2024 AMC 10B Problem 14

Attempt Problem 14 of the 2024 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

A dartboard is the region BB in the coordinate plane consisting of points (x,y)(x, y) such that ∣x∣+∣y∣≤8.|x| + |y| \le 8. A target TT is the region where (x2+y2−25)2≤49.(x^2 + y^2 - 25)^2 \le 49. A dart is thrown and lands at a random point in B.B. The probability that the dart lands in TT can be expressed as mn⋅π,\dfrac{m}{n} \cdot \pi, where mm and nn are relatively prime positive integers. What is m+n?m + n?
3939
7171
7373
7575
135135
Answer: B
Concepts:geometric probabilityannulustangent line
Small hint:
BB is a square of diagonal 16,16, and TT is the annulus 18≤x2+y2≤3218 \le x^2 + y^2 \le 32
Big hint:
The outer radius 32=42\sqrt{32} = 4\sqrt2 equals the distance from the origin to each edge of B,B, so TT sits entirely inside BB
Solution:
BB is the square ∣x∣+∣y∣≤8,|x| + |y| \le 8, with area 2⋅82=128.2 \cdot 8^2 = 128. The target condition (x2+y2−25)2≤49(x^2 + y^2 - 25)^2 \le 49 unpacks to ∣x2+y2−25∣≤7,|x^2 + y^2 - 25| \le 7, that is 18≤x2+y2≤32,18 \le x^2 + y^2 \le 32, an annulus of area π(32−18)=14π.\pi(32 - 18) = 14\pi. Does it fit inside B?B? The distance from the origin to an edge x+y=8x + y = 8 is 82=42=32,\tfrac{8}{\sqrt2} = 4\sqrt2 = \sqrt{32}, exactly the outer radius, so yes, the annulus sits inside the square. The probability is 14π128=764π,\tfrac{14\pi}{128} = \tfrac{7}{64}\pi, giving m+n=71.m + n = 71. Therefore, the answer is B.
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