Difference between revisions of "2024 AMC 8 Problems/Problem 18"
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==Solution 3 (Ruler Tool)== | ==Solution 3 (Ruler Tool)== | ||
− | As I remember correctly, 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>20</math> degrees short of <math>x</math> itself, so <math>x-20=180-x</math>, solving gives <math>x=100</math>, therefore the closest answer is <math>\boxed\textbf{(A)}\,108</math> | + | As I remember correctly, 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>20</math> degrees short of <math>x</math> itself, so <math>x-20=180-x</math>, solving gives <math>x=100</math>, therefore the closest answer is <math>\boxed{\textbf{(A)}\,108}</math> |
==Video Solution 1 (super clear!) by Power Solve== | ==Video Solution 1 (super clear!) by Power Solve== |
Revision as of 18:44, 3 February 2024
Contents
- 1 Problem
- 2 Solution
- 3 Solution 2
- 4 Solution 3 (Ruler Tool)
- 5 Video Solution 1 (super clear!) by Power Solve
- 6 Video Solution 2 by Math-X (First fully understand the problem!!!)
- 7 Video Solution 3 by SpreadTheMathLove
- 8 Video Solution by NiuniuMaths (Easy to understand!)
- 9 Video Solution 2 by OmegaLearn.org
- 10 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 11 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 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
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 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)
As I remember correctly, 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 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
Video Solution 1 (super clear!) by Power Solve
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.