, 2022

Solution:

The first thing to note is that the area of $ABCD$ does not matter in this problem, so for the sake of convenience, introduce coordinates so that $A = (0, 0)$, $B = (1, 0)$, and $C = (0, 1)$.

Suppose $A$ and $B$ lie on the same side of $\ell_{2}$. Then, by symmetry, $C$ and $D$ lie on the same side of $\ell_{3}$. Now suppose $BC$ intersects $\ell_{2}$ and $\ell_{3}$ at $X$ and $Y$, respectively, and that $DA$ intersects $\ell_{2}$ and $\ell_{3}$ at $U$ and $V$, respectively. Note that $XYVU$ is a parallelogram. Since $BC = BX + XY + YC = BX + 2XY > 2XY$, we have that $XY$ is less than half the side length of the square, so the area of $XYVU$ is at most half of the area of square $ABCD$. However, since $0.53 > \frac{1}{2}$, this can't happen. Similar reasoning applies if $B$ and $C$ lie on the same side of $\ell_{3}$. Therefore, points $B$ and $D$ lie between $\ell_{2}$ and $\ell_{3}$.

Let $AB$ and $AD$ intersect $\ell_{2}$ at points $M$ and $N$, respectively. Let $r = AM$ and $s = AN$. By symmetry, $[AMN] = 0.235$, so $rs = 0.47$. Additionally, in coordinates line $\ell_{2}$ is just $\frac{x}{r} + \frac{y}{s} = 1$. Therefore line $\ell_{4}$ is given by $\frac{x}{r} + \frac{y}{s} = 3$. Since $C = (1, 1)$ lies on this line, $\frac{1}{r} + \frac{1}{s} = 3$.

The answer that we want is
$$1 - \frac{2r + 2s}{4} = 1 - \frac{r + s}{2}$$
On the other hand, the condition $\frac{1}{r} + \frac{1}{s} = 3$ rearranges to $3rs = r + s$, so $r + s = 1.41$. Thus the answer is $1 - \frac{1.41}{2} = 0.295 = \frac{59}{200}$.