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Geometry Difficulty 4.8 AIME Find the answer United States

Problem:
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 $100a + b$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

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
$\operatorname{perimeter}\left(X_{0}\right) + (2n-1) \pi$
Since $X_{0}$ is a $(3,4,5)$ triangle, and $n=21$ , this is $12 + 41\pi$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.