath>a_4 = a_1+2a_2</cmath>
 
<cmath>a_4 = a_1+2a_2</cmath>
 
<cmath>b_4 = b_1+2b_2</cmath>
 
<cmath>b_4 = b_1+2b_2</cmath>

Revision as of 18:01, 9 December 2020

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: \[a_3b_2a_1-a_2b_3a_1 = a_1\] \[a_3b_1a_2-a_1b_3a_2 = -a_2\] \[a_3(a_1b_2-a_2b_1) = a_1+a_2\] \[a_3 = a_1+a_2\] \[a_3b_2b_1-a_2b_3b_1 = b_1\] \[a_3b_1b_2-a_1b_3b_2 = -b_2\] \[b_3(a_1b_2-a_2b_1) = b_1+b_2\]

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

See also

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

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