2024 AMC 12A 第 20 题

先试着解答 2024 AMC 12A 第 20 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 12A 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。

20.

在等边三角形 △ABC\triangle ABC 中,点 PP 和 QQ 分别在边 AB‾\overline{AB} 与 AC‾\overline{AC} 上独立均匀随机选取。下面哪个区间包含 △APQ\triangle APQ 的面积小于 △ABC\triangle ABC 面积一半的概率?
Points PP and QQ are chosen uniformly and independently at random on sides AB‾\overline{AB} and AC‾,\overline{AC}, respectively, of equilateral triangle △ABC.\triangle ABC. Which of the following intervals contains the probability that the area of △APQ\triangle APQ is less than half the area of △ABC?\triangle ABC?
[38,12]\left[\tfrac38,\tfrac12\right]
(12,23]\left(\tfrac12,\tfrac23\right]
(23,34]\left(\tfrac23,\tfrac34\right]
(34,78]\left(\tfrac34,\tfrac78\right]
(78,1]\left(\tfrac78,1\right]
答案:D
知识点:几何概率面积比
小提示:
令 x=APABx=\dfrac{AP}{AB}、y=AQACy=\dfrac{AQ}{AC},它们在 [0,1][0,1] 上均匀分布;面积比为 xyxy。
Let x=APABx=\dfrac{AP}{AB} and y=AQACy=\dfrac{AQ}{AC} be uniform on [0,1];[0,1]; the area ratio is xyxy
大提示:
计算 P(xy≥12)=∫121(1−12x)dxP(xy\ge\tfrac12)=\displaystyle\int_{\frac{1}{2}}^{1}\left(1-\tfrac{1}{2x}\right)dx,再用 11 减去它
Compute P(xy≥12)=∫121(1−12x)dxP(xy\ge\tfrac12)=\displaystyle\int_{\frac{1}{2}}^{1}\left(1-\tfrac{1}{2x}\right)dx and subtract from 11
解答:
令 x=APABx=\tfrac{AP}{AB},y=AQACy=\tfrac{AQ}{AC},二者在 [0,1][0,1] 上均匀分布,则面积比 [APQ][ABC]=xy\tfrac{[APQ]}{[ABC]}=xy。补事件 xy≥12xy\ge\tfrac12 要求 x≥12x\ge\tfrac12,且 y∈[12x,1]y\in[\tfrac{1}{2x},1],其概率为 ∫121(1−12x)dx=12−ln⁡22≈0.153。 \begin{aligned} &\int_{\frac{1}{2}}^{1}\left(1-\frac{1}{2x}\right)dx \\ &=\frac12-\frac{\ln2}{2}\approx0.153 \end{aligned}\text{。} 因此 P(xy<12)≈1−0.153=0.847P(xy\lt\tfrac12)\approx1-0.153=0.847,落在 (34,78]\left(\tfrac34,\tfrac78\right] 中。因此正确答案是 D。
With x=APABx=\tfrac{AP}{AB} and y=AQACy=\tfrac{AQ}{AC} uniform on [0,1],[0,1], the area ratio [APQ][ABC]=xy.\tfrac{[APQ]}{[ABC]}=xy. The complementary event xy≥12xy\ge\tfrac12 requires x≥12x\ge\tfrac12 and y∈[12x,1],y\in[\tfrac{1}{2x},1], with probability ∫121(1−12x)dx=12−ln⁡22≈0.153. \begin{aligned} &\int_{\frac{1}{2}}^{1}\left(1-\frac{1}{2x}\right)dx \\ &=\frac12-\frac{\ln2}{2}\approx0.153. \end{aligned} Therefore P(xy<12)≈1−0.153=0.847,P(xy\lt\tfrac12)\approx1-0.153=0.847, which lies in (34,78].\left(\tfrac34,\tfrac78\right]. Thus, the correct answer is D.
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