AMC12 2005 B

AMC12 2005 B · Q16

AMC12 2005 B · Q16. It mainly tests Distance / midpoint, 3D geometry (volume).

Eight spheres of radius 1, one per octant, are each tangent to the coordinate planes. What is the radius of the smallest sphere, centered at the origin, that contains these eight spheres? $\mathrm {

有8个半径为$1$的球，每个球位于一个八分体中，且每个球都与坐标平面相切。求以原点为中心的最小球的半径，该球包含这8个球。 $\mathrm {

(A) \sqrt{2} \sqrt{2}

(B) \sqrt{3} \sqrt{3}

(D) 1 + \sqrt{3} 1 + \sqrt{3}

(E) 3 3

Answer

Correct choice: (D)

正确答案：（D）

Solution

The eight spheres are formed by shifting spheres of radius $1$ and center $(0, 0, 0)$ $\pm 1$ in the $x, y, z$ directions. Hence, the centers of the spheres are $(\pm 1, \pm 1, \pm 1)$. For a sphere centered at the origin to contain all eight spheres, its radius must be greater than or equal to the longest distance from the origin to one of these spheres. This length is the sum of the distance from $(\pm 1, \pm 1, \pm 1)$ to the origin and the radius of the spheres, or $\sqrt{3} + 1$. To verify this is the longest length, we can see from the triangle inequality that the length from the origin to any other point on the spheres is strictly smaller. Thus, the answer is $\boxed{\mathrm{D}}$.

这8个球可看作将半径为$1$、球心在$(0, 0, 0)$的球分别在$x,y,z$方向各平移$\pm 1$得到。因此，这些球的球心为$(\pm 1, \pm 1, \pm 1)$。要使以原点为中心的球包含这8个球，其半径必须大于等于从原点到这些球中最远处的距离。该距离等于从原点到某个球心$(\pm 1, \pm 1, \pm 1)$的距离再加上小球半径，即$\sqrt{3} + 1$。为验证这是最大值，可由三角不等式看出，从原点到球面上任意其他点的距离严格更小。因此答案为$\boxed{\mathrm{D}}$。

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