# 1985 AHSME Problems/Problem 25

## Problem

The volume of a certain rectangular solid is $8 \text{cm}^3$, its total surface area is $32 \text{cm}^2$, 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

Let the side lengths be $\frac{b}{r}, b, br$. Thus, the volume is $\left(\frac{b}{r}\right)(b)(br)=b^3=8$, so $b=2$ and the side lengths can be written as $\frac{2}{r}, 2, 2r$.

The surface area is $2\left(\frac{2}{r}\right)(2)+2\left(\frac{2}{r}\right)(2r)+2(2)(2r)=32$ $\frac{8}{r}+8+8r=32$ $r+\frac{1}{r}=3$ $r^2-3r+1=0$ $r=\frac{3\pm\sqrt{5}}{2}$

Both values of $r$ give the same side length, the only difference is that one makes them count up and one makes them count down. We pick $r=\frac{3+\sqrt{5}}{2}$. (The solution proceeds the same had we picked $r=\frac{3-\sqrt{5}}{2}$). Thus, the side lengths are $\frac{2}{\frac{3+\sqrt{5}}{2}}, 2, 2\left(\frac{3+\sqrt{5}}{2}\right)$ $\frac{4}{3+\sqrt{5}}, 2, 3+\sqrt{5}$ $3-\sqrt{5}, 2, 3+\sqrt{5}$

We have the sum of the distinct side lengths is $3-\sqrt{5}+2+3+\sqrt{5}=8$, and since each side length repeats $4$ times, the total sum is $4(8)=32, \boxed{\text{B}}$.

### Solution 2

We see let the side lengths be $b$, $rb$, and $r^2b$ because they form an arithmetic progression. Therefore, we have that $8 = r^3b^3 = (rb)^3$. Therefore, $rb = 2$. The next piece of information tells us that $$2rb^2 + 2r^2b^2 + 2r^3b^2 = 32.$$ Dividing both sides by $2rb = 4$, which is clearly nonzero, as it is a side length, we see that $b + rb + r^2b = 8$. For finding the sum of the side lengths, we will need to obtain $4b + 4rb + 4r^2b$ though, and so the answer is $4 \times 8 = 32$, $\boxed{B}$.

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