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

(→‎Solution 1)
(→‎Solution 1)
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==Solution 1==
 
==Solution 1==
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<cmath>|x|+|y| \le 8</cmath>
+
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>.
−
Inequalities of this form 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  
 
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,
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<cmath>r^2 - 25 \le 7</cmath>
+
<cmath>r^2 - 25 \le 7 \implies r \le \sqrt{32},</cmath>
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<cmath>r \le \sqrt{32}</cmath>
+
and
−
And
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<cmath>r^2 - 25 \ge -7\implies r\ge \sqrt{18}.</cmath>
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<cmath>r^2 - 25 \ge -7</cmath>
 
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<cmath>r \ge \sqrt{18}</cmath>
 
  
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This corresponds to a ring in space with outer radius <math>\sqrt{32}</math> and inner radius <math>\sqrt{18}</math>.
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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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Note that the outer circle is inscribed within the square, meaning it completely lies within the square.
 
  
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Our probability, then, is <cmath>\frac {\pi(32 - 18)}{128}</cmath>
+
-anonymous, countmath1
−
<cmath>= \frac{7\pi}{64}</cmath>
 
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<math>m = 7</math> and <math>n = 64</math>
 
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<cmath>m + n = 71</cmath>
 
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So <math>\boxed{\textbf{(B) }71}</math>
 
  
 
==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.

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 $m + n$?

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

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

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

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