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2024 AMC 8 Problems/Problem 18: Difference between revisions

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==Problem==
==Problem==


Three concentric circles centered at <math>O</math> have radii of <math>1</math>, <math>2</math>, and <math>3</math>. Points <math>B</math> and <math>C</math> 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 angle <math>BOC</math>, as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of <math>\angle{BOC}</math> in degrees?
Three concentric circles centered at <math>O</math> have radii of <math>1</math>, <math>2</math>, and <math>3</math>. Points <math>B</math> and <math>C</math> 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 <math>BOC</math>, as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of <math>\angle{BOC}</math> in degrees?


<asy>
<asy>
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</asy>
</asy>
<math>\textbf{(A) } 108\qquad\textbf{(B) } 120\qquad\textbf{(C) } 135\qquad\textbf{(D) } 144\qquad\textbf{(E) } 150</math>
<math>\textbf{(A) } 108\qquad\textbf{(B) } 120\qquad\textbf{(C) } 135\qquad\textbf{(D) } 144\qquad\textbf{(E) } 150</math>


==Solution 1==
==Solution 1==
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Let <math>x=\angle{BOC}</math>.
Let <math>x=\angle{BOC}</math>.


We see that the shaded region is the inner ring plus a sector <math>x^\circ</math> of the outer ring. The area of this in terms of <math>x</math> is <math>\left( 4 \pi - \pi \right)+\frac{x}{360} \left( 9 \pi - 4 \pi \right)</math>. This simplifies to <math>3 \pi + \frac{x}{360}(5 \pi)</math>.
We see that the shaded region is the inner ring plus a sector <math>x^\circ</math> of the outer ring. Using the formula for the area of a circle (<math>A = \pi r^2</math>), we find that the area of <math>x</math> is <math>\left( 4 \pi - \pi \right)+\frac{x}{360} \left( 9 \pi - 4 \pi \right)</math>. This simplifies to <math>3 \pi + \frac{x}{360}(5 \pi)</math>.


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


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


~MrThinker
~MrThinker
~Dash11 (Edited coefficient of pi. Credit goes to MrThinker for the solution and explanation.)


==Solution 2==
==Solution 2==


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


The inner most circle has radius <math>1</math>, and the second circle has radius 2. Therefore, the first shaded area has <math>4 \pi - \pi = 3 \pi</math> area. The circle has total area <math>9 \pi</math>, so the other shaded region must have <math>1.5 \pi</math> area, as the non-shaded and shaded area is equivalent. So for the 3rd outer ring, the total area is <math>9 \pi - 4 \pi = 5 \pi</math>, so the non-shaded part of the outer ring is <math>5 \pi - 1.5 \pi = 3.5 \pi</math>.  
The inner most circle has radius <math>1</math>, and the second circle has radius 2. Therefore, the first shaded area has <math>4 \pi - \pi = 3 \pi</math> area. The circle has total area <math>9 \pi</math>, so the other shaded region must have <math>1.5 \pi</math> area, as the non-shaded and shaded area is equivalent. So for the 3rd outer ring, the total area is <math>9 \pi - 4 \pi = 5 \pi</math>, so the non-shaded part of the outer ring is <math>5 \pi - 1.5 \pi = 3.5 \pi</math>.  
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==Solution 3 (Ruler Tool)==
==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 <math>180</math> degrees), and we let the angle of desire be <math>x</math>. We can estimate that <math>180-x</math> is just about <math>30</math> degrees short of <math>x</math> itself, so <math>x-30=180-x</math>, solving gives <math>x=105</math>, therefore the closest answer is <math>\boxed{\textbf{(A)}\,108}</math>.  
The AMC 8 test allows a ruler tool that you can rotate and drag. You can use the tool to make a straight segment (which we know is <math>180</math> degrees), and we let the angle of desire be <math>x</math>. We can estimate that <math>180-x</math> is just about <math>30</math> degrees short of <math>x</math> itself, so <math>x-30=180-x</math>, solving gives <math>x=105</math>, therefore the closest answer is <math>\boxed{\textbf{(A)}\,108}</math>.  


