2014 AMC 12A Problems/Problem 17

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Problem

A $4\times 4\times h$ rectangular box contains a sphere of radius $2$ and eight smaller spheres of radius $1$. The smaller spheres are each tangent to three sides of the box, and the larger sphere is tangent to each of the smaller spheres. What is $h$?

[asy] import graph3; import solids; real h=2+2*sqrt(7); currentprojection=orthographic((0.75,-5,h/2+1),target=(2,2,h/2)); currentlight=light(4,-4,4); draw((0,0,0)--(4,0,0)--(4,4,0)--(0,4,0)--(0,0,0)^^(4,0,0)--(4,0,h)--(4,4,h)--(0,4,h)--(0,4,0)); draw(shift((1,3,1))*unitsphere,gray(0.85)); draw(shift((3,3,1))*unitsphere,gray(0.85)); draw(shift((3,1,1))*unitsphere,gray(0.85)); draw(shift((1,1,1))*unitsphere,gray(0.85)); draw(shift((2,2,h/2))*scale(2,2,2)*unitsphere,gray(0.85)); draw(shift((1,3,h-1))*unitsphere,gray(0.85)); draw(shift((3,3,h-1))*unitsphere,gray(0.85)); draw(shift((3,1,h-1))*unitsphere,gray(0.85)); draw(shift((1,1,h-1))*unitsphere,gray(0.85)); draw((0,0,0)--(0,0,h)--(4,0,h)^^(0,0,h)--(0,4,h)); [/asy]

$\textbf{(A) }2+2\sqrt 7\qquad \textbf{(B) }3+2\sqrt 5\qquad \textbf{(C) }4+2\sqrt 7\qquad \textbf{(D) }4\sqrt 5\qquad \textbf{(E) }4\sqrt 7\qquad$

Solution

Let $A$ be the point in the same plane as the centers of the top spheres equidistant from said centers. Let $B$ be the analogous point for the bottom spheres, and let $C$ be the midpoint of $\overline{AB}$ and the center of the large sphere. Let $D$ and $E$ be the points at which line $AB$ intersects the top of the box and the bottom, respectively.

Let $O$ be the center of any of the top spheres (you choose!). We have $AO=1\cdot\sqrt{2}$, and $CO=3$, so $AC=\sqrt{3^2-\sqrt2^2}=\sqrt{7}$. Similarly, $BC=\sqrt{7}$. $\overline{AD}$ and $\overline{BE}$ are clearly equal to the radius of the small spheres, $1$. Thus the total height is $AD+AC+BC+BE=2+2\sqrt7$, or $\boxed{\textbf{(A)}}$.

(Solution by AwesomeToad)

See Also

2014 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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

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