AMC12 2025 A

AMC12 2025 A · Q22

AMC12 2025 A · Q22. It mainly tests Probability (basic), Geometry misc.

Three real numbers are chosen independently and uniformly at random between $0$ and $1$. What is the probability that the greatest of these three numbers is greater than $2$ times each of the other two numbers? (In other words, if the chosen numbers are $a \geq b \geq c$, then $a > 2b$.)

独立均匀随机地在 $[0,1]$ 中选择三个实数。求这三个数中最大的那个大于另外两个的 $2$ 倍的概率？（换言之，若所选数字为 $a \geq b \geq c$，则 $a > 2b$）。

(A) \frac{1}{12} \frac{1}{12}

(B) \frac19 \frac19

(D) \frac16 \frac16

(E) \frac14 \frac14

Answer

Correct choice: (E)

正确答案：（E）

Solution

We can solve the problem by approaching it geometrically by mapping each possible triple to a coordinate $(a, b, c)$ in a unit cube. Now, we just have to find the volume of the solution set over $1$. WLOG, assume that $a$ is the greatest number. The intersection between $a>2b, a>2c$ and the unit cube is an oblique square pyramid with apex $(0,0,0)$ and base vertices $(1,0,0), (1,0,\frac12), (1,\frac12,0), \text{ and } (1,\frac12, \frac12)$. It helps to visualize the two planes cutting into the cube and leaving triangular traces in the $xy$ and $xz$ planes. The square base $b=s^2=(\frac12) ^2=\frac14$, and the height $h=1$, so the volume $v=\frac13 bh=\frac13(\frac14)(1)=\frac{1}{12}$. From here, we can multiply this volume by $3$ to account for $b,c$ being the greatest (all mutually exclusive). Alternatively, we can find $P(a>2b, 2c|a>b,c)$ by taking $\frac{1}{12}$ over the volume of the interesction between $a>b, a>c$ and the unit cube (a pyramid wth base $1$ and height $1$). Either way, we get the final probability of $\boxed{\text{(E) } \frac{1}{4}}$.

可以通过几何方法解决，将每个可能的有序三元组映射到单位立方体中的坐标 $(a, b, c)$。只需求解集的体积除以 $1$。不失一般性，假设 $a$ 是最大的数。$a>2b, a>2c$ 与单位立方体的交集是一个倾斜的方锥，顶点为 $(0,0,0)$，底面顶点为 $(1,0,0), (1,0,\frac12), (1,\frac12,0), (1,\frac12, \frac12)$。可视化两个平面切割立方体，在 $xy$ 和 $xz$ 平面上留下三角形痕迹。方形底面面积 $s^2=(\frac12) ^2=\frac14$，高度 $h=1$，体积 $v=\frac13 bh=\frac13(\frac14)(1)=\frac{1}{12}$。由此乘以 $3$ 以考虑 $b,c$ 分别为最大的情况（所有互斥）。或者，求 $P(a>2b, 2c|a>b,c)$ 为 $\frac{1}{12}$ 除以 $a>b, a>c$ 与单位立方体交集的体积（底面 $1$ 高度 $1$ 的锥体）。总之，最终概率为 $\boxed{\text{(E) } \frac{1}{4}}$。

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