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== Problem ==
For certain ordered pairs <math>(a,b)\,</math> of [[real number]]s, the system of equations
<center><math>ax+by=1\,</math></center>
<center><math>x^2+y^2=50\,</math></center>
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 ==
<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>). 

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{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.

== See also ==
{{AIME box|year=1994|num-b=6|num-a=8}}

[[Category:Intermediate Combinatorics Problems]]