2014 AMC 12A Problems/Problem 17

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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 1

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 $AC=\sqrt{3^2-\left(\sqrt2\right)^2}=\sqrt{7}$. 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)

Solution 2


Let $A$ be the center of the large sphere and $C$ be the center of any small sphere. Let $D$ be a vertex of the rectangular prism closest to point $C$. Let $F$ be the point on the edge of the prism such that $\overline{DF}$ and $\overline{AF}$ are perpendicular. Let points $B$ and point $E$ lie on $\overline{AF}$ and $\overline{DF}$ respectively such that $\overline{CE}$ and $\overline{CB}$ are perpendicular at $C$.

$AC$ is the radii of the spheres, so $AC=2+1=3$. $CE$ is the shortest length between the center of a small sphere and the edge of the prism, so $CE=\sqrt{2}$. Similarly, $AF=2\sqrt{2}$. Since $CEFB$ is a rectangle, $BF=CE=\sqrt{2}$. Since $AF=2\sqrt{2}$, $AB=AF - BF = \sqrt{2}$. Then, $BC=\sqrt{3^2-\left(\sqrt2\right)^2}=\sqrt{7}=EF$. $DE$ is the length from $C$ to the top of the prism or $1$. Thus, $DF=DE+EF=1+\sqrt{7}$. The prism is symmetrical, so $h=2DF=\boxed{\textbf{(A)}}$

(Solution by BJHHar)

Solution 3

take a cross section and see that h is made up of two radii of the circle plus some radical expression. the only choice satisfying this condition is (a)

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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