Difference between revisions of "1994 AIME Problems/Problem 7"

(Solution)
(Solution)
Line 6: Line 6:
  
 
== 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>.
+
bro whoever came up with this problem must've been drunk tbh, there are literally infinite solutions...
 
 
<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>).
 
 
 
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.
 
 
 
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.
 
 
 
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.  
 
 
 
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.  
 
 
 
Thus, there are a total of <math>66-6+12=\boxed{72}</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.
 
  
 
== See also ==
 
== See also ==

Revision as of 21:46, 15 September 2023

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

bro whoever came up with this problem must've been drunk tbh, there are literally infinite solutions...

See also

1994 AIME (ProblemsAnswer KeyResources)
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