 # Putnam Puzzler

### 2010

Question:
Consider a set S and a binary operation *, i.e. for each a, b in S, a*b is also in S. Assume that (a*b)*a = b for all a, b in S. Prove that a*(b*a) = b for all a, b in S.

Solution:
The hypothesis implies that ((b*a)*b)*(b*a) = b for all a, b in S by replacing a with b*a. Hence, a*(b*a) = b for all a,b in S using (b*a)*b = a.

### 2009

Question:
What is the maximum number of rational points that can lie on a circle in the plan whose center is not a rational point? (a rational point is a point both of whose coordinates are rational numbers.)

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), and 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 (c-a)/(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 then must be:

g = (CE - BD)/(AE-BD)h = (AF - BC)/(AE-BD)

by elementary algebra or Cramer's rule. So the center of the circle is rational. This proves the desired result.

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