1985 AMC 12 Problem 25

Attempt Problem 25 of the 1985 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1985 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

25.

The volume of a certain rectangular solid is 8 cm3,8\text{ cm}^3, its total surface area is 32 cm2,32\text{ cm}^2, and its three dimensions are in geometric progression. The sum of the lengths in cm of all the edges of this solid is
2828
3232
3636
4040
4444
Answer: B
Concepts:geometric sequencerectangular prismsurface areavolume
Small hint:
Write the three dimensions as tr,t,tr\frac{t}{r},t,tr
Big hint:
Use the volume to find t,t, then use the surface area to find r+1rr+\frac{1}{r}
Solution:
Write the dimensions as 2r,2,2r,\frac{2}{r},2,2r, since their product is 8.8. Half the surface area is the sum of the three pairwise products, so 4r+4+4r=16, \frac4r+4+4r=16, giving r+1r=3.r+\frac{1}{r}=3. The sum of the dimensions is 2(r+1+1r)=8.2(r+1+\frac{1}{r})=8. Since each dimension occurs on four edges, the sum of all edge lengths is 4⋅8=32.4\cdot8=32.
Thus the correct answer is B.
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