Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
m (→‎Solution 1)
 
(8 intermediate revisions by 5 users not shown)
Line 28: Line 28:
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
−
==Solution==
+
==Solution 1==
  
 
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 center 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>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 center 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>.
+
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-\left(\sqrt2\right)^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)
+
~AwesomeToad
  
 
==Solution 2==
 
==Solution 2==
Line 42: Line 42:
  
  
−
<math>AC</math> is the radii of the spheres, so <math>AC=2+1=3</math>. <math>CE</math> is the shortest length between the center of a small sphere and the edge of the prism, so <math>CE=\sqrt{2}</math>. Similarly, <math>AF=2\sqrt{2}</math>. Since <math>CEFB</math> is a rectangle, <math>CE=BF=2\sqrt{2}</math>. Since <math>AF=2\sqrt{2}</math>, <math>AB=\sqrt{2}</math>. Then, <math>BC=\sqrt{3^2-\sqrt2^2}=\sqrt{7}=EF</math>. <math>DE</math> is the length from <math>C</math> to the top of the prism or <math>1</math>. Thus, <math>DF=DE+EF=1+\sqrt{7}</math>. The prism is symmetrical, so <math>h=2DF=\boxed{\textbf{(A)}}</math>
+
<math>AC</math> is the radii of the spheres, so <math>AC=2+1=3</math>. <math>CE</math> is the shortest length between the center of a small sphere and the edge of the prism, so <math>CE=\sqrt{2}</math>. Similarly, <math>AF=2\sqrt{2}</math>. Since <math>CEFB</math> is a rectangle, <math>BF=CE=\sqrt{2}</math>. Since <math>AF=2\sqrt{2}</math>, <math>AB=AF - BF = \sqrt{2}</math>. Then, <math>BC=\sqrt{3^2-\left(\sqrt2\right)^2}=\sqrt{7}=EF</math>. <math>DE</math> is the length from <math>C</math> to the top of the prism or <math>1</math>. Thus, <math>DF=DE+EF=1+\sqrt{7}</math>. The prism is symmetrical, so <math>h=2DF=\boxed{\textbf{(A)}}</math>
  
−
(Solution by BJHHar)
+
~BJHHar
  
 
==Solution 3==
 
==Solution 3==
  
−
Use the 4 bottom spheres. (In progress)
+
Take a cross section and you will see that h is made up of the two radii of the circles plus some radical expression. The only choice satisfying this condition is <math>\boxed{\textbf{(A)}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2014|ab=A|num-b=16|num-a=18}}
 
{{AMC12 box|year=2014|ab=A|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:42, 13 June 2022

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 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 $\boxed{\textbf{(A)}}$.

~AwesomeToad

Solution 2

caption

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 $h=2DF=\boxed{\textbf{(A)}}$

~BJHHar

Solution 3

Take a cross section and you will see that h is made up of the two radii of the circles plus some radical expression. The only choice satisfying this condition is $\boxed{\textbf{(A)}}$.

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12A_Problems/Problem_17&oldid=174783"