Difference between revisions of "2024 AMC 8 Problems/Problem 18"
Countmath1 (talk | contribs) (→Problem) |
Niuniumaths (talk | contribs) (→Video Solution 1 (super clear!) by Power Solve) |
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==Video Solution 1 (super clear!) by Power Solve== | ==Video Solution 1 (super clear!) by Power Solve== | ||
https://youtu.be/TlTN7EQcFvE | https://youtu.be/TlTN7EQcFvE | ||
+ | ==Video Solution by NiuniuMaths (Easy to understand!)== | ||
+ | https://www.youtube.com/watch?v=V-xN8Njd_Lc | ||
+ | |||
+ | ~NiuniuMaths | ||
==Video Solution 2 by Math-X (First fully understand the problem!!!)== | ==Video Solution 2 by Math-X (First fully understand the problem!!!)== |
Revision as of 01:11, 27 January 2024
Contents
[hide]Problem
Three concentric circles centered at have radii of , , and . Points and lie on the largest cirlce. 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 , as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of in degrees?
Solution
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
Video Solution 1 (super clear!) by Power Solve
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
Video Solution 2 by Math-X (First fully understand the problem!!!)
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=Svibu3nKB7E
~Math-X
Video Solution 2 by OmegaLearn.org
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.