Putnam 2008, Problem B-1

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? (A rational point is a point both of whose coordinates are rational numbers.)

Solution:
There are at most two such points. For example, the points $(0,0)$ and $(1,0)$ lie on a circle with center $(1/2,x)$ for any real number $x$, not necessarily rational.

On the other hand, suppose $P = (a,b), Q = (c,d), R = (e,f)$ are three rational points that lie on a circle. The midpoint $M$ of the side $PQ$ is $((a+c)/2, (b+d)/2)$, which is again rational. Moreover, the slope of the line $PQ$ is $(d-b)/(c-a)$, so the slope of the line through $M$ perpendicular to $PQ$ is $(a-c)/(b-d)$, which is rational or infinite.

Similarly, if $N$ is the midpoint of $QR$, then $N$ is a rational point and the line through $N$ perpendicular to $QR$ has rational slope. The center of the circle lies on both of these lines, so its coordinates $(g,h)$ satisfy two linear equations with rational coefficients, say $Ag + Bh = C$ and $Dg + Eh = F$. Moreover, these equations have a unique solution. That solution must then be
$$g = \frac{CE - BD}{AE - BD}$$
$$h = \frac{AF - BC}{AE - BD}$$

(by elementary algebra, or Cramer's rule), so the center of the circle is rational. This proves the desired result.

**Remark:** The above solution is deliberately more verbose than is really necessary. A shorter way to say this is that any two distinct rational points determine a rational line (a line of the form $ax + by + c = 0$ with $a,b,c$ rational), while any two nonparallel rational lines intersect at a rational point. A similar statement holds with the rational numbers replaced by any field.

**Remark:** A more explicit argument is to show that the equation of the circle through the rational points $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is
$$0 = \det \begin{pmatrix} x_1^2 + y_1^2 & x_1 & y_1 & 1 \\ x_2^2 + y_2^2 & x_2 & y_2 & 1 \\ x_3^2 + y_3^2 & x_3 & y_3 & 1 \\ x^2 + y^2 & x & y & 1 \end{pmatrix}$$
which has the form $a(x^2 + y^2) + dx + ey + f = 0$ for $a,d,e,f$ rational. The center of this circle is $(-d/(2a), -e/(2a))$, which is again a rational point.