2005 AMC 12B Problems/Problem 16


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?

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


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 $\boxed{\mathrm{D}}$.

Solution 2

Simplify the problem - in $2$ dimensions, place $4$ circles in each quadrant in the coordinate plane. To find the circle enclosing these $4$ circles, notice that if you connect the $4$ centers as a square, the diameter of the large circle will be the diagonal of the square plus two small radii.

Similarly, in $3$ dimensions, consider the cube created by connecting the $8$ centers of the spheres. The diameter of the large spheres will just be the space diagonal of this cube, plus the two radiuses on either side. The side length of the cube is $2$, so the space diagonal will be $2\sqrt{3}$. Then the diameter of the large sphere is $2\sqrt{3} + 2$ and the radius is $\sqrt{3} + 1$ or $\boxed{\mathrm{D}}$.


Video Solution by OmegaLearn


~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png