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Difference between revisions of "2024 AMC 8 Problems/Problem 18"
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==Solution 3 (Ruler Tool)==
 
==Solution 3 (Ruler Tool)==
  
−
The AMC 8's 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 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>.  
  
−
*Note: This isn't the most accurate method to use on AMC 8's, 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
 
~Tacos_are_yummy_1

Revision as of 13:06, 31 March 2024

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 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?

[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

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 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 $3x = 108, \boxed{A}$.

~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 (super clear!) by Power Solve

https://youtu.be/TlTN7EQcFvE

Video Solution 2 by Math-X (First fully understand the problem!!!)

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

~Math-X

Video Solution 3 by SpreadTheMathLove

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

Video Solution by NiuniuMaths (Easy to understand!)

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

~NiuniuMaths

Video Solution 2 by OmegaLearn.org

https://youtu.be/b_pfNdmLp8A

Video Solution by CosineMethod [🔥Fast and Easy🔥]

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

Video Solution by Interstigation

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

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

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

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