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Difference between revisions of "2024 AMC 10B Problems/Problem 14"

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==Problem==
 
==Problem==
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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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\textbf{(E) }135 \qquad
 
\textbf{(E) }135 \qquad
 
</math>
 
</math>
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 +
==Diagram==
 +
<asy>
 +
// By Elephant200
 +
// Feel free to adjust the code
 +
 +
size(10cm);
 +
 +
pair A = (8, 0);
 +
pair B = (0, 8);
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pair C = (-8, 0);
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pair D = (0, -8);
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draw(A--B--C--D--cycle, linewidth(1.5));
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 +
label("$(8,0)$", A, NE);
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label("$(0,8)$", B, NE);
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label("$(-8,0)$", C, NW);
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label("$(0,-8)$", D, SE);
 +
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filldraw(circle((0,0),4*sqrt(2)), gray, linewidth(1.5));
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filldraw(circle((0,0),3*sqrt(2)), white, linewidth(1.5));
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draw((-10, 0)--(10,0),EndArrow(5));
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draw((10, 0)--(-10,0),EndArrow(5));
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draw((0,-10)--(0,10), EndArrow(5));
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draw((0,10)--(0,-10),EndArrow(5));
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</asy>
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~Elephant200
  
 
==Solution 1==
 
==Solution 1==
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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 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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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
 +
 +
==Solution 2 (Calculus)==
 +
Expressing the Area of Region \( B \)
 +
 +
Region \( B \) consists of points where \( |x| + |y| \le 8 \)
 +
 +
In each quadrant, this can be expressed by the following functions:
 +
 +
First quadrant: \( y = 8 - x \)
 +
Second quadrant: \( y = 8 + x \)
 +
Third quadrant: \( y = -8 - x \)
 +
Fourth quadrant: \( y = -8 + x \)
 +
 +
In the first quadrant, \( x \) ranges from 0 to 8, and \( y \) ranges from 0 to \( 8 - x \). Thus, the area in the first quadrant is:
 +
<cmath>
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\text{Area of first quadrant} = \int_0^8 \int_0^{8 - x} \, dy \, dx
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</cmath>
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<cmath>
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= \int_0^8 [y]_{y=0}^{y=8-x} \, dx = \int_0^8 (8 - x) \, dx
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</cmath>
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<cmath>
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= \left[ 8x - \frac{x^2}{2} \right]_0^8 = 64 - 32 = 32
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</cmath>
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The total area of region \( B \) is:
 +
<cmath>
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\text{Area of } B = 4 \times 32 = 128
 +
</cmath>
 +
 +
Expressing the Area of Region \( T \)
 +
Region \( T \) is defined by the inequality \( (x^2 + y^2 - 25)^2 \le 49 \), which can be rewritten as:
 +
<cmath>
 +
18 \le x^2 + y^2 \le 32
 +
</cmath>
 +
 +
To find the area, we switch to polar coordinates with \( x = r \cos \theta \) and \( y = r \sin \theta \), where \( x^2 + y^2 = r^2 \). Here, \( r \) ranges from \( \sqrt{18} \) to \( \sqrt{32} \), and \( \theta \) ranges from 0 to \( 2\pi \).
 +
 +
The area of \( T \) can then be found by:
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<cmath>
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\text{Area of } T = \int_0^{2\pi} \int_{\sqrt{18}}^{\sqrt{32}} r \, dr \, d\theta
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</cmath>
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<cmath>
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= \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>
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= \int_0^{2\pi} 7 \, d\theta = 14\pi
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</cmath>
 +
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The probability \( P \) that a dart lands in region \( T \) is the area of \( T \) divided by the area of \( B \):
 +
<cmath>
 +
P = \frac{\text{Area of } T}{\text{Area of } B} = \frac{14\pi}{128} = \frac{7\pi}{64}
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</cmath>
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So the probability is of the form \( \frac{m}{n} \pi \), where \( m = 7 \) and \( n = 64 \), so \( m + n = 7 + 64 = 71 \).
 +
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<cmath>
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\boxed{\textbf{(B)}\ 71}
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</cmath>
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~[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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~ Pi Academy
 
~ Pi Academy
 +
 +
==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=24EZaeAThuE
 +
  
 
==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.

