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

(Solution 1)
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\textbf{(E) }135 \qquad
 
\textbf{(E) }135 \qquad
 
</math>
 
</math>
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 +
==Diagram==
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<asy>
 +
// By Elephant200
 +
// Feel free to adjust the code
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size(10cm);
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draw((-12, 0)--(12,0),EndArrow(5));
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draw((12, 0)--(-12,0),EndArrow(5));
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draw((0,-12)--(0,12), EndArrow(5));
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draw((0,12)--(0,-12),EndArrow(5));
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 +
pair A = (8, 0);
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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);
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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, SW);
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label("$(0,-8)$", D, SW);
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 +
draw(circle((0,0),3*sqrt(2)));
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draw(circle((0,0),4*sqrt(2)));
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</asy>
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~Elephant200
  
 
==Solution 1==
 
==Solution 1==

Revision as of 13:02, 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 $m + n$?

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

Diagram

[asy] // By Elephant200 // Feel free to adjust the code size(10cm); draw((-12, 0)--(12,0),EndArrow(5)); draw((12, 0)--(-12,0),EndArrow(5)); draw((0,-12)--(0,12), EndArrow(5)); draw((0,12)--(0,-12),EndArrow(5));  pair A = (8, 0); pair B = (0, 8); pair C = (-8, 0); pair D = (0, -8); draw(A--B--C--D--cycle);  label("$(8,0)$", A, NE); label("$(0,8)$", B, NE); label("$(-8,0)$", C, SW); label("$(0,-8)$", D, SW);  draw(circle((0,0),3*sqrt(2))); draw(circle((0,0),4*sqrt(2))); [/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

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/YqKmvSR1Ckk?feature=shared

~ Pi Academy

See also

2024 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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