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 "1994 AIME Problems/Problem 7"

m (minor)
m
 
(18 intermediate revisions by 6 users not shown)
Line 5: Line 5:
 
has at least one solution, and each solution is an ordered pair <math>(x,y)\,</math> of integers.  How many such ordered pairs <math>(a,b)\,</math> are there?
 
has at least one solution, and each solution is an ordered pair <math>(x,y)\,</math> of integers.  How many such ordered pairs <math>(a,b)\,</math> are there?
  
−
== Solution ==
+
==Solution==
−
<math>x^2+y^2=50</math> is the equation of a circle of radius <math>\sqrt{50}</math>, centered at the origin. The [[lattice points]] on this circle are <math>(\pm1,\pm7)</math>, <math>(\pm5,\pm5)</math>, and <math>(\pm7,\pm1)</math>.
 
  
−
<math>ax+by=1</math> is the equation of a line that does not pass through the origin. (Since <math>(x,y)=(0,0)</math> yields <math>a(0)+b(0)=0 \neq 1</math>).  
+
The equation <math>x^2+y^2=50</math> is that of a circle of radius <math>\sqrt{50}</math>, centered at the origin. By testing integers until the left side becomes too big, we see that the lattice points on this circle are <math>(\pm1,\pm7)</math>, <math>(\pm5,\pm5)</math>, and <math>(\pm7,\pm1)</math> where the signs are all independent of each other, for a total of <math>3\cdot 2\cdot 2=12</math> lattice points. They are indicated by the blue dots below.
  
−
So, we are looking for the number of lines which pass through either one or two of the <math>12</math> lattice points on the circle, but do not pass through the origin.
+
<asy> size(150); draw(circle((0,0),sqrt(50)));
  
−
It is clear that if a line passes through two opposite points, then it passes through the origin, and if a line passes through two non-opposite points, the it does not pass through the origin.
+
draw((1,7)--(-1,-7),red); draw((7,1)--(5,-5), green);
  
−
There are <math>\binom{12}{2}=66</math> ways to pick two distinct lattice points, and thus <math>66</math> distinct lines which pass through two lattice points on the circle. However, <math>\frac{12}{2}=6</math> of these lines pass through the origin.
+
dot((0,0));
  
−
Since there is a unique tangent line to the circle at each of these lattice points, there are <math>12</math> distinct lines which pass through exactly one lattice point on the circle.
+
dot((1,7),blue); dot((1,-7),blue); dot((-1,7),blue); dot((-1,-7),blue);
  
−
Thus, there are a total of <math>66-6+12=\boxed{072}</math> distinct lines which pass through either one or two of the <math>12</math> lattice points on the circle, but do not pass through the origin.
+
dot((5,5),blue); dot((5,-5),blue); dot((-5,5),blue); dot((-5,-5),blue);
 +
 
 +
dot((7,1),blue); dot((7,-1),blue); dot((-7,1),blue); dot((-7,-1),blue); </asy>
 +
 
 +
Since <math>(x,y)=(0,0)</math> yields <math>a\cdot 0+b\cdot 0=0 \neq 1</math>, we know that <math>ax+by=1</math> is the equation of a line that does not pass through the origin. So, we are looking for the number of lines which pass through at least one of the <math>12</math> lattice points on the circle, but do not pass through the origin or through any non-lattice point on the circle. An example is the green line above. It is straightforward to show that a line passes through the origin precisely when there exist two opposite points <math>(p,q)</math> and <math>(-p,-q)</math> through which it passes. And example is the red line above.
 +
 
 +
There are <math>\binom{12}{2}=66</math> ways to pick two distinct lattice points, and subsequently <math>66</math> distinct lines which pass through two distinct lattice points on the circle. Then we subtract the lines which pass through the origin by noting that the lattice points on the circle can be grouped into opposite pairs <math>(p,q)</math> and <math>(-p,-q)</math>, for a total of <math>\frac{12}{2}=6</math> lines. Finally, we add the <math>12</math> unique tangent lines to the circle at each of the lattice points.
 +
 
 +
Therefore, our final count of distinct lines which pass through one or two of the lattice points on the circle, but do not pass through the origin, is<cmath>66-6+12=\boxed{72}.</cmath>
  
 
== See also ==
 
== See also ==
Line 24: Line 31:
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 18:51, 8 January 2024

Problem

For certain ordered pairs $(a,b)\,$ of real numbers, the system of equations

$ax+by=1\,$
$x^2+y^2=50\,$

has at least one solution, and each solution is an ordered pair $(x,y)\,$ of integers. How many such ordered pairs $(a,b)\,$ are there?

Solution

The equation $x^2+y^2=50$ is that of a circle of radius $\sqrt{50}$, centered at the origin. By testing integers until the left side becomes too big, we see that the lattice points on this circle are $(\pm1,\pm7)$, $(\pm5,\pm5)$, and $(\pm7,\pm1)$ where the signs are all independent of each other, for a total of $3\cdot 2\cdot 2=12$ lattice points. They are indicated by the blue dots below.

[asy] size(150); draw(circle((0,0),sqrt(50))); draw((1,7)--(-1,-7),red); draw((7,1)--(5,-5), green); dot((0,0)); dot((1,7),blue); dot((1,-7),blue); dot((-1,7),blue); dot((-1,-7),blue); dot((5,5),blue); dot((5,-5),blue); dot((-5,5),blue); dot((-5,-5),blue); dot((7,1),blue); dot((7,-1),blue); dot((-7,1),blue); dot((-7,-1),blue); [/asy]

Since $(x,y)=(0,0)$ yields $a\cdot 0+b\cdot 0=0 \neq 1$, we know that $ax+by=1$ is the equation of a line that does not pass through the origin. So, we are looking for the number of lines which pass through at least one of the $12$ lattice points on the circle, but do not pass through the origin or through any non-lattice point on the circle. An example is the green line above. It is straightforward to show that a line passes through the origin precisely when there exist two opposite points $(p,q)$ and $(-p,-q)$ through which it passes. And example is the red line above.

There are $\binom{12}{2}=66$ ways to pick two distinct lattice points, and subsequently $66$ distinct lines which pass through two distinct lattice points on the circle. Then we subtract the lines which pass through the origin by noting that the lattice points on the circle can be grouped into opposite pairs $(p,q)$ and $(-p,-q)$, for a total of $\frac{12}{2}=6$ lines. Finally, we add the $12$ unique tangent lines to the circle at each of the lattice points.

Therefore, our final count of distinct lines which pass through one or two of the lattice points on the circle, but do not pass through the origin, is\[66-6+12=\boxed{72}.\]

See also

1994 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems/Problem_7&oldid=210227"