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

(Solution)
(Solution)
Line 13: Line 13:
 
== Solution ==
 
== Solution ==
  
The eight spheres are formed by shifting spheres of radius <math>1</math> and center <math>(0, 0, 0)</math> <math>\pm 1</math> in the <math>x, y, z</math> directions. Hence, the centers of the spheres are <math>(\pm 1, \pm 1, \pm 1)</math>. For a sphere 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 <math>(\pm 1, \pm 1, \pm 1)</math> to the origin and the radius of the sphere, or <math>\sqrt{3} + 1</math>. To verify this is the longest length from the origin to a point on any of the spheres, we can see from the triangle inequality that the length from the origin to any other point on the sphere is strictly smaller.
+
The eight spheres are formed by shifting spheres of radius <math>1</math> and center <math>(0, 0, 0)</math> <math>\pm 1</math> in the <math>x, y, z</math> directions. Hence, the centers of the spheres are <math>(\pm 1, \pm 1, \pm 1)</math>. For a sphere 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 <math>(\pm 1, \pm 1, \pm 1)</math> to the origin and the radius of the sphere, or <math>\sqrt{3} + 1</math>. 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 sphere is strictly smaller.
  
 
== See also ==
 
== See also ==
 
* [[2005 AMC 12B Problems]]
 
* [[2005 AMC 12B Problems]]

Revision as of 18:32, 12 September 2010

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 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 sphere, 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 sphere is strictly smaller.

See also