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Difference between revisions of "1996 AHSME Problems/Problem 23"
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Latest revision as of 15:24, 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

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

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 $\boxed{(\text{B})}$.

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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