Difference between revisions of "2014 AMC 12A Problems/Problem 17"
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[[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 | + | ~AwesomeToad |
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | [[File:2014 AMC 12A Problem 17.JPG|none|500px|caption]] | ||
+ | Let <math>A</math> be the center of the large sphere and <math>C</math> be the center of any small sphere. 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>. | ||
+ | |||
+ | |||
+ | <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> | ||
+ | |||
+ | ~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 <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 02:42, 13 June 2022
Problem
A rectangular box contains a sphere of radius
and eight smaller spheres of radius
. 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
?
![[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]](http://latex.artofproblemsolving.com/9/d/8/9d894dbf42883508d51bf4973c78d1368570c6fa.png)
Solution 1
Let be the point in the same plane as the centers of the top spheres equidistant from said centers. Let
be the analogous point for the bottom spheres, and let
be the midpoint of
and the center of the large sphere. Let
and
be the points at which line
intersects the top of the box and the bottom, respectively.
Let be the center of any of the top spheres (you choose!). We have
, and
, so
. Similarly,
.
and
are clearly equal to the radius of the small spheres,
. Thus the total height is
, or
.
~AwesomeToad
Solution 2
Let be the center of the large sphere and
be the center of any small sphere. Let
be a vertex of the rectangular prism closest to point
. Let
be the point on the edge of the prism such that
and
are perpendicular. Let points
and point
lie on
and
respectively such that
and
are perpendicular at
.
is the radii of the spheres, so
.
is the shortest length between the center of a small sphere and the edge of the prism, so
. Similarly,
. Since
is a rectangle,
. Since
,
. Then,
.
is the length from
to the top of the prism or
. Thus,
. The prism is symmetrical, so
~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 .
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.