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

(Created page with "==Problem== A <math>4\times 4\times h</math> rectangular box contains a sphere of radius <math>2</math> and eight smaller spheres of radius <math>1</math>. The smaller spheres a...")
 
(Problem)
Line 26: Line 26:
 
\textbf{(D) }4\sqrt 5\qquad
 
\textbf{(D) }4\sqrt 5\qquad
 
\textbf{(E) }4\sqrt 7\qquad</math>
 
\textbf{(E) }4\sqrt 7\qquad</math>
 +
 +
==Solution==
 +
 +
Let <math>A</math> be the point in the same plane as the centers of the top spheres equidistant from said centers. Let <math>B</math> be the analogous point for the bottom spheres, and let <math>C</math> be the midpoint of <math>\overline{AB}</math> and the midpoint of the large sphere. Let <math>D</math> and <math>E</math> be the points at which line <math>AB</math> intersects the top of the box and the bottom, respectively.
 +
 +
Let <math>O</math> be the center of any of the top spheres (you choose!). We have <math>AO=1\cdot\sqrt{2}</math>, and <math>CO=3</math>, so <math>AC=\sqrt{3^2-\sqrt2^2}=\sqrt{7}</math>. Similarly, <math>BC=\sqrt{7}</math>. <math>\overline{AD}</math> and <math>\overline{BE}</math> are clearly equal to the radius of the small spheres, <math>1</math>. Thus the total height is <math>AD+AC+BC+BE=2+2\sqrt7</math>, or <math>\boxed{\textbf{(A)}}</math>.
 +
 +
(Solution by AwesomeToad)

Revision as of 20:41, 7 February 2014

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 three; import solids; real h=2+2*sqrt(7); currentprojection=orthographic((0.75,-5,h/2+1),target=(2,2,h/2)); currentlight=light(1,0,3); 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,black); draw(shift((3,3,1))*unitsphere,black); draw(shift((3,1,1))*unitsphere,black); draw(shift((1,1,1))*unitsphere,black); draw(shift((2,2,h/2))*scale(2,2,2)*unitsphere,black); draw(shift((1,3,h-1))*unitsphere,black); draw(shift((3,3,h-1))*unitsphere,black); draw(shift((3,1,h-1))*unitsphere,black); draw(shift((1,1,h-1))*unitsphere,black); 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 midpoint 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)