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Difference between revisions of "1996 AHSME Problems/Problem 23"

(→‎Problem 23)
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<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>
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 +
==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>
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 +
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>
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<math>x^2 + y^2 + z^2 = 21^2</math>
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 +
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>
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 +
Substituting in our long diagonal and surface area expressions, we get:
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 +
<math>21^2 + S = 35^2</math>
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<math>S = (35 + 21)(35 - 21)</math>
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<math>S = 56\cdot 14</math>
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<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 $140$, and the distance from one corner of the box to the farthest corner is $21$. The total surface area of the box is

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

Solution

Let $x$, $y$, and $z$ be the unique lengths of the edges of the box. Each box has $4$ edges of each length, so:

$4x + 4y + 4z = 140$

$x + y + z = 35$

The spacial diagonal (longest distance) is given by $\sqrt{x^2 + y^2 + z^2}$. Thus, we have:

$\sqrt{x^2 + y^2 + z^2} = 21$

$x^2 + y^2 + z^2 = 21^2$

Our target expression is the surface area of the box:

$S = 2xy + 2xz + 2yz$

Since $S$ is a symmetric polynomial of degree $2$, we try squaring the first equation to get:

$(x + y + z)^2 = 35^2$

$x^2 + y^2 + z^2 + 2xy +2yz + 2xz = 35^2$

Substituting in our long diagonal and surface area expressions, we get:

$21^2 + S = 35^2$

$S = (35 + 21)(35 - 21)$

$S = 56\cdot 14$

$S = 784$, which is option $\boxed{B}$

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
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