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

m (→‎Diagram)
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-anonymous, countmath1
 
-anonymous, countmath1
 +
 +
==Solution 2 (Calculus)==
 +
Expressing the Area of Region \( B \)
 +
 +
Region \( B \) consists of points where \( |x| + |y| \le 8 \), which forms a diamond centered at the origin with a side length of 16. We can rewrite this as the inequality:
 +
<cmath>
 +
|x| + |y| \le 8
 +
</cmath>
 +
In each quadrant, this can be expressed by the following functions:
 +
1. First quadrant: \( y = 8 - x \)
 +
2. Second quadrant: \( y = 8 + x \)
 +
3. Third quadrant: \( y = -8 - x \)
 +
4. 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>
 +
\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 \( B \) is:
 +
<cmath>
 +
\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:
 +
<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 \( P \) 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}
 +
</cmath>
 +
 +
So the probability is of the form \( \frac{m}{n} \pi \), where \( m = 7 \) and \( n = 64 \), so \( m + n = 7 + 64 = 71 \).
 +
 +
<cmath>
 +
\boxed{\textbf{(B)}\ 71}
 +
</cmath>
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx Athmyx]
  
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==

Revision as of 00:15, 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 at a random point in B. The probability that the dart lands in T can be expressed as $\frac{m}{n} \pi$, 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 union of these inequalities is the circular region $\mathcal{R}$ for which every circle in $\mathcal{R}$ 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 \( |x| + |y| \le 8 \), which forms a diamond centered at the origin with a side length of 16. We can rewrite this as the inequality: \[|x| + |y| \le 8\] In each quadrant, this can be expressed by the following functions: 1. First quadrant: \( y = 8 - x \) 2. Second quadrant: \( y = 8 + x \) 3. Third quadrant: \( y = -8 - x \) 4. 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

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

Solution 2

2024 AMC 12B P09.jpeg

~Kathan

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

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