Difference between revisions of "1996 AHSME Problems/Problem 23"
Talkinaway (talk | contribs) (→Problem 23) |
Talkinaway (talk | contribs) |
||
Line 6: | Line 6: | ||
<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</math>, <math>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: | ||
+ | |||
+ | <math>4x + 4y + 4z = 140</math> | ||
+ | |||
+ | <math>x + y + z = 35</math> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | <math>x^2 + y^2 + z^2 = 21^2</math> | ||
+ | |||
+ | Our target expression is the surface area of the box: | ||
+ | |||
+ | <math>S = 2xy + 2xz + 2yz</math> | ||
+ | |||
+ | Since <math>S</math> is a symmetric polynomial of degree <math>2</math>, we try squaring the first equation to get: | ||
+ | |||
+ | <math>(x + y + z)^2 = 35^2</math> | ||
+ | |||
+ | <math>x^2 + y^2 + z^2 + 2xy +2yz + 2xz = 35^2</math> | ||
+ | |||
+ | Substituting in our long diagonal and surface area expressions, we get: | ||
+ | |||
+ | <math>21^2 + S = 35^2</math> | ||
+ | |||
+ | <math>S = (35 + 21)(35 - 21)</math> | ||
+ | |||
+ | <math>S = 56\cdot 14</math> | ||
+ | |||
+ | <math>S = 784</math>, which is option <math>\boxed{B}</math> | ||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=22|num-a=24}} | {{AHSME box|year=1996|num-b=22|num-a=24}} |
Revision as of 14:18, 20 August 2011
Problem
The sum of the lengths of the twelve edges of a rectangular box is , and the distance from one corner of the box to the farthest corner is . The total surface area of the box is
Solution
Let , , and be the unique lengths of the edges of the box. Each box has edges of each length, so:
The spacial diagonal (longest distance) is given by . Thus, we have:
Our target expression is the surface area of the box:
Since is a symmetric polynomial of degree , we try squaring the first equation to get:
Substituting in our long diagonal and surface area expressions, we get:
, which is option
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 |