AMC10 2009 A

AMC10 2009 A · Q24

AMC10 2009 A · Q24. It mainly tests Combinations, 3D geometry (volume).

Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?

从立方体的三个不同顶点中随机选择。确定的平面包含立方体内点的概率是多少？

(A) $\frac{1}{4}$ $\frac{1}{4}$

(B) $\frac{3}{8}$ $\frac{3}{8}$

(D) $\frac{5}{7}$ $\frac{5}{7}$

(E) $\frac{3}{4}$ $\frac{3}{4}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): A plane that intersects at least three vertices of a cube either cuts into the cube or is coplanar with a cube face. Therefore the three randomly chosen vertices result in a plane that does not contain points inside the cube if and only if the three vertices come from the same face of the cube. There are 6 cube faces, so the number of ways to choose three vertices on the same cube face is $6\cdot\binom{4}{3}=24$. The total number of ways to choose the distinct vertices without restriction is $\binom{8}{3}=56$. Hence the probability is $1-\frac{24}{56}=\frac{4}{7}$.

答案（C）：一个平面若与立方体至少三个顶点相交，要么会切入立方体内部，要么与立方体的某个面共面。因此，随机选取的三个顶点所确定的平面不包含立方体内部点，当且仅当这三个顶点来自立方体的同一个面。立方体有 6 个面，所以在同一个面上选取三个顶点的方式数为 $6\cdot\binom{4}{3}=24$。不加限制从 8 个顶点中选取 3 个不同顶点的总方式数为 $\binom{8}{3}=56$。因此概率为 $1-\frac{24}{56}=\frac{4}{7}$。

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