*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.
*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==
==Solution 4==
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~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]


==Video Solution 1 (super clear!) by Power Solve==
https://youtu.be/TlTN7EQcFvE


==Video Solution 2 by Math-X (First fully understand the problem!!!)==
 
==Video Solution 1 by Math-X (First fully understand the problem!!!)==
https://youtu.be/BaE00H2SHQM?si=lg7OGcJ7OwdDFHAn&t=4872
https://youtu.be/BaE00H2SHQM?si=lg7OGcJ7OwdDFHAn&t=4872


~Math-X
~Math-X


==Video Solution 3 by SpreadTheMathLove==
==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
https://www.youtube.com/watch?v=Svibu3nKB7E


==Video Solution 4 by NiuniuMaths (Easy to understand!)==
==Video Solution 5 by NiuniuMaths (Easy to understand!)==
https://www.youtube.com/watch?v=V-xN8Njd_Lc
https://www.youtube.com/watch?v=V-xN8Njd_Lc
==Video Solution by Central Valley Math Circle (Goes through full thought process)==
https://youtu.be/JZx2-QX--Os
~mr_mathman


~NiuniuMaths
~NiuniuMaths


==Video Solution 5 by OmegaLearn.org==
==Video Solution 6 by OmegaLearn.org==
https://youtu.be/b_pfNdmLp8A
https://youtu.be/b_pfNdmLp8A


== Video Solution 6 by CosineMethod [🔥Fast and Easy🔥]==
== Video Solution 7 by CosineMethod [🔥Fast and Easy🔥]==


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


==Video Solution by Interstigation==
==Video Solution 8 by Interstigation==
https://youtu.be/ktzijuZtDas&t=2045
https://youtu.be/ktzijuZtDas&t=2045
==Video Solution 9 by Dr. David==
https://youtu.be/ySuCpQZtsZY
==Video Solution by WhyMath==
https://youtu.be/v0Aba87nnA4
==Official Video Solution (Forming a trivial algebraic equation!)==
https://www.youtube.com/watch?v=0bqpjCvqJsE
~TheMathGeek (Subscribe!)
==Video Solution by Daily Dose of Math (Simple, Certified, and Logical)==
https://youtu.be/OwJvuq6F7sQ
~Thesmartgreekmathdude
== Video solution by TheNeuralMathAcademy ==
https://youtu.be/f63MY1T2MgI&t=1822s


==See Also==
==See Also==
{{AMC8 box|year=2024|num-b=17|num-a=19}}
{{AMC8 box|year=2024|num-b=17|num-a=19}}
{{MAA Notice}}
{{MAA Notice}}
[[Category:Introductory Geometry Problems]]

Latest revision as of 14:38, 18 August 2025

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. Using the formula for the area of a circle ($A = \pi r^2$), we find that the area of $x$ is $\left( 4 \pi - \pi \right)+\frac{x}{360} \left( 9 \pi - 4 \pi \right)$. This simplifies to $3 \pi + \frac{x}{360}(5 \pi)$.

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

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

~MrThinker

~Dash11 (Edited coefficient of pi. Credit goes to MrThinker for the solution and explanation.)

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 8 test allows 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.

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

Video Solution by Central Valley Math Circle (Goes through full thought process)

https://youtu.be/JZx2-QX--Os

~mr_mathman

~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


Official Video Solution (Forming a trivial algebraic equation!)

https://www.youtube.com/watch?v=0bqpjCvqJsE ~TheMathGeek (Subscribe!)

Video Solution by Daily Dose of Math (Simple, Certified, and Logical)

https://youtu.be/OwJvuq6F7sQ

~Thesmartgreekmathdude

Video solution by TheNeuralMathAcademy

https://youtu.be/f63MY1T2MgI&t=1822s

See Also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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