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1994 AIME Problems/Problem 7

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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.

See also

1994 AIME (ProblemsResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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