2024 AMC 12B Problem 9

Attempt Problem 9 of the 2024 AMC 12B 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 12B solutions, or check the answer key.

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

9.

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 probabilityannulus
Small hint:
The region BB is a square with diagonals of length 16,16, and (x2+y2−25)2≤49(x^2+y^2-25)^2 \le 49 means 18≤x2+y2≤3218 \le x^2 + y^2 \le 32
Big hint:
The annulus TT has area π(32−18);\pi(32 - 18); check that its outer radius 32\sqrt{32} exactly reaches the sides of B,B, so TT lies entirely inside BB
Solution:
The dartboard ∣x∣+∣y∣≤8|x| + |y| \le 8 is a square with diagonals 16,16, so its area is 12⋅16⋅16=128.\tfrac12 \cdot 16 \cdot 16 = 128. The target condition (x2+y2−25)2≤49(x^2 + y^2 - 25)^2 \le 49 means −7≤x2+y2−25≤7,-7 \le x^2 + y^2 - 25 \le 7, i.e. 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.
The distance from the origin to a side of the square (for instance x+y=8x + y = 8) is 82=32,\dfrac{8}{\sqrt2} = \sqrt{32}, exactly the annulus’s outer radius. So the annulus is tangent to the square and lies entirely within B.B. The probability is 14π128=764π,\dfrac{14\pi}{128} = \dfrac{7}{64}\pi, giving m+n=7+64=71.m + n = 7 + 64 = 71.
Thus, the correct answer is B.
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