Problem

A dartboard is the region B in the coordinate plane consisting of points $(x, y)$ such that $|x| + |y| \le 8$. A target T is the region where $(x^2 + y^2 - 25)^2 \le 49$. A dart is thrown and lands at a random point in B. The probability that the dart lands in T can be expressed as $\frac{m}{n} \cdot \pi$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$?

$\textbf{(A) }39 \qquad \textbf{(B) }71 \qquad \textbf{(C) }73 \qquad \textbf{(D) }75 \qquad \textbf{(E) }135 \qquad$

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 $|x|+|y| \le 8$ are well-known and correspond to a square in space with centre at origin and vertices at $(8, 0)$, $(-8, 0)$, $(0, 8)$, $(0, -8)$. The diagonal length of this square is clearly $16$, so it has an area of \[\frac{1}{2} \cdot 16 \cdot 16 = 128\] Now, \[(x^2 + y^2 - 25)^2 \le 49\] Converting to polar form, \[r^2 - 25 \le 7 \implies r \le \sqrt{32},\] and \[r^2 - 25 \ge -7\implies r\ge \sqrt{18}.\]

The intersection of these inequalities is the circular region $T$ for which every circle in $T$ has a radius between $\sqrt{18}$ and $\sqrt{32}$, inclusive. The area of such a region is thus $\pi(32-18)=14\pi.$ The requested probability is therefore $\frac{14\pi}{128} = \frac{7\pi}{64},$ yielding $(m,n)=(7,64).$ We have $m+n=7+64=\boxed{\textbf{(B)}\ 71}.$

-anonymous, countmath1

Solution 2 (Calculus)

Expressing the Area of Region \( B \)

Region \( B \) consists of points where \( |x| + |y| \le 8 \)

In each quadrant, this can be expressed by the following functions:

First quadrant: \( y = 8 - x \) Second quadrant: \( y = 8 + x \) Third quadrant: \( y = -8 - x \) Fourth quadrant: \( y = -8 + x \)

In the first quadrant, \( x \) ranges from 0 to 8, and \( y \) ranges from 0 to \( 8 - x \). Thus, the area in the first quadrant is: \[\text{Area of first quadrant} = \int_0^8 \int_0^{8 - x} \, dy \, dx\] \[= \int_0^8 [y]_{y=0}^{y=8-x} \, dx = \int_0^8 (8 - x) \, dx\] \[= \left[ 8x - \frac{x^2}{2} \right]_0^8 = 64 - 32 = 32\] The total area of region \( B \) is: \[\text{Area of } B = 4 \times 32 = 128\]

Expressing the Area of Region \( T \) Region \( T \) is defined by the inequality \( (x^2 + y^2 - 25)^2 \le 49 \), which can be rewritten as: \[18 \le x^2 + y^2 \le 32\]

To find the area, we switch to polar coordinates with \( x = r \cos \theta \) and \( y = r \sin \theta \), where \( x^2 + y^2 = r^2 \). Here, \( r \) ranges from \( \sqrt{18} \) to \( \sqrt{32} \), and \( \theta \) ranges from 0 to \( 2\pi \).

The area of \( T \) can then be found by: \[\text{Area of } T = \int_0^{2\pi} \int_{\sqrt{18}}^{\sqrt{32}} r \, dr \, d\theta\] \[= \int_0^{2\pi} \left[ \frac{r^2}{2} \right]_{r=\sqrt{18}}^{r=\sqrt{32}} \, d\theta = \int_0^{2\pi} \left( \frac{32}{2} - \frac{18}{2} \right) \, d\theta\] \[= \int_0^{2\pi} 7 \, d\theta = 14\pi\]

The probability \( P \) that a dart lands in region \( T \) is the area of \( T \) divided by the area of \( B \): \[P = \frac{\text{Area of } T}{\text{Area of } B} = \frac{14\pi}{128} = \frac{7\pi}{64}\]

So the probability is of the form \( \frac{m}{n} \pi \), where \( m = 7 \) and \( n = 64 \), so \( m + n = 7 + 64 = 71 \).

\[\boxed{\textbf{(B)}\ 71}\]

~Athmyx

Solution 3

2024 AMC 12B P09.jpeg

~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. AMC logo.png

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