Difference between revisions of "1985 AHSME Problems/Problem 25"
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<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== |
Let the side lengths be <math> \frac{b}{r}, b, br </math>. Thus, the volume is <math> \left(\frac{b}{r}\right)(b)(br)=b^3=8 </math>, so <math> b=2 </math> and the side lengths can be written as <math> \frac{2}{r}, 2, 2r </math>. | Let the side lengths be <math> \frac{b}{r}, b, br </math>. Thus, the volume is <math> \left(\frac{b}{r}\right)(b)(br)=b^3=8 </math>, so <math> b=2 </math> and the side lengths can be written as <math> \frac{2}{r}, 2, 2r </math>. | ||
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We have the sum of the distinct side lengths is <math> 3-\sqrt{5}+2+3+\sqrt{5}=8 </math>, and since each side length repeats <math> 4 </math> times, the total sum is <math> 4(8)=32, \boxed{\text{B}} </math>. | We have the sum of the distinct side lengths is <math> 3-\sqrt{5}+2+3+\sqrt{5}=8 </math>, and since each side length repeats <math> 4 </math> times, the total sum is <math> 4(8)=32, \boxed{\text{B}} </math>. | ||
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+ | ===Solution 2=== | ||
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+ | We see let the side lengths be <math>b</math>, <math>rb</math>, and <math>r^2b</math> because they form an arithmetic progression. Therefore, we have that <math>8 = r^3b^3 = (rb)^3</math>. Therefore, <math>rb = 2</math>. The next piece of information tells us that <cmath>2rb^2 + 2r^2b^2 + 2r^3b^2 = 32.</cmath> Dividing both sides by <math>2rb = 4</math>, which is clearly nonzero, as it is a side length, we see that <math>b + rb + r^2b = 8</math>. For finding the sum of the side lengths, we will need to obtain <math>4b + 4rb + 4r^2b</math> though, and so the answer is <math>4 \times 8 = 32</math>, <math>\boxed{B}</math>. | ||
==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}} | {{MAA Notice}} |
Revision as of 10:24, 3 October 2017
Contents
[hide]Problem
The volume of a certain rectangular solid is , its total surface area is , 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
Let the side lengths be . Thus, the volume is , so and the side lengths can be written as .
The surface area is
Both values of give the same side length, the only difference is that one makes them count up and one makes them count down. We pick . (The solution proceeds the same had we picked ). Thus, the side lengths are
We have the sum of the distinct side lengths is , and since each side length repeats times, the total sum is .
Solution 2
We see let the side lengths be , , and because they form an arithmetic progression. Therefore, we have that . Therefore, . The next piece of information tells us that Dividing both sides by , which is clearly nonzero, as it is a side length, we see that . For finding the sum of the side lengths, we will need to obtain though, and so the answer 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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