Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(Improved solutions, wording, and formatting)
 
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
The [[volume]] of a certain rectangular solid is <math> 8 \text{cm}^3 </math>, its total [[surface area]] is <math> 32 \text{cm}^2 </math>, and its three dimensions are in [[geometric progression]]. The sums of the lengths in cm of all the edges of this solid is
+
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 1==
 
==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>.
+
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>
  
−
The surface area is <math> 2\left(\frac{2}{r}\right)(2)+2\left(\frac{2}{r}\right)(2r)+2(2)(2r)=32 </math>
+
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> \frac{8}{r}+8+8r=32 </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>
  
−
<math> r+\frac{1}{r}=3 </math>
+
==Solution 2==
  
−
<math> r^2-3r+1=0 </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>
−
 
 
−
<math> r=\frac{3\pm\sqrt{5}}{2} </math>
 
−
 
 
−
Both values of <math> r </math> give the same side length, the only difference is that one makes them count up and one makes them count down. We pick <math> r=\frac{3+\sqrt{5}}{2} </math>. (The solution proceeds the same had we picked <math> r=\frac{3-\sqrt{5}}{2} </math>). Thus, the side lengths are
 
−
 
 
−
<math> \frac{2}{\frac{3+\sqrt{5}}{2}}, 2, 2\left(\frac{3+\sqrt{5}}{2}\right) </math>
 
−
 
 
−
<math> \frac{4}{3+\sqrt{5}}, 2, 3+\sqrt{5} </math>
 
−
 
 
−
<math> 3-\sqrt{5}, 2, 3+\sqrt{5} </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>.
 
−
 
 
−
===Solution 2===
 
−
 
 
−
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}}

Latest revision as of 00:37, 20 March 2024

Problem

The volume of a certain rectangular solid is $8$ cm3, its total surface area is $32$ cm2, 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

As the dimensions are in geometric progression, let them be $\frac{b}{r}$, $b$, and $br$ cm, so the volume is $\left(\frac{b}{r}\right)(b)(br) = b^3$, giving $b^3 = 8$ and thus $b = 2$. The surface area condition now yields \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*}

Since \[\frac{3-\sqrt{5}}{2} = \frac{1}{\left(\frac{3+\sqrt{5}}{2}\right)},\] the two possible values of $r$ do not actually give different dimensions, but merely determine whether they are in increasing or decreasing order. Therefore, without loss of generality, we take $r = \frac{3+\sqrt{5}}{2}$, giving the dimensions (in cm) as \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*}

As there are $4$ edges with each of these distinct lengths, it follows that the sum of all the edge lengths (in cm) is \begin{align*}4\left(3-\sqrt{5}+2+3+\sqrt{5}\right) &= 4(8) \\ &= \boxed{\text{(B)} \ 32}.\end{align*}

Solution 2

Similarly to in Solution 1, we let the dimensions (in cm) be $b$, $br$, and $br^2$, so that the volume condition gives $8 = b^3r^3 = (br)^3$ and thus $br = 2$. The surface area condition now becomes \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*} so substituting $br = 2$ from above immediately gives \[b+br+br^2 = \frac{16}{2} = 8,\] and hence, without needing to actually compute the dimensions, we deduce that the sum of the edge lengths (in cm) is \[4b + 4br + 4br^2 = 4(8) = \boxed{\text{(B)} \ 32}.\]

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1985_AHSME_Problems/Problem_25&oldid=217174"