Difference between revisions of "2024 AMC 10B Problems/Problem 14"
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==Problem== | ==Problem== | ||
− | A dartboard is the region B in the coordinate plane consisting of points <math>(x, y)</math> such that <math>|x| + |y| \le 8</math>. A target T is the region where <math>(x^2 + y^2 - 25)^2 \le 49</math>. A dart is thrown at a random point in B. The probability that the dart lands in T can be expressed as <math>\frac{m}{n} \pi</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m + n</math>? | + | A dartboard is the region B in the coordinate plane consisting of points <math>(x, y)</math> such that <math>|x| + |y| \le 8</math>. A target T is the region where <math>(x^2 + y^2 - 25)^2 \le 49</math>. A dart is thrown and lands at a random point in B. The probability that the dart lands in T can be expressed as <math>\frac{m}{n} \cdot \pi</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m + n</math>? |
<math> | <math> | ||
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<cmath>r^2 - 25 \ge -7\implies r\ge \sqrt{18}.</cmath> | <cmath>r^2 - 25 \ge -7\implies r\ge \sqrt{18}.</cmath> | ||
− | The | + | The intersection of these inequalities is the circular region <math>T</math> for which every circle in <math>T</math> has a radius between <math>\sqrt{18}</math> and <math>\sqrt{32}</math>, inclusive. The area of such a region is thus <math>\pi(32-18)=14\pi.</math> The requested probability is therefore <math>\frac{14\pi}{128} = \frac{7\pi}{64},</math> yielding <math>(m,n)=(7,64).</math> We have <math>m+n=7+64=\boxed{\textbf{(B)}\ 71}.</math> |
-anonymous, countmath1 | -anonymous, countmath1 | ||
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Expressing the Area of Region | Expressing the Area of Region | ||
− | Region | + | Region |
− | + | ||
− | |||
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In each quadrant, this can be expressed by the following functions: | In each quadrant, this can be expressed by the following functions: | ||
− | + | ||
− | + | First quadrant: | |
− | + | Second quadrant: | |
− | + | Third quadrant: | |
+ | Fourth quadrant: | ||
In the first quadrant, | In the first quadrant, | ||
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~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx Athmyx] | ~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx Athmyx] | ||
+ | |||
+ | ==Solution 3== | ||
+ | [[Image: 2024_AMC_12B_P09.jpeg|thumb|center|600px|]] | ||
+ | ~Kathan | ||
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ||
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https://www.youtube.com/watch?v=24EZaeAThuE | https://www.youtube.com/watch?v=24EZaeAThuE | ||
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==See also== | ==See also== |
Latest revision as of 18:49, 15 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 and lands 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 intersection 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
Solution 2 (Calculus)
Expressing the Area of Region
Region
In each quadrant, this can be expressed by the following functions:
First quadrant:
In the first quadrant,
Expressing the Area of Region
To find the area, we switch to polar coordinates with
The area of
The probability
So the probability is of the form
Solution 3
~Kathan
Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
https://youtu.be/YqKmvSR1Ckk?feature=shared
~ Pi Academy
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=24EZaeAThuE
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.