# What is the maximum number of rational points that can lie on a circle in $\mathbb{R}^2$ whose center is not a rational point?

This is a Putnam I found a few weeks ago, I believe I solved it, but something just seems off with my approach, can someone check wether my solution is right or not?

So, since I'm searching for the maximum I can stablish ideal conditions, for example, a circumference of a radii such that $$(0,1)$$ lies on the circumference, then, if I choose any secant with rational slope that goes through $$(0,1)$$, it has to cut through a rational point, since we would have a sistem of equations of degree $$2$$, and we already know one of the solutions, mainly, $$(0,1)$$, which is a rational point, and we know that if a cuadratic equation has a rational solution, the other solution must be rational as well. Then, we just need to take every secant with rational slope that goes through $$(0,1)$$ and we would cut in a rational point that lies on the circumference, hence, there are infinitely many rational points lying on that circumference.

What just does't seem right to me, is that this isn't something new, I saw a very similar idea in my number theory class when we found all pythagorean triplets. But this is a Putnam, I wouldn't expect to find such an answer. Could someone please tell me if my procedure is correct or not?

• Why must a secant cut the circle at a second rational point? – Lord Shark the Unknown Oct 26 '18 at 15:04

## 1 Answer

Given two points with rational coordinates, the equation of the perpendicular bisector to the segment they bound has rational coefficients. So given a triangle with rational point vertices, the circumcentre is is also a rational point, since it's the intersection of the perpendicular bisectors of the sides (which have equations with rational coefficients).

Therefore a circle with an irrational point as its centre can have at most two rational points.

• How do you prove that you're able to have 2 points? – Bruno Andrades Oct 26 '18 at 18:02
• Take any point on their perpendicular bisector as the centre of the circle. @BrunoAndrades – Lord Shark the Unknown Oct 26 '18 at 18:33