Second Round

Let us analyze the process step by step.

Let $C_n$ be the circumference after $n$ minutes.

At the start ($n=0$): The figure is a square with side $1$, so the circumference is $4\times 1=4$.

Minute 1: On each vertical side, a square of side $\frac{1}{3}$ is added in the middle. Each such square adds $2$ new vertical segments of length $\frac{1}{3}$ (the left and right sides of the small square), and replaces a segment of length $\frac{1}{3}$ on the original side with the top and bottom of the small square (which are also length $\frac{1}{3}$ each). The total length added is $4\times 4\times \frac{1}{3}$ (since $4$ squares are added, each with $4$ sides of length $\frac{1}{3}$), but we must subtract the length that is replaced (the $\frac{1}{3}$ segments on the original sides, $2$ per square, so $4\times 2\times \frac{1}{3}$).

But let's look for a pattern.

Let us focus on the vertical segments: - At step $0$, there are $2$ vertical segments (the left and right sides of the square), each of length $1$. - At step $1$, on each vertical segment, a square is added in the middle, so each vertical segment is split into $2$ segments of length $\frac{1}{3}$ (the parts not covered by the new square), and $2$ vertical sides of the new square (each $\frac{1}{3}$).

But the key is that at each step, every vertical segment is replaced by $2$ vertical segments, each $\frac{1}{3}$ the length of the previous segment.

Let us formalize this: Let $V_n$ be the number of vertical segments after $n$ steps, and $L_n$ be the length of each vertical segment.

At $n=0$: $V_0=2$, $L_0=1$ At $n=1$: $V_1=4$, $L_1=\frac{1}{3}$ At $n=2$: $V_2=8$, $L_2=\frac{1}{9}$ In general: $V_n=2^{n+1}$, $L_n=\frac{1}{3^n}$

The total length of vertical segments at step $n$ is $V_n\times L_n=2^{n+1}\times \frac{1}{3^n}$

Now, for the horizontal segments: At each step, the number of horizontal segments increases as well, but the process is similar.

But notice that the total circumference at each step is: - At $n=0$: $4$ - At $n=1$: $4+4\times (\frac{4}{3}-\frac{2}{3})=4+4\times \frac{2}{3}=4+\frac{8}{3}$ But let's use the recursive pattern.

Alternatively, notice that at each step, the total added length is $2^{n+1}\times \frac{2}{3^n}$ (since each new square adds $4$ sides of length $\frac{1}{3^n}$, but $2$ of these replace existing segments).

But the process is self-similar, and the total circumference at step $n$ is:

$C_n=4\left(1+\frac{2}{3}+\frac{2}{3^2}+\cdots +\frac{2}{3^n}\right)$

This is a geometric series:

$C_n=4\left(1+2{\sum }_{k=1}^n\frac{1}{3^k}\right)$

The sum ${\sum }_{k=1}^n\frac{1}{3^k}=\frac{1/3(1-(1/3{)}^n)}{1-1/3}=\frac{1}{2}(1-(1/3{)}^n)$

So: $C_n=4\left(1+2\times \frac{1}{2}(1-(1/3{)}^n)\right)=4(1+1-(1/3{)}^n)=4(2-(1/3{)}^n)=8-4(1/3{)}^n$

After $60$ minutes, $n=60$:

$C_{60}=8-4\left(\frac{1}{3^{60}}\right)$

Since $\frac{1}{3^{60}}$ is extremely small, the circumference is very close to $8$.

Answer:

The circumference of the figure after one hour is $8-4\left(\frac{1}{3^{60}}\right)$.