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

Revision as of 13:08, 5 July 2013

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: $4x + 4y + 4z = 140 \ \Longrightarrow \ x + y + z = 35.$ The spacial diagonal (longest distance) is given by $\sqrt{x^2 + y^2 + z^2}$ . Thus, we have $\sqrt{x^2 + y^2 + z^2} = 21$ , so $x^2 + y^2 + z^2 = 21^2$ .

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:

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

Substituting in our long diagonal and surface area expressions, we get: $21^2 + S = 35^2$ , so $S = (35 + 21)(35 - 21) = 56\cdot 14 = 784$ , which is option $\boxed{(\text{B})}$ .

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 [[Category:Introductory Algebra Problems]]
 [[Category:Algebraic Manipulations Problems]]
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Difference between revisions of "1996 AHSME Problems/Problem 23"

Revision as of 13:08, 5 July 2013

Problem

Solution

See also