Difference between revisions of "1985 AHSME Problems/Problem 25"
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==Problem== | ==Problem== | ||
− | The | + | The volume of a certain rectangular solid is <math>8</math> cm<sup>3</sup>, its total surface area is <math>32</math> cm<sup>2</sup>, and its three dimensions are in geometric progression. The sums of the lengths in cm of all the edges of this solid is |
<math> \mathrm{(A)\ } 28 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 40 \qquad \mathrm{(E) \ }44 </math> | <math> \mathrm{(A)\ } 28 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 40 \qquad \mathrm{(E) \ }44 </math> | ||
− | ==Solution== | + | ==Solution 1== |
− | + | As the dimensions are in geometric progression, let them be <math>\frac{b}{r}</math>, <math>b</math>, and <math>br</math> cm, so the volume is <math>\left(\frac{b}{r}\right)(b)(br) = b^3</math>, giving <math>b^3 = 8</math> and thus <math>b = 2</math>. The surface area condition now yields <cmath>\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*}</cmath> | |
− | + | Since <cmath>\frac{3-\sqrt{5}}{2} = \frac{1}{\left(\frac{3+\sqrt{5}}{2}\right)},</cmath> the two possible values of <math>r</math> do not actually give different dimensions, but merely determine whether they are in increasing or decreasing order. Therefore, without loss of generality, we take <math>r = \frac{3+\sqrt{5}}{2}</math>, giving the dimensions (in cm) as <cmath>\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*}</cmath> | |
− | <math> \ | + | As there are <math>4</math> edges with each of these distinct lengths, it follows that the sum of all the edge lengths (in cm) is <cmath>\begin{align*}4\left(3-\sqrt{5}+2+3+\sqrt{5}\right) &= 4(8) \ &= \boxed{\text{(B)} \ 32}.\end{align*}</cmath> |
− | + | ==Solution 2== | |
− | <math> | + | Similarly to in Solution 1, we let the dimensions (in cm) be <math>b</math>, <math>br</math>, and <math>br^2</math>, so that the volume condition gives <math>8 = b^3r^3 = (br)^3</math> and thus <math>br = 2</math>. The surface area condition now becomes <cmath>\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*}</cmath> so substituting <math>br = 2</math> from above immediately gives <cmath>b+br+br^2 = \frac{16}{2} = 8,</cmath> and hence, without needing to actually compute the dimensions, we deduce that the sum of the edge lengths (in cm) is <cmath>4b + 4br + 4br^2 = 4(8) = \boxed{\text{(B)} \ 32}.</cmath> |
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==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=24|num-a=26}} | {{AHSME box|year=1985|num-b=24|num-a=26}} | ||
+ | {{MAA Notice}} |
Latest revision as of 00:37, 20 March 2024
Contents
[hide]Problem
The volume of a certain rectangular solid is cm3, its total surface area is cm2, and its three dimensions are in geometric progression. The sums of the lengths in cm of all the edges of this solid is
Solution 1
As the dimensions are in geometric progression, let them be , , and cm, so the volume is , giving and thus . The surface area condition now yields
Since the two possible values of do not actually give different dimensions, but merely determine whether they are in increasing or decreasing order. Therefore, without loss of generality, we take , giving the dimensions (in cm) as
As there are edges with each of these distinct lengths, it follows that the sum of all the edge lengths (in cm) is
Solution 2
Similarly to in Solution 1, we let the dimensions (in cm) be , , and , so that the volume condition gives and thus . The surface area condition now becomes so substituting from above immediately gives and hence, without needing to actually compute the dimensions, we deduce that the sum of the edge lengths (in cm) is
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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