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Difference between revisions of "2024 AMC 10B Problems/Problem 14"
(→‎Solution 1)
Line 21: Line 21:
 
Converting to polar form,
 
Converting to polar form,
 
<cmath>r^2 - 25 \le 7</cmath>
 
<cmath>r^2 - 25 \le 7</cmath>
−
<cmath>r \le \sqrt32</cmath>
+
<cmath>r \le \sqrt{32}</cmath>
 
And
 
And
 
<cmath>r^2 - 25 \ge -7</cmath>
 
<cmath>r^2 - 25 \ge -7</cmath>
−
<cmath>r \ge \sqrt18</cmath>
+
<cmath>r \ge \sqrt{18}</cmath>
  
−
This corresponds to a ring in space with outer radius <math>\sqrt32</math> and inner radius <math>\sqrt18</math>.
+
This corresponds to a ring in space with outer radius <math>\sqrt{32}</math> and inner radius <math>\sqrt{18}</math>.
 
Note that the outer circle is inscribed within the square, meaning it completely lies within the square.
 
Note that the outer circle is inscribed within the square, meaning it completely lies within the square.
  

Revision as of 10:49, 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, \[r^2 - 25 \le 7\] \[r \le \sqrt{32}\] And \[r^2 - 25 \ge -7\] \[r \ge \sqrt{18}\]

This corresponds to a ring in space with outer radius $\sqrt{32}$ and inner radius $\sqrt{18}$. 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

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