2021 AMC 10 A Fall

Complete problem set with solutions and individual problem pages

Problem 19 Hard

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 2 A. The value of s can be written as a+\frac{b \pi}{c}, where a, b, and c are positive integers and b and c are relatively prime. What is a+b+c ?(2021 AMC Fall 10A, Question #19)

  • A.

    10

  • B.

    11

  • C.

    12

  • D.

    13

  • E.

    14

Answer:A

The side length of the inner square traced out by the disk with radius 1 is s-4. However, there is a piece at each corner (bounded by two line segments and one 90^{\circ} arc) where the disk never sweeps out. The combined area of these four pieces is (1+1)^{2}-\pi \cdot 1^{2}=4-\pi. As a result, we have A=s^{2}-(s-4)^{2}-(4-\pi)=8 s-20+\pi . Now, we consider the second disk. The part it sweeps is comprised of four quarter circles with radius 2 and four rectangles with side lengths of 2 and s. When we add it all together, we have 2 A=8 s+4 \pi, or A=4 s+2 \pi \text {. } We equate the expressions for A, and then solve for s : 8 s-20+\pi=4 s+2 \pi . We get s=5+\frac{\pi}{4}, so the answer is 5+1+4= (A) 10 .

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