Austin Math Circle

(As usual, put solutions/suggestions in comments. A solution will be posted on June 12.) Let k be an arbitrary circle. We then have four smaller circles k 1 , k 2 , k 3 , and k 4 , with all four circles having their centers O 1 , O 2 , O 3 , and O 4 (respectively) on k . Suppose: k 1 and k 2 intersect at two points A 1 and B 1 , with A 1 on circle k .  k 2 and k 3 intersect at A 2 and B 2 , with A 2 on k .  k 3 and k 4 intersect at A 3 and B 3 , with A 3 on k .  k 4 and k 1 intersect at A 4 and B 4 , with A 4 on k.   Finally, suppose that O 1 , A 1 , O 2 , A 2 , O 3 , A 3 , O 4 , and A 4 lie in that order on k , and are all distinct. Prove that B 1 B 2 B 3 B 4 is a rectangle. (Source: Middle European Mathematical Olympiad 2007, Q4. But don't look! Solve it on your own.)

First, a reminder of the problem: Problem : The arithmetic, geometric, and harmonic mean of two positive integers are themselves three distinct positive integers. What is the smallest possible value of the arithmetic mean. The solution Maeve gave in the comments works, but it relies on a lot of searching through cases. It would be nice to have a more general approach, especially since we might some day encounter a similar problem for which the solution isn’t so conveniently small (which would make testing cases much more impractical.). So let’s consider: Solution : Let’s start by reminding ourselves of the definitions of arithmetic, geometric, and harmonic means. Let our two positive integers be a and b . Then: Arithmetic Mean of a and b : (a+b)/2 Geometric Mean of a and b : √   ab   Harmonic Mean of a and b : 2ab/(a+b) We need each of these to be an integer, and then to minimize the arithmetic mean. Let’s start by considering a specia...