Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (Protected "2024 AMC 10B Problems/Problem 19" ([Edit=Allow only administrators] (expires 04:59, 14 November 2024 (UTC)) [Move=Allow only administrators] (expires 04:59, 14 November 2024 (UTC))))
Line 1: Line 1:
 +
==Solution 1==
 +
Let's look at zero slope first. All lines of such form will be expressed in the form <math>y=k</math>, where <math>k</math> is some real number. If <math>k</math> is an integer, the line passes through infinitely many lattice points. One such example is <math>y=1</math>. If <math>k</math> is not an integer, the line passes through <math>0</math> lattice points. One such example is <math>y=1.1</math>. So we have <math>2</math> cases.
  
 +
 +
Let's now look at the second case. The line has slope <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime integers. The line has the form <math>y = \frac{p}{q}x + m</math>. If the line passes through lattice point <math>(a,b)</math>, then the line must also pass through the lattice point <math>(a+kq, b+kp)</math>, where <math>a,b,k</math> are all integers. Therefore, the line can pass through infinitely many lattice points but it cannot pass through exactly <math>1</math> or <math>2</math>. The line can pass through <math>0</math> lattice points, such as <math>y=x+0.5</math>. This contributes <math>2</math> more cases.
 +
 +
 +
If the line has an irrational slope, it can never pass through more than <math>2</math> lattice points. We prove this using contradiction. Let's say the line passes through lattice points <math>(a,b)</math> and <math>(c,d)</math>. Then the line has slope <math>\frac{d-b}{c-a}</math>, which is rational. However, the slope of the line is irrational. Therefore, the line can pass through at most <math>1</math> lattice point. One example of this is <math>y=\sqrt{2}x</math>. This line contributes <math>2</math> final cases.
 +
 +
Our answer is therefore <math>\boxed{6}</math>.
 +
 +
 +
~lprado

Revision as of 00:44, 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10B_Problems/Problem_19&oldid=233517"