2024 AMC 12B 第 9 题

先试着解答 2024 AMC 12B 第 9 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2024 AMC 12B 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

9.

一个飞镖靶是坐标平面中的区域 BB,由满足 ∣x∣+∣y∣≤8|x| + |y| \le 8 的点 (x,y)(x, y) 组成。目标区域 TT 由满足 (x2+y2−25)2≤49(x^2 + y^2 - 25)^2 \le 49 的点组成。飞镖随机落在 BB 中一点。飞镖落在 TT 中的概率可表示为 mn⋅π\dfrac{m}{n} \cdot \pi,其中 mm 和 nn 是互质正整数。求 m+nm + n。
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?
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答案:B
知识点:几何概率圆环
小提示:
区域 BB 是对角线长为 1616 的正方形,且 (x2+y2−25)2≤49(x^2+y^2-25)^2 \le 49 等价于 18≤x2+y2≤3218 \le x^2 + y^2 \le 32。
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
大提示:
环形区域 TT 的面积为 π(32−18)\pi(32 - 18);检查其外半径 32\sqrt{32} 恰好到达 BB 的边,所以 TT 完全位于 BB 内。
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
解答:
飞镖靶 ∣x∣+∣y∣≤8|x| + |y| \le 8 是一个对角线长为 1616 的正方形,面积为 12⋅16⋅16=128\tfrac12 \cdot 16 \cdot 16 = 128。目标条件 (x2+y2−25)2≤49(x^2 + y^2 - 25)^2 \le 49 等价于 −7≤x2+y2−25≤7-7 \le x^2 + y^2 - 25 \le 7,即 18≤x2+y2≤3218 \le x^2 + y^2 \le 32,这是一个面积为 π(32−18)=14π\pi(32 - 18) = 14\pi 的圆环。
原点到正方形一条边(如 x+y=8x + y = 8)的距离为 82=32\dfrac{8}{\sqrt2} = \sqrt{32},正好等于环形区域外半径,所以整个环形区域都在 BB 内。概率为 14π128=764π\dfrac{14\pi}{128} = \dfrac{7}{64}\pi,因此 m+n=7+64=71m + n = 7 + 64 = 71。
所以正确答案是 B。
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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