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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>
  
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==Solution==
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==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

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 $4 \times 8 = 32$, $\boxed{B}$.

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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