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+ | {{duplicate|[[2024 AMC 10B Problems/Problem 14|2024 AMC 10B #14]] and [[2024 AMC 12B Problems/Problem 9|2024 AMC 12B #9]]}} | ||
+ | ==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 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> | ||
+ | \textbf{(A) }39 \qquad | ||
+ | \textbf{(B) }71 \qquad | ||
+ | \textbf{(C) }73 \qquad | ||
+ | \textbf{(D) }75 \qquad | ||
+ | \textbf{(E) }135 \qquad | ||
+ | </math> | ||
+ | |||
+ | ==Diagram== | ||
+ | <asy> | ||
+ | // By Elephant200 | ||
+ | // Feel free to adjust the code | ||
+ | |||
+ | size(10cm); | ||
+ | |||
+ | pair A = (8, 0); | ||
+ | pair B = (0, 8); | ||
+ | pair C = (-8, 0); | ||
+ | pair D = (0, -8); | ||
+ | draw(A--B--C--D--cycle, linewidth(1.5)); | ||
+ | |||
+ | label("$(8,0)$", A, NE); | ||
+ | label("$(0,8)$", B, NE); | ||
+ | label("$(-8,0)$", C, NW); | ||
+ | label("$(0,-8)$", D, SE); | ||
+ | |||
+ | filldraw(circle((0,0),4*sqrt(2)), gray, linewidth(1.5)); | ||
+ | filldraw(circle((0,0),3*sqrt(2)), white, linewidth(1.5)); | ||
+ | |||
+ | draw((-10, 0)--(10,0),EndArrow(5)); | ||
+ | draw((10, 0)--(-10,0),EndArrow(5)); | ||
+ | draw((0,-10)--(0,10), EndArrow(5)); | ||
+ | draw((0,10)--(0,-10),EndArrow(5)); | ||
+ | </asy> | ||
+ | ~Elephant200 | ||
+ | |||
+ | ==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>. | ||
+ | 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> | ||
+ | Now, | ||
+ | <cmath>(x^2 + y^2 - 25)^2 \le 49</cmath> | ||
+ | Converting to polar form, | ||
+ | <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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | -AbhiSood1234, countmath1 | ||
+ | |||
+ | ==Solution 2 (Calculus)== | ||
+ | Expressing the Area of Region | ||
+ | |||
+ | Region | ||
+ | |||
+ | In each quadrant, this can be expressed by the following functions: | ||
+ | |||
+ | First quadrant: | ||
+ | Second quadrant: | ||
+ | Third quadrant: | ||
+ | Fourth quadrant: | ||
+ | |||
+ | In the first quadrant, | ||
+ | <cmath> | ||
+ | \text{Area of first quadrant} = \int_0^8 \int_0^{8 - x} \, dy \, dx | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | = \int_0^8 [y]_{y=0}^{y=8-x} \, dx = \int_0^8 (8 - x) \, dx | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | = \left[ 8x - \frac{x^2}{2} \right]_0^8 = 64 - 32 = 32 | ||
+ | </cmath> | ||
+ | The total area of region | ||
+ | <cmath> | ||
+ | \text{Area of } B = 4 \times 32 = 128 | ||
+ | </cmath> | ||
+ | |||
+ | Expressing the Area of Region | ||
+ | Region | ||
+ | <cmath> | ||
+ | 18 \le x^2 + y^2 \le 32 | ||
+ | </cmath> | ||
+ | |||
+ | To find the area, we switch to polar coordinates with | ||
+ | |||
+ | The area of | ||
+ | <cmath> | ||
+ | \text{Area of } T = \int_0^{2\pi} \int_{\sqrt{18}}^{\sqrt{32}} r \, dr \, d\theta | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | = \int_0^{2\pi} \left[ \frac{r^2}{2} \right]_{r=\sqrt{18}}^{r=\sqrt{32}} \, d\theta = \int_0^{2\pi} \left( \frac{32}{2} - \frac{18}{2} \right) \, d\theta | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | = \int_0^{2\pi} 7 \, d\theta = 14\pi | ||
+ | </cmath> | ||
+ | |||
+ | The probability | ||
+ | <cmath> | ||
+ | P = \frac{\text{Area of } T}{\text{Area of } B} = \frac{14\pi}{128} = \frac{7\pi}{64} | ||
+ | </cmath> | ||
+ | |||
+ | So the probability is of the form | ||
+ | |||
+ | <cmath> | ||
+ | \boxed{\textbf{(B)}\ 71} | ||
+ | </cmath> | ||
+ | |||
+ | ~[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 ⚡🚀)== | ||
+ | |||
+ | https://youtu.be/YqKmvSR1Ckk?feature=shared | ||
+ | |||
+ | ~ Pi Academy | ||
+ | |||
+ | ==Video Solution 2 by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=24EZaeAThuE | ||
+ | |||
+ | |||
+ | ==See also== | ||
+ | {{AMC10 box|year=2024|ab=B|num-b=13|num-a=15}} | ||
+ | {{AMC12 box|year=2024|ab=B|num-b=8|num-a=10}} | ||
+ | {{MAA Notice}} |
Latest revision as of 03:14, 30 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
-AbhiSood1234, 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.