, 2020

Solution:

If we note that $x = \{x\} + \lfloor x \rfloor$, then we can rewrite our given inequality as $\{x\} > \frac{\alpha}{1-\alpha} \lfloor x \rfloor$. However, since $\{x\} < 1$, we know that we must have $\frac{\alpha}{1-\alpha} \lfloor x \rfloor < \{x\} < 1$, so each interval is of the form $\left(n + \frac{\alpha}{1-\alpha} n, n+1\right)$ for some integer $n$, which has length $\frac{1 - (n+1) \alpha}{1-\alpha}$. If we let $k$ be the smallest integer such that $\frac{1 - (k+1) \alpha}{1-\alpha} < 0$, then the total length of all our intervals is the sum
$$\sum_{n=0}^{k-1} \frac{1 - (n+1) \alpha}{1-\alpha} = \frac{k - \frac{k(k+1)}{2} \alpha}{1-\alpha}$$
If we set this to $20.2$, we can solve for $\alpha$ to get
$$\alpha = \frac{k - 20.2}{\frac{k(k+1)}{2} - 20.2}.$$
Since we defined $k$ to be the smallest integer such that $1 - (k+1) \alpha < 0$, we know that $k$ is the largest integer such that $k \alpha < 1$. If we plug in our value for $\alpha$, we get that this is equivalent to
$$\frac{k^2 - 20.2 k}{\frac{k(k+1)}{2} - 20.2} < 1 \Longrightarrow k < 40.4$$
Thus, we have $k = 40$, and plugging this in for our formula for $\alpha$ gives us
$$\alpha = \frac{40 - 20.2}{\frac{40 \cdot 41}{2} - 20.2} = \frac{33}{1333}.$$