Difference between revisions of "2014 AMC 12A Problems/Problem 17"

Revision as of 19:42, 11 April 2020

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)

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, $BC=\sqrt{3^2-\sqrt2^2}=\sqrt{7}=EF$ . $DE$ 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)

@@ Line 48: / Line 48: @@
 ==Solution 3==
-Use the 4 bottom spheres. (In progress)
+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==
 {{AMC12 box|year=2014|ab=A|num-b=16|num-a=18}}
 {{MAA Notice}}

2014 AMC 12A (Problems • Answer Key • Resources)
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