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

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

~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

@@ Line 113: / Line 113: @@
 ~[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 ⚡🚀)==
@@ Line 123: / Line 127: @@
 https://www.youtube.com/watch?v=24EZaeAThuE
-==Solution 2==
-[[Image: 2024_AMC_12B_P09.jpeg|thumb|center|600px|]]
-~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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