Difference between revisions of "2024 AMC 10B Problems/Problem 19"

Line 14: Line 14:
  
 
==See also==
 
==See also==
{{AMC10 box|year=2024|ab=B|num-b=14|num-a=16}}
+
{{AMC10 box|year=2024|ab=B|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:31, 14 November 2024

Solution 1

Let's look at zero slope first. All lines of such form will be expressed in the form $y=k$, where $k$ is some real number. If $k$ is an integer, the line passes through infinitely many lattice points. One such example is $y=1$. If $k$ is not an integer, the line passes through $0$ lattice points. One such example is $y=1.1$. So we have $2$ cases.


Let's now look at the second case. The line has slope $\frac{p}{q}$, where $p$ and $q$ are relatively prime integers. The line has the form $y = \frac{p}{q}x + m$. If the line passes through lattice point $(a,b)$, then the line must also pass through the lattice point $(a+kq, b+kp)$, where $a,b,k$ are all integers. Therefore, the line can pass through infinitely many lattice points but it cannot pass through exactly $1$ or $2$. The line can pass through $0$ lattice points, such as $y=x+0.5$. This contributes $2$ more cases.


If the line has an irrational slope, it can never pass through more than $2$ lattice points. We prove this using contradiction. Let's say the line passes through lattice points $(a,b)$ and $(c,d)$. Then the line has slope $\frac{d-b}{c-a}$, which is rational. However, the slope of the line is irrational. Therefore, the line can pass through at most $1$ lattice point. One example of this is $y=\sqrt{2}x$. This line contributes $2$ final cases.

Our answer is therefore $\boxed{6}$.


~lprado

See also

2024 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png