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Difference between revisions of "1985 AHSME Problems/Problem 25"

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Revision as of 13:01, 5 July 2013

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

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 $4(8)=32, \boxed{\text{B}}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 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

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