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

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<cmath>m + n = 71</cmath>
 
<cmath>m + n = 71</cmath>
 
So <math>\boxed{\textbf{(B) }71}</math>
 
So <math>\boxed{\textbf{(B) }71}</math>
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 +
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
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 +
https://youtu.be/YqKmvSR1Ckk?feature=shared
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~ Pi Academy
  
 
==See also==
 
==See also==

Revision as of 10:03, 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

\[|x|+|y| \le 8\] Inequalities of this form 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, \[r^2 - 25 \le 7\] \[r \le \sqrt32\] And \[r^2 - 25 \ge -7\] \[r \ge \sqrt18\]

This corresponds to a ring in space with outer radius $\sqrt32$ and inner radius $\sqrt18$. Note that the outer circle is inscribed within the square, meaning it completely lies within the square.

Our probability, then, is \[\frac {\pi(32 - 18)}{128}\] \[= \frac{7\pi}{64}\] $m = 7$ and $n = 64$ \[m + n = 71\] So $\boxed{\textbf{(B) }71}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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