2024 AMC 8 Problems/Problem 18
Contents
- 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 See Also
Problem
Three concentric circles centered at have radii of , , and . Points and 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 , as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of in degrees?
Solution 1
Let .
We see that the shaded region is the inner ring plus a sector of the outer ring. The area of this in terms of is . This simplifies to .
Also, the unshaded portion is comprised of the smallest circle plus the sector of the outer ring. The area of this is .
We are told these are equal, therefore . Solving for reveals .
~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 to . With that, all we need to do is solve for the shaded region.
The inner most circle has radius , and the second circle has radius 2. Therefore, the first shaded area has area. The circle has total area , so the other shaded region must have area, as the non-shaded and shaded area is equivalent. So for the 3rd outer ring, the total area is , so the non-shaded part of the outer ring is .
Now as said before, the ratio of these two areas is the ratio of and . So, . We have where , , so our answer is .
~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 degrees), and we let the angle of desire be . We can estimate that is just about degrees short of itself, so , solving gives , therefore the closest answer is .
- 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 of the total degree measure of the circle. We now compile a list of the shaded and unshaded areas. The inner circle of radius is completely unshaded, so it contributes to the unshaded area. (Everything will be a multiple of , so we omit it.) The inner annulus has area , which it contributes to the shaded area. The outer annulus has a total area of ; the fraction is shaded, so the shaded portion of the outer annulus contributes to the shaded area, while the other fraction is unshaded, so the unshaded portion contributes to the unshaded area. We now equate and solve. Upon solving, we find that , so the degree measure is .
~ 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
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
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
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.