Difference between revisions of "2024 AMC 10B Problems/Problem 14"
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==Solution 1== | ==Solution 1== | ||
− | < | + | Inequalities of the form <math>|x|+|y| \le 8</math> are well-known and correspond to a square in space with centre at origin and vertices at <math>(8, 0)</math>, <math>(-8, 0)</math>, <math>(0, 8)</math>, <math>(0, -8)</math>. |
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The diagonal length of this square is clearly <math>16</math>, so it has an area of | The diagonal length of this square is clearly <math>16</math>, so it has an area of | ||
<cmath>\frac{1}{2} \cdot 16 \cdot 16 = 128</cmath> | <cmath>\frac{1}{2} \cdot 16 \cdot 16 = 128</cmath> | ||
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<cmath>(x^2 + y^2 - 25)^2 \le 49</cmath> | <cmath>(x^2 + y^2 - 25)^2 \le 49</cmath> | ||
Converting to polar form, | Converting to polar form, | ||
− | <cmath>r^2 - 25 \le 7 | + | <cmath>r^2 - 25 \le 7 \implies r \le \sqrt{32},</cmath> |
− | + | and | |
− | + | <cmath>r^2 - 25 \ge -7\implies r\ge \sqrt{18}.</cmath> | |
− | <cmath>r^2 - 25 \ge -7 | ||
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− | + | The union of these inequalities is the circular region <math>\mathcal{R}</math> for which every circle in <math>\mathcal{R}</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> | |
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− | + | -anonymous, countmath1 | |
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==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== | ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)== |
Revision as of 12:29, 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 ?
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.