Difference between revisions of "2024 AMC 10B Problems/Problem 14"
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// Feel free to adjust the code | // Feel free to adjust the code | ||
size(10cm); | size(10cm); | ||
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pair A = (8, 0); | pair A = (8, 0); | ||
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label("$(0,-8)$", D, SW); | label("$(0,-8)$", D, SW); | ||
− | + | filldraw(circle((0,0),4*sqrt(2)), gray); | |
− | + | filldraw(circle((0,0),3*sqrt(2)), white); | |
+ | |||
+ | draw((-12, 0)--(12,0),EndArrow(5)); | ||
+ | draw((12, 0)--(-12,0),EndArrow(5)); | ||
+ | draw((0,-12)--(0,12), EndArrow(5)); | ||
+ | draw((0,12)--(0,-12),EndArrow(5)); | ||
</asy> | </asy> | ||
~Elephant200 | ~Elephant200 |
Revision as of 13:40, 14 November 2024
- The following problem is from both the 2024 AMC 10B #14 and 2024 AMC 12B #9, so both problems redirect to this page.
Contents
[hide]Problem
A dartboard is the region B in the coordinate plane consisting of points such that . A target T is the region where . A dart is thrown at a random point in B. The probability that the dart lands in T can be expressed as , where and are relatively prime positive integers. What is ?
Diagram
~Elephant200
Solution 1
Inequalities of the form are well-known and correspond to a square in space with centre at origin and vertices at , , , . The diagonal length of this square is clearly , so it has an area of Now, Converting to polar form, and
The union of these inequalities is the circular region for which every circle in has a radius between and , inclusive. The area of such a region is thus The requested probability is therefore yielding We have
-anonymous, countmath1
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/YqKmvSR1Ckk?feature=shared
~ Pi Academy
See also
2024 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 AMC 10 Problems and Solutions |
2024 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.