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Difference between revisions of "2024 AMC 8 Problems/Problem 18"

(→‎Video Solution 2 by OmegaLearn.org)
(→‎Solution)
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~MrThinker
 
~MrThinker
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==Solution 2==
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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.
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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 = 3pi</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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Now as said before, the ratio of these two areas is the ratio of <math>\angle{BOC}</math> and <math>360 - \angle{BOC}</math>. So, <math>\frac{3.5}{1.5} = \frac{7}{3}</math>. We have <math>7x:3x</math> where <math>7x+3x = 360</math>, <math>x = 36</math>, so our answer is <math>3x = 108, \boxed{A}</math>.
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~MaxyMoosy
  
 
==Video Solution 1 (super clear!) by Power Solve==
 
==Video Solution 1 (super clear!) by Power Solve==

Revision as of 18:24, 27 January 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 $\pi + \frac{x}{360}(5 \pi)$.

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

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 $4 \pi - \pi = 3pi$ 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

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/4Y8GUzNEAJQ

Video Solution 3 by SpreadTheMathLove

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

~Math-X

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

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

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