2012 AMC 10A Problem 25

The point $(x, y, z)$ is uniform in a cube of side $n$, so the probability is the volume of the good region divided by $n^3$.

By symmetry the six orderings of $x, y, z$ are equally likely, so restrict to $x < y < z$. This slice has volume $\frac{n^3}{6}$, and within it the condition is just $y - x \ge 1$ and $z - y \ge 1$ (the pair $x, z$ is then automatically at least $2$ apart).

Remove the mandatory gaps: set $x' = x$, $y' = y - 1$, $z' = z - 2$. The conditions become
$$0 \le x' \le y' \le z' \le n - 2,$$
which is the ordered slice of a cube of side $n - 2$. Shifting does not change volume, so the good ordered volume is $\frac{(n-2)^3}{6}$, and
$$P(n) = \frac{(n-2)^3/6}{n^3/6} = \left(\frac{n-2}{n}\right)^3.$$

We need $P(n) > \frac12$. Since $P(n)$ increases with $n$, test the choices upward: $n = 9$ gives $\left(\frac{7}{9}\right)^3 = \frac{343}{729} < \frac12$, and $n = 10$ gives $\left(\frac{8}{10}\right)^3 = \frac{512}{1000} > \frac12$.

The answer is $\boxed{\textbf{(D)}\ 10}$.

"No two within 1 unit" is the continuous version of choosing integers with required gaps: sort the values, subtract the forced spacing, and the constraint vanishes. Sorting costs a factor of $6$ that cancels, and translation preserves volume, so the probability is a pure scaling ratio $\left(\frac{n-2}{n}\right)^3$. The transferable move is the gap-removing substitution, which works for any number of variables and any gap size.

Sanity-check the method one dimension down. For two numbers in $[0, n]$, the shift trick predicts probability $\left(\frac{n-1}{n}\right)^2$; drawing the $n \times n$ square and removing the strip $|x - y| < 1$ (area $n^2 - (n-1)^2$) confirms it. The three-variable region is hard to draw, which is exactly why the ordering-plus-shift argument is preferable to direct volume computation.

• Treating the three pairwise conditions as independent, or extrapolating the two-variable answer ((n-1)/n)^2, which gives a formula with n-1 instead of n-2.
• Dividing only the numerator by $6$ (giving $\frac{(n-2)^3}{6n^3}$) or only the denominator, instead of noting the ordering factor cancels.
• Stopping at $n = 9$ because $\left(\frac79\right)^3 \approx 0.47$ "looks close," or forgetting the problem asks for strictly greater than $\frac12$.