Difference between revisions of "2005 AMC 12B Problems/Problem 16"

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== See also ==
 
== See also ==
* [[2005 AMC 12B Problems]]
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{{AMC12 box|year=2005|ab=B|num-b=15|num-a=17}}

Revision as of 14:51, 4 July 2011

Problem

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?

$\mathrm (A)\ \sqrt{2}  \qquad \mathrm (B)\ \sqrt{3}  \qquad \mathrm (C)\ 1+\sqrt{2}\qquad \mathrm (D)\ 1+\sqrt{3}\qquad \mathrm (E)\ 3$

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 $\boxed{D}$.

See also

2005 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions