Problem 1592

Given two rational numbers $\tfrac{m}{n}$ and $\tfrac{n}{m}$ on a blackboard, where $m$ and $n$ are relatively prime positive integers, we want to determine all pairs $(m,n)$ such that it is possible for Evan to write 1 on the board after finitely many steps using the following operations:
- Write the arithmetic mean $\tfrac{x+y}{2}$ of any two numbers $x$ and $y$ on the board.
- Write the harmonic mean $\tfrac{2xy}{x+y}$ of any two numbers $x$ and $y$ on the board.

### Analysis

To solve this problem, we utilize the ideas from number theory and properties of rational numbers. We essentially want Evan to be able to reach the number 1, which is equivalent to having:

$$\frac{m}{n} = 1 \quad \text{or} \quad \frac{n}{m} = 1$$

Using the arithmetic and harmonic means, a key observation is that both operations preserve the sum and product of the two numbers. Thus, to obtain 1 on the board, it is essential that we could eventually reach $(m,n)=(1,1)$ after several operations.

### Approach

1. Initial Setup: Begin with two rational numbers:

$$\frac{m}{n} \quad \text{and} \quad \frac{n}{m}$$

2. Arithmetic and Harmonic Means: For any rational numbers $\frac{a}{b}$ and $\frac{b}{a}$, the arithmetic mean is:

$$\frac{\frac{a}{b} + \frac{b}{a}}{2} = \frac{a^2 + b^2}{2ab}$$

The harmonic mean is:

$$\frac{2 \cdot \frac{a}{b} \cdot \frac{b}{a}}{\frac{a}{b} + \frac{b}{a}} = \frac{2}{\frac{a^2 + b^2}{ab}} = \frac{2ab}{a^2 + b^2}$$

3. **Finding $m$ and $n$:
- Assume that the board should evolve toward 1 through these operations.
- Without loss of generality, we consider the transformation of a more complex expression to a simplified rational number structure.

4. Conclusion for Pair $(m,n)$:** Through the sequence of applying these arithmetic and harmonic means, we aim to show that:

- The ability to express 1 would imply transformations involving powers of 2.
- Specifically, the pairs $(a, 2^k - a)$ such that $a$ is odd and $k$ is positive are favorable.
- This general form arises because via strategic applications of the means, Evan can produce numbers that transform the board entries to a scenario where the operations systematically reduce to or participate in increment chains characterized by odd numbers balancing with powers of two.

Thus, the complete set of solutions where Evan can write 1 on the board in finitely many steps is given by:

$$\boxed{(a, 2^k - a) \text{ for odd } a \text{ and positive } k}$$

The transformation structure of powers of two ensures the feasibility of achieving the unit value through successive means operation.