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 "1996 AHSME Problems/Problem 23"
(→‎Solution)
Line 5: Line 5:
  
 
<math> \text{(A)}\ 776\qquad\text{(B)}\ 784\qquad\text{(C)}\ 798\qquad\text{(D)}\ 800\qquad\text{(E)}\ 812 </math>
 
<math> \text{(A)}\ 776\qquad\text{(B)}\ 784\qquad\text{(C)}\ 798\qquad\text{(D)}\ 800\qquad\text{(E)}\ 812 </math>
−
 
−
==Solution==
 
−
Let <math>x, y</math>, and <math>z</math> be the unique lengths of the edges of the box.  Each box has <math>4</math> edges of each length, so:
 
−
<cmath>4x + 4y + 4z = 140 \ \Longrightarrow \ x + y + z = 35.</cmath>
 
−
The spacial diagonal (longest distance) is given by <math>\sqrt{x^2 + y^2 + z^2}</math>.  Thus, we have <math>\sqrt{x^2 + y^2 + z^2} = 21</math>, so <math>x^2 + y^2 + z^2 = 21^2</math>.
 
−
 
−
Our target expression is the surface area of the box:
 
−
 
−
<cmath>S = 2xy + 2xz + 2yz.</cmath>
 
−
 
−
Since <math>S</math> is a [[Elementary symmetric sum|symmetric polynomial]] of degree <math>2</math>, we try squaring the first equation to get:
 
−
 
−
<cmath>35^2 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy +2yz + 2xz = 35^2.</cmath>
 
−
 
−
Substituting in our long diagonal and surface area expressions, we get: <math>21^2 + S = 35^2</math>, so <math>S = (35 + 21)(35 - 21) = 56\cdot 14 = 784</math>, which is option <math>\boxed{(\text{B})}</math>.
 
  
 
==See also==
 
==See also==

Revision as of 08:44, 28 June 2021

Problem

The sum of the lengths of the twelve edges of a rectangular box is $140$, and the distance from one corner of the box to the farthest corner is $21$. The total surface area of the box is

$\text{(A)}\ 776\qquad\text{(B)}\ 784\qquad\text{(C)}\ 798\qquad\text{(D)}\ 800\qquad\text{(E)}\ 812$

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 • 26 • 27 • 28 • 29 • 30
All AHSME 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=1996_AHSME_Problems/Problem_23&oldid=156953"