2024 AMC 10B Problems/Problem 14

Revision as of 01:50, 14 November 2024 by Abhisood1234 (talk | contribs) (Simple Coordinate Geometry)
The following problem is from both the 2024 AMC 10B #14 and 2024 AMC 12B #9, so both problems redirect to this page.

Problem

Simple Coordinate Geometry

\[|x|+|y| \le 8\] 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 $\boxed{\textbf{(B) }71}$

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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