1. Initial Setup and Definitions:
- Let $\mathcal{P}$ be a polygon with area $1$.
- Define the function $f(\vec{v}) = 1 - |\mathcal{P} \cap (\mathcal{P} + \vec{v})|$ for a translation vector $\vec{v}$.
- We aim to show that there exists a constant $c < 1$ such that for some translation vector $\vec{v}$ with $|\vec{v}| = \frac{1}{100}$, the area of the intersection of $\mathcal{P}$ and its translation $\mathcal{Q} = \mathcal{P} + \vec{v}$ is at most $c$.

2. **Subadditivity of $f$:
- Claim:** $f(\vec{u} + \vec{v}) \leq f(\vec{u}) + f(\vec{v})$ for any vectors $\vec{u}$ and $\vec{v}$.
- Proof:
$$\begin{aligned} f(\vec{u} + \vec{v}) &= 1 - |\mathcal{P} \cap (\mathcal{P} + \vec{u} + \vec{v})| \\ &= |\mathcal{P}| - |\mathcal{P} \cap (\mathcal{P} + \vec{u} + \vec{v})| \\ &= |\mathcal{P} \setminus (\mathcal{P} + \vec{u} + \vec{v})| \\ &= |\mathcal{P} \setminus (\mathcal{P} + \vec{u})| + |(\mathcal{P} \cap (\mathcal{P} + \vec{u})) \setminus (\mathcal{P} + \vec{u} + \vec{v})| \\ &\leq |\mathcal{P} \setminus (\mathcal{P} + \vec{u})| + |(\mathcal{P} + \vec{u}) \setminus (\mathcal{P} + \vec{u} + \vec{v})| \\ &= f(\vec{u}) + f(\vec{v}). \end{aligned}$$
This proves the subadditivity of $f$. $\blacksquare$

3. Reduction to Smaller Translations:
- For some $\varepsilon > 0$ and a vector $\vec{u}$ with $|\vec{u}| \leq \frac{1}{50}$, if $f(\vec{u}) > \varepsilon$, we can find vectors $\vec{v_1}$ and $\vec{v_2}$, both with $|\vec{v_1}| = |\vec{v_2}| = \frac{1}{100}$, such that $\vec{u} = \vec{v_1} + \vec{v_2}$.
- By the subadditivity of $f$, we have:
$$f(\vec{u}) \leq f(\vec{v_1}) + f(\vec{v_2}).$$
If $f(\vec{u}) > \varepsilon$, then $f(\vec{v_1}) + f(\vec{v_2}) > \varepsilon$, implying that at least one of $f(\vec{v_1})$ or $f(\vec{v_2})$ is greater than $\frac{\varepsilon}{2}$.

4. Main Lemma:
- Lemma: Let $S$ be a finite subset of $\mathbb{Z}^+$, and $n \in \mathbb{Z}^+$. Then
$$\frac{1}{n} \sum_{a=1}^n |S \setminus (S + a)| \geq \frac{\min \{|S|, n\}}{2}.$$
- Proof:
- Consider the number of pairs $(i, j)$ such that $0 < j - i \leq n$, $i \in S$, and $j \notin S$ (call such pairs *good*).
- Let $k = \min \{|S|, n\}$ and let $a_1 < a_2 < \ldots < a_k$ be the $k$ greatest elements of $S$.
- For each $a_i$, the number of good pairs $(a_i, u)$ is at least $n - (k - i)$.
- Summing over all $k$ elements, we get:
$$\sum_{i=1}^k (n - (k - i)) = nk - \frac{k(k-1)}{2} = \frac{k(2n - k + 1)}{2} \geq \frac{nk}{2}.$$
- Thus, the number of good pairs is at least $\frac{nk}{2}$, proving the lemma. $\blacksquare$

5. Approximation by a Grid:
- Grid the plane so that a $1 \times 1$ square is divided into a $50N \times 50N$ grid, where $N$ is very large.
- Define rows and columns as *weird* if they have at most $N$ cells and *normal* otherwise.
- Consider three cases:
1. If weird rows have at least $49N^2$ cells, translating the polygon rightward by $\frac{k}{N}$ for some $k$ will subtract an area of at least $\frac{49N^2}{2}$ cells.
2. If there are at least $49N$ normal rows, translating by $\frac{k}{N}$ for some $k$ will subtract an area of at least $49N \cdot \frac{N}{2}$ cells.
3. If neither of the above cases holds, the number of cells in normal columns is at most $2450N^2 + o(N^2)$, which contradicts the total number of cells being $2500N^2 + o(N)$.

6. Conclusion:
- The problem is solved for $\varepsilon < \frac{49}{5000}$, implying $c \geq 1 - \frac{\varepsilon}{2} > 1 - \frac{49}{10000}$.

The final answer is $c \geq 1 - \frac{49}{10000}$.