2024 AMC 8 Problems/Problem 18

Problem

Three concentric circles centered at $O$ have radii of $1$ , $2$ , and $3$ . Points $B$ and $C$ lie on the largest cirlce. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle $BOC$ , as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of $\angle{BOC}$ in degrees?

-figure-

$\textbf{(A) } 108\qquad\textbf{(B) } 120\qquad\textbf{(C) } 135\qquad\textbf{(D) } 144\qquad\textbf{(E) } 150$

Solution

Let $x=\angle{BOC}$ .

We see that the shaded region is the inner ring plus a sector $x^\circ$ of the outer ring. The area of this in terms of $x$ is $\left( 4 \pi - \pi \right)+\frac{x}{360} \left( 9 \pi - 4 \pi \right)$ . This simplifies to $3 \pi + \frac{x}{360}(5 \pi)$ .

Also, the unshaded portion is comprised of the smallest circle plus the sector $(360-x)^\circ$ of the outer ring. The area of this is $\pi + \frac{x}{360}(5 \pi)$ .

We are told these are equal, therefore $\pi + \frac{x}{360}(5 \pi) = 3 \pi + \frac{x}{360}(5 \pi)$ . Solving for $x$ reveals $x=\boxed{\textbf{(A) } 108}$ .

~MrThinker

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Solution 1

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2024 AMC 8 Problems/Problem 18

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Problem

Solution

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Solution 1

Video Solution 1 (super clear!) by Power Solve