2024 AMC 8 Problems/Problem 18

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3 (Ruler Tool)
5 Solution 4
6 Video Solution 1 by Math-X (First fully understand the problem!!!)
7 Video Solution 2 (A Clever Explanation You’ll Get Instantly)
8 Video Solution 3 (super clear!) by Power Solve
9 Video Solution 4 by SpreadTheMathLove
10 Video Solution 5 by NiuniuMaths (Easy to understand!)
11 Video Solution 6 by OmegaLearn.org
12 Video Solution 7 by CosineMethod [🔥Fast and Easy🔥]
13 Video Solution 8 by Interstigation
14 Video Solution 9 by Dr. David
15 Video Solution by WhyMath
16 See Also

Problem

Three concentric circles centered at $O$ have radii of $1$ , $2$ , and $3$ . Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angles $BOC$ , 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?

$[asy] size(150); import graph; draw(circle((0,0),3)); real radius = 3; real angleStart = -54; // starting angle of the sector real angleEnd = 54; // ending angle of the sector label("$O$",(0,0),W); pair O = (0, 0); filldraw(arc(O, radius, angleStart, angleEnd)--O--cycle, gray); filldraw(circle((0,0),2),gray); filldraw(circle((0,0),1),white); draw((1.763,2.427)--(0,0)--(1.763,-2.427)); label("$B$",(1.763,2.427),NE); label("$C$",(1.763,-2.427),SE); [/asy]$ $\textbf{(A) } 108\qquad\textbf{(B) } 120\qquad\textbf{(C) } 135\qquad\textbf{(D) } 144\qquad\textbf{(E) } 150$

Solution 1

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 $\pi + \frac{360-x}{360}(5 \pi)$ .

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

~MrThinker

Solution 2

Notice that for the 3rd most outer ring of the circle, the ratio of the shaded region to non-shaded region is the ratio of $\angle{BOC}$ to $360-\angle{BOC}$ . With that, all we need to do is solve for the shaded region.

The inner most circle has radius $1$ , and the second circle has radius 2. Therefore, the first shaded area has $4 \pi - \pi = 3 \pi$ area. The circle has total area $9 \pi$ , so the other shaded region must have $1.5 \pi$ area, as the non-shaded and shaded area is equivalent. So for the 3rd outer ring, the total area is $9 \pi - 4 \pi = 5 \pi$ , so the non-shaded part of the outer ring is $5 \pi - 1.5 \pi = 3.5 \pi$ .

Now as said before, the ratio of these two areas is the ratio of $\angle{BOC}$ and $360 - \angle{BOC}$ . So, $\frac{3.5}{1.5} = \frac{7}{3}$ . We have $7x:3x$ where $7x+3x = 360$ , $x = 36$ , so our answer is $3x = 108, \boxed{(A) 108}$ .

~MaxyMoosy

Solution 3 (Ruler Tool)

The AMC 8s allow a ruler tool that you can rotate and drag. You can use the tool to make a straight segment (which we know is $180$ degrees), and we let the angle of desire be $x$ . We can estimate that $180-x$ is just about $30$ degrees short of $x$ itself, so $x-30=180-x$ , solving gives $x=105$ , therefore the closest answer is $\boxed{\textbf{(A)}\,108}$ .

Note: This isn't the most accurate method to use on AMC 8s, this is just a quick method if time is short or you do not know how to do the problem and want to guess at it.

~Tacos_are_yummy_1

Solution 4

Suppose the desired angle is some fraction $x$ of the total degree measure of the circle. We now compile a list of the shaded and unshaded areas. The inner circle of radius $1$ is completely unshaded, so it contributes $1$ to the unshaded area. (Everything will be a multiple of $\pi$ , so we omit it.) The inner annulus has area $2^2 - 1^2 = 3$ , which it contributes to the shaded area. The outer annulus has a total area of $3^2 - 2^2 = 5$ ; the fraction $x$ is shaded, so the shaded portion of the outer annulus contributes $5x$ to the shaded area, while the other $1 - x$ fraction is unshaded, so the unshaded portion contributes $5(1-x)$ to the unshaded area. We now equate and solve. $1 + 5(1-x) = 3 + 5x$ Upon solving, we find that $x = \frac{3}{10}$ , so the degree measure is $360 \cdot \frac{3}{10} = \boxed{\textbf{(A)} 108}$ .

~ cxsmi

Video Solution 1 by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=lg7OGcJ7OwdDFHAn&t=4872

~Math-X

Video Solution 2 (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=A8VVtUidVZlXDyrN&t=2517

~hsnacademy

Video Solution 3 (super clear!) by Power Solve

https://youtu.be/TlTN7EQcFvE

Video Solution 4 by SpreadTheMathLove

https://www.youtube.com/watch?v=Svibu3nKB7E

Video Solution 5 by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=V-xN8Njd_Lc

~NiuniuMaths

Video Solution 6 by OmegaLearn.org

https://youtu.be/b_pfNdmLp8A

Video Solution 7 by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=ahVNjSlwKmA

Video Solution 8 by Interstigation

https://youtu.be/ktzijuZtDas&t=2045

Video Solution 9 by Dr. David

https://youtu.be/ySuCpQZtsZY

Video Solution by WhyMath

https://youtu.be/v0Aba87nnA4

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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