February, 2017

Solution:

Note that the product of the integers on the board is a constant. Indeed, we have that
$$\frac{x+y}{2} \cdot 2\left(\frac{1}{x}+\frac{1}{y}\right)^{-1}=x y$$
Therefore, we expect that the answer to the problem is approximately $\sqrt{1 \cdot 10^{8}}=10^{4}$.
To be more rigorous, we have to show that the process indeed converges quickly enough. To show this, we bound the difference between the integers on the board at time $i$. Say that at time $i$, the integers on the board are $a_{i}<b_{i}$. Note that
$$\begin{aligned} d_{i+1}=b_{i+1}-a_{i+1}= & \frac{a_{i}+b_{i}}{2}-2\left(\frac{1}{a_{i}}+\frac{1}{b_{i}}\right)^{-1}=\frac{\left(a_{i}-b_{i}\right)^{2}}{2\left(a_{i}+b_{i}\right)} \\ & <\frac{b_{i}-a_{i}}{2}=\frac{d_{i}}{2} \end{aligned}$$
The inequality at the end follows from that obvious fact that $b_{i}-a_{i}<b_{i}+a_{i}$. Therefore, $d_{i+1} \leq \frac{d_{i}}{2}$, so $d_{2017}<\frac{10^{8}}{2^{2017}}$, which is extremely small. So the difference is essentially $0$ at time $2017$, which completes the argument.