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 "2005 AMC 12A Problems/Problem 22"

m (→‎Solution 2)
(→‎Problem)
 
(3 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
A rectangular box <math> P </math> is [[inscribe]]d in a [[sphere]] of [[radius]] <math>r</math>. The [[surface area]] of <math>P</math> is 384, and the sum of the lengths of its 12 edges is 112. What is <math>r</math>?
+
A rectangular box <math> P </math> is inscribed in a sphere of radius <math>r</math>. The surface area of <math>P</math> is 384, and the sum of the lengths of its 12 edges is 112. What is <math>r</math>?
  
 
<math>\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16</math>
 
<math>\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16</math>
Line 7: Line 7:
  
 
Box P has dimensions <math>l</math>, <math>w</math>, and <math>h</math>.  
 
Box P has dimensions <math>l</math>, <math>w</math>, and <math>h</math>.  
−
Its surface area is <cmath>2lw+2lh+2wh=384,</cmath> *I fixed a typo here*
+
Its surface area is <cmath>2lw+2lh+2wh=384,</cmath>  
 
and the sum of all its edges is <cmath>l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.</cmath>
 
and the sum of all its edges is <cmath>l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.</cmath>
  

Latest revision as of 09:20, 16 September 2022

Problem

A rectangular box $P$ is inscribed in a sphere of radius $r$. The surface area of $P$ is 384, and the sum of the lengths of its 12 edges is 112. What is $r$?

$\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16$

Solution

Box P has dimensions $l$, $w$, and $h$. Its surface area is \[2lw+2lh+2wh=384,\] and the sum of all its edges is \[l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.\]

The diameter of the sphere is the space diagonal of the prism, which is \[\sqrt{l^2 + w^2 +h^2}.\] Notice that \[(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,\] so the diameter is \[\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.\] The radius is half of the diameter, so \[r=\frac{20}{2} = \boxed{\textbf{(B)} 10}.\]

Solution 2

As in the previous solution, we have that $2lw+2lh+2wh=384$ and $l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28$, and the diameter of the sphere is the space diagonal of the prism, $\sqrt{l^2 + w^2 + h^2}$.


Now, since this is competition math, we only need to find the space diagonal of any one box that fits the requirements, so assume that $h=0$. (This essentially means that we have an infinitesimally thin box.) We now have that $2lw = 384$ and $l + w = 28$, and we are solving for $\sqrt{l^2 + w^2}$. Because \[(l + w)^2 - 2lw = l^2 + 2lw + w^2 - 2lw = l^2 + w^2,\] this means that \[l^2 + w^2 = 28^2 - 384 = 400,\] so the space diagonal is $\sqrt{400} = 20$. Since the diameter of the sphere is $20$, the radius is $\boxed{\textbf{(B) } 10}$. ~emerald_block

See also

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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=2005_AMC_12A_Problems/Problem_22&oldid=178250"