1994 AIME Problems/Problem 7

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

$x^2+y^2=50$ is the equation of a circle of radius $\sqrt{50}$ , centered at the origin. The lattice points on this circle are $(\pm1,\pm7)$ , $(\pm5,\pm5)$ , and $(\pm7,\pm1)$ .

$ax+by=1$ is the equation of a line that does not pass through the origin. (Since $(x,y)=(0,0)$ yields $a(0)+b(0)=0 \neq 1$ ).

So, we are looking for the number of lines which pass through either one or two of the $12$ 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 $\binom{12}{2}=66$ ways to pick two distinct lattice points, and thus $66$ distinct lines which pass through two lattice points on the circle. However, $\frac{12}{2}=6$ 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 $12$ distinct lines which pass through exactly one lattice point on the circle.

Thus, there are a total of $66-6+12=\boxed{072}$ distinct lines which pass through either one or two of the $12$ lattice points on the circle, but do not pass through the origin.

1994 AIME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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1994 AIME Problems/Problem 7

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