2. Let $X_{0}$ be the interior of a triangle with side lengths 3,4 , and 5 . For all positive integers $n$, define $X_{n}$ to be the set of points within 1 unit of some point in $X_{n-1}$. The area of the region outside $X_{20}$ but inside $X_{21}$ can be written as $a \pi+b$, for integers $a$ and $b$. Compute $100 a+b$.
Answer: 4112
Solution:
$X_{n}$ is the set of points within $n$ units of some point in $X_{0}$. The diagram above shows $X_{0}, X_{1}, X_{2}$, and $X_{3}$. As seen above it can be verified that $X_{n}$ is the union of
- $X_{0}$,
- three rectangles of height $n$ with the sides of $X_{0}$ as bases, and
- three sectors of radius $n$ centered at the vertices and joining the rectangles

Therefore the total area of $X_{n}$ is
$$
\left[X_{0}\right]+n \cdot \operatorname{perimeter}\left(X_{0}\right)+n^{2} \pi
$$

Since $X_{n-1}$ is contained entirely within $X_{n}$, the area within $X_{n}$ but not within $X_{n-1}$ is
$$
\text { perimeter }\left(X_{0}\right)+(2 n-1) \pi \text {. }
$$

Since $X_{0}$ is a $(3,4,5)$ triangle, and $n=21$, this is $12+41 \pi$.