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

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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