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

Revision as of 21:39, 26 August 2019

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 sphere of radius $2$ and $C$ be the center of any of the spheres. 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 the small sphere and the edge of the prism, so $CE=\sqrt{2}$ . Similarly, $AF=2\sqrt{2}$ . Since $CEFB$ is a rectangle, $CE=BF=2\sqrt{2}$ . Since $AF=2\sqrt{2}$ , $AB=\sqrt{2}$ . 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, $DF=DE+EF=1+\sqrt{7}$ . Since the prism is symmetrical, $h=2DF=\boxed{\textbf{(A)}}$

(Solution by BJHHar)

Solution 3

Use the 4 bottom spheres. (In progress)

@@ Line 38: / Line 38: @@
 ==Solution 2==
-[[File:2014 AMC 12A Problem 17.JPG|none|250px|caption]]
+[[File:2014 AMC 12A Problem 17.JPG|none|500px|caption]]
 Let <math>A</math> be the center of the sphere of radius <math>2</math> and <math>C</math> be the center of any of the spheres. Let <math>D</math> be a vertex of the rectangular prism closest to point <math>C</math>. Let <math>F</math> be the point on the edge of the prism such that <math>\overline{DF}</math> and <math>\overline{AF}</math> are perpendicular. Let points <math>B</math> and point <math>E</math> lie on <math>\overline{AF}</math> and <math>\overline{DF}</math> respectively such that <math>\overline{CE}</math> and <math>\overline{CB}</math> are perpendicular at <math>C</math>.

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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