Answer: 6100
Solution: The first thing to note is that the area of $A B C D$ 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 $B C$ intersects $\ell_{2}$ and $\ell_{3}$ at $X$ and $Y$, respectively, and that $D A$ intersects $\ell_{2}$ and $\ell_{3}$ at $U$ and $V$, respectively. Note that $X Y V U$ is a parallelogram. Since $B C=B X+X Y+Y C=B X+2 X Y>2 X Y$, we have that $X Y$ is less than half the side length of the square, so the area of $X Y V U$ is at most half of the area of square $A B C D$. 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 $A B$ and $A D$ intersect $\ell_{2}$ at points $M$ and $N$, respectively. Let $r=A M$ and $s=A N$. By symmetry, $[A M N]=0.235$, so $r s=0.47$. 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
$$
1-\frac{2 r+2 s}{4}=1-\frac{r+s}{2} .
$$

On the other hand, the condition $\frac{1}{r}+\frac{1}{s}=3$ rearranges to $3 r s=r+s$, so $r+s=1.41$. Thus the answer is $1-\frac{1.41}{2}=0.295=\frac{59}{200}$.
4. Parallel lines $\ell_{1}, \ell_{2}, \ell_{3}, \ell_{4}$ are evenly spaced in the plane, in that order. Square $A B C D$ has the property that $A$ lies on $\ell_{1}$ and $C$ lies on $\ell_{4}$. Let $P$ be a uniformly random point in the interior of $A B C D$ and let $Q$ be a uniformly random point on the perimeter of $A B C D$. Given that the probability that $P$ lies between $\ell_{2}$ and $\ell_{3}$ is $\frac{53}{100}$, the probability that $Q$ lies between $\ell_{2}$ and $\ell_{3}$ can be expressed as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Compute $100 a+b$.