# 1985 AHSME Problems/Problem 25

## Problem

The volume of a certain rectangular solid is $8$ cm3, its total surface area is $32$ cm2, and its three dimensions are in geometric progression. The sums of the lengths in cm of all the edges of this solid is

$\mathrm{(A)\ } 28 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 40 \qquad \mathrm{(E) \ }44$

## Solution 1

As the dimensions are in geometric progression, let them be $\frac{b}{r}$, $b$, and $br$ cm, so the volume is $\left(\frac{b}{r}\right)(b)(br) = b^3$, giving $b^3 = 8$ and thus $b = 2$. The surface area condition now yields \begin{align*}2\left(\frac{2}{r}\right)(2)+2(2)(2r)+2(2r)\left(\frac{2}{r}\right) = 32 &\iff \frac{8}{r}+8+8r = 32 \\ &\iff r+\frac{1}{r} = 3 \\ &\iff r^2-3r+1 = 0 \\ &\iff r = \frac{3 \pm \sqrt{5}}{2}.\end{align*}

Since $$\frac{3-\sqrt{5}}{2} = \frac{1}{\left(\frac{3+\sqrt{5}}{2}\right)},$$ the two possible values of $r$ do not actually give different dimensions, but merely determine whether they are in increasing or decreasing order. Therefore, without loss of generality, we take $r = \frac{3+\sqrt{5}}{2}$, giving the dimensions (in cm) as \begin{align*}&\frac{2}{\left(\frac{3+\sqrt{5}}{2}\right)}, 2, \text{ and } 2\left(\frac{3+\sqrt{5}}{2}\right) \\ &= \frac{4}{3+\sqrt{5}}, 2, \text{ and } 3+\sqrt{5} \\ &= 3-\sqrt{5}, 2, \text{ and } 3+\sqrt{5}.\end{align*}

As there are $4$ edges with each of these distinct lengths, it follows that the sum of all the edge lengths (in cm) is \begin{align*}4\left(3-\sqrt{5}+2+3+\sqrt{5}\right) &= 4(8) \\ &= \boxed{\text{(B)} \ 32}.\end{align*}

## Solution 2

Similarly to in Solution 1, we let the dimensions (in cm) be $b$, $br$, and $br^2$, so that the volume condition gives $8 = b^3r^3 = (br)^3$ and thus $br = 2$. The surface area condition now becomes \begin{align*}2(b)(br)+2(br)\left(br^2\right)+2\left(br^2\right)(b) = 32 &\iff b^2r+b^2r^2+b^2r^3 = 16 \\&\iff br\left(b+br+br^2\right) = 16,\end{align*} so substituting $br = 2$ from above immediately gives $$b+br+br^2 = \frac{16}{2} = 8,$$ and hence, without needing to actually compute the dimensions, we deduce that the sum of the edge lengths (in cm) is $$4b + 4br + 4br^2 = 4(8) = \boxed{\text{(B)} \ 32}.$$

 1985 AHSME (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byProblem 26 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