AMC10 2012 A

AMC10 2012 A · Q25

AMC10 2012 A · Q25. It mainly tests Probability (basic), Geometric probability (basic).

Real numbers x, y, and z are chosen independently and at random from the interval [0, n] for some positive integer n. The probability that no two of x, y, and z are within 1 unit of each other is greater than $\frac{1}{2}$. What is the smallest possible value of n?

实数 x, y, z 独立均匀随机从区间 [0, n] 中选取，其中 n 为正整数。x, y, z 中任意两个不相距 1 个单位以内的概率大于 $\frac{1}{2}$。n 的最小可能值是多少？

(A) 7 7

(B) 8 8

(D) 10 10

(E) 11 11

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): It may be assumed that $x\le y\le z$. Because there are six possible ways of permuting the triple $(x,y,z)$, it follows that the set of all triples $(x,y,z)$ with $0\le x\le y\le z\le n$ is a region whose volume is $\frac16$ of the volume of the cube $[0,n]^3$, that is $\frac16 n^3$. Let $S$ be the set of triples meeting the required condition. For every $(x,y,z)\in S$ consider the translation $(x,y,z)\mapsto (x',y',z')=(x,y-1,z-2)$. Note that $y'=y-1>x=x'$ and $z'=z-2>y-1=y'$. Thus the image of $S$ under this translation is equal to $\{(x',y',z'):0\le x'<y'<z'\le n-2\}$. Again by symmetry of the possible permutations of the triples $(x',y',z')$, the volume of this set is $\frac16 (n-2)^3$. Because $\frac{7^3}{9^3}=\frac{343}{729}<\frac12$ and $\frac{8^3}{10^3}=\frac{512}{1000}>\frac12$, the smallest possible value of $n$ is $10$.

答案（D）：可以假设 $x\le y\le z$。由于三元组 $(x,y,z)$ 有六种可能的排列方式，因此所有满足 $0\le x\le y\le z\le n$ 的三元组 $(x,y,z)$ 构成的区域，其体积是立方体 $[0,n]^3$ 体积的 $\frac16$，即 $\frac16 n^3$。设 $S$ 为满足所需条件的三元组集合。对每个 $(x,y,z)\in S$，考虑平移 $(x,y,z)\mapsto (x',y',z')=(x,y-1,z-2)$。注意到 $y'=y-1>x=x'$ 且 $z'=z-2>y-1=y'$。因此，$S$ 在该平移下的像为 $\{(x',y',z'):0\le x'<y'<z'\le n-2\}$。同样利用三元组 $(x',y',z')$ 的排列对称性，该集合的体积为 $\frac16 (n-2)^3$。因为 $\frac{7^3}{9^3}=\frac{343}{729}<\frac12$ 且 $\frac{8^3}{10^3}=\frac{512}{1000}>\frac12$，所以 $n$ 的最小可能值是 $10$。

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