Question

A disk of radius $1$ rolls all the way around the inside of a square of side length $s > 4$ and sweeps out a region of area $A.$ A second disk of radius $1$ rolls all the way around the outside of the same square and sweeps out a region of area $2A.$ The value of $s$ can be written as $a+\dfrac{b\pi}{c},$ where $a,b,$ and $c$ are positive integers and $b$ and $c$ are relatively prime. What is $a+b+c?$

$10$

$11$

$12$

$13$

$14$

Solution:

The side length of the inner square traced out by the inner circle is $s - 4.$

There are also the small pieces remaining in the corner. These form a total area of $(1 + 1)^2 - \pi 1^2 = 4 - \pi.$

Therefore, $A = s^2 - (s - 4)^2 - (4 - \pi)$ $= 8s - 20 + \pi.$

The outer disk traces out an area that is comprised of $4$ rectangles and $4$ quarter-circles. The rectangles have area $s \cdot 2 = 2s$ and the quarter-circles form a circle with radius $2$ and area $4 \pi.$

This gives us $2A = 8s + 4 \pi.$

Equating the two equations we get $8s + 4 \pi = 2(8s - 20 + \pi).$ Solving yields $8s = 40 + 2 \pi$ $s = 5 + \dfrac{\pi}{4}.$

Thus, A is the correct answer.

2021 AMC 10A Fall Problem 19

Solution:

Problem 19 in Other Years