Problem: Show that there is an absolute constant $c < 1$
with the following property:
whenever $\mathcal P$ is a polygon with area $1$ in the plane,
one can translate it by a distance of $\frac{1}{100}$ in some direction
to obtain a polygon $\mathcal Q$, for which
the intersection of the interiors of $\mathcal P$ and $\mathcal Q$
has total area at most $c$.

\end{enumerate}

Solution: The following solution is due to Brian Lawrence.
We will prove the result with the generality
of any measurable set $\mathcal{P}$ (rather than a polygon).
For a vector $v$ in the plane,
write $\mathcal{P} + v$ for the translate of $\mathcal{P}$ by $v$.

Suppose $\mathcal{P}$ is a polygon of area $1$,
and $\eps > 0$ is a constant,
such that for any translate $\mathcal{Q} = \mathcal{P} + v$,
where $v$ has length exactly $\frac{1}{100}$,
the intersection of $\mathcal{P}$ and $\mathcal{Q}$ has area at least $1 - \eps$.
The problem asks us to prove a lower bound on $\eps$.

\begin{lemma*}
  Fix a sequence of $n$ vectors
  $v_1$, $v_2$, \dots, $v_n$, each of length $\frac{1}{100}$.
  A grasshopper starts at a random point $x$ of $\mathcal{P}$,
  and makes $n$ jumps to $x + v_1 + \dots + v_n$.
  Then it remains in $\mathcal{P}$ with probability at least $1 - n\eps$.
\end{lemma*}

\begin{proof}
  In order for the grasshopper to leave $\mathcal{P}$ at step $i$,
  the grasshopper's position before step $i$
  must be inside the difference set $\mathcal{P} \backslash (\mathcal{P} - v_i)$.
  Since this difference set has area at most $\eps$,
  the probability the grasshopper leaves $\mathcal{P}$
  at step $i$ is at most $\eps$.
  Summing over the $n$ steps,
  the probability that the grasshopper
  ever manages to leave $\mathcal{P}$ is at most $n \eps$.
\end{proof}

\begin{corollary*}
  Fix a vector $w$ of length at most $8$.
  A grasshopper starts at a random point $x$ of $\mathcal{P}$,
  and jumps to $x + w$.
  Then it remains in $\mathcal{P}$ with probability at least $1 - 800 \eps$.
\end{corollary*}
\begin{proof}
  Apply the previous lemma with $800$ jumps.
  Any vector $w$ of length at most $8$
  can be written as $w = v_1 + v_2 + \dotsb + v_{800}$,
  where each $v_i$ has length exactly $\frac{1}{100}$.
\end{proof}

Now consider the process where we select a random
starting point $x \in \mathcal P$ for our grasshopper,
\emph{and} a random vector $w$ of length at most $8$
(sampled uniformly from the closed disk of radius $8$).
Let $q$ denote the probability of staying inside $\mathcal P$
we will bound $q$ from above and below.

\begin{itemize}
\ii On the one hand, suppose we pick $w$ first.
By the previous corollary, $q \ge 1 - 800\eps$
(irrespective of the chosen $w$).

\ii On the other hand, suppose we pick $x$ first.
Then the possible landing points $x + w$ are uniformly
distributed over a closed disk of radius $8$,
which has area $64\pi$.
The probability of landing in $\mathcal P$
is certainly at most $\frac{[\mathcal P]}{64\pi}$.
\end{itemize}

Consequently, we deduce
\[ 1 - 800\eps \le q \le \frac{[\mathcal P]}{64\pi}
  \implies \eps > \frac{1 - \frac{[\mathcal P]}{64\pi}}{800}
  > 0.001
\]
as desired.

\begin{remark*}
  The choice of $800$ jumps is only for concreteness;
  any constant $n$ for which $\pi(n/100)^2 > 1$ works.
  I think $n = 98$ gives the best bound following this approach.
\end{remark*}