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Difference between revisions of "2015 AMC 12B Problems/Problem 23"

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{{duplicate|[[2015 AMC 10B Problems#Problem 25|2015 AMC 10B #25]] and [[2015 AMC 12B Problems#Problem 23|2015 AMC 12B #23]]}}
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==Problem==
 
==Problem==
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A rectangular box measures <math>a \times b \times c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>1\leq a \leq b \leq c</math>. The volume and the surface area of the box are numerically equal. How many ordered triples <math>(a,b,c)</math> are possible?
 
A rectangular box measures <math>a \times b \times c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>1\leq a \leq b \leq c</math>. The volume and the surface area of the box are numerically equal. How many ordered triples <math>(a,b,c)</math> are possible?
  
 
<math>\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26</math>
 
<math>\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26</math>
  
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==Solution==
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==Solution 1==
  
 
We need <cmath>abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).</cmath>
 
We need <cmath>abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).</cmath>
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Since <math>ab, ac \le bc</math>, we get <math>abc \le 6bc</math>. Thus <math>a\le 6</math>. From the second equation we see that <math>a > 2</math>. Thus <math>a\in \{3, 4, 5, 6\}</math>.  
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Since <math>a\le b</math> and <math>a,b,c</math> are all positive<math>,ac \le bc</math>. From the first equation we get <math>abc \le 6bc</math>. Thus <math>a\le 6</math>. From the second equation we see that <math>a > 2</math>. Thus <math>a\in \{3, 4, 5, 6\}</math>.
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*If <math>a=3</math> we need <math>bc = 6(b+c) \Rightarrow (b-6)(c-6)=36</math>. We get '''five''' roots <math>\{(3, 7, 42), (3, 8, 24), (3,9,18), (3, 10, 15), (3,12,12)\}</math>.
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*If <math>a=4</math> we need <math>2bc = 8(b+c) \Rightarrow bc = 4(b+c) \Rightarrow (b-4)(c-4)=16</math>. We get '''three''' roots <math>\{(4,5,20), (4,6,12), (4,8,8)\}</math>.
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*If <math>a=5</math> we need <math>3bc = 10(b+c) \Rightarrow 9bc=30(b+c) \Rightarrow (3b-10)(3c-10)=100</math>. We get '''one''' root <math>\{(5,5,10)\}</math>.
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*If <math>a=6</math> we need <math>4bc = 12(b+c) \Rightarrow bc = 3(b+c) \Rightarrow (b-3)(c-3)=9</math>. We get ''' one''' root <math>\{(6,6,6)\}</math>.
  
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*If <math>a=3</math> we need <math>bc = 6(b+c) \Rightarrow (b-6)(c-6)=36</math>. We get '''five''' roots <math>\{(3, 7, 42), (3, 8, 24), (3,9,18), (3, 10, 15), (3,12,12)\}.</math>
 
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*If <math>a=4</math> we need <math>bc = 4(b+c) \Rightarrow (b-4)(c-4)=16</math>. We get '''three''' roots <math>\{(4,5,20), (4,6,12), (4,8,8)\}</math>.
 
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*If <math>a=5</math> we need <math>3bc = 10(b+c)</math>, which is the same as <math>9bc=30(b+c)\Rightarrow (3b-10)(3c-10)=100</math>. We get only '''one''' root (corresponding to <math>100=5\cdot 20</math>) <math>(5,5,10)</math>.
 
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*If <math>a=6</math> we need <math>4bc = 12(b+c)</math>. Then <math>(b-3)(c-3)=9</math>. We get ''' one''' root <math>(6,6,6)</math>.
 
 
Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions.
 
Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions.
  
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Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions.
 
Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions.
  
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==Note==
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==Simplification of Solution 2==
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This is also problem 25 of the AMC 10B Problem 25, but they don't connect. Someone should fix this.
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The surface area is <math>2(ab+bc+ca)</math>, the volume is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>.
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 +
Divide both sides by <math>2abc</math>, we have: <cmath>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.</cmath>
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First consider the bound of the variable <math>a</math>. Since <math>\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},</math> we have <math>a>2</math>, or <math>a\ge 3</math>.
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Also note that <math>c\ge b\ge a>0</math>, we have <math>\frac{1}{a}\ge \frac{1}{b}\ge \frac{1}{c}</math>. Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{3}{a}</math>, so <math>a\le 6</math>.
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So we have <math>a=3, 4, 5</math> or <math>6</math>.
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 +
 
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We can say <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{q}</math>, where <math>\frac{1}{q} = \frac{1}{2}-\frac{1}{a}</math>.
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Notice <math>\emph{\text{immediately}}</math> that <math>b, c > q</math>. This is our key step.
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Then we can say <math>b=q+d</math>, <math>c=q+e</math>. If we clear the fraction about b and c (do the math), our immediate result is that <math>de = q^2</math>. Realize also that <math>d \leq e</math>.
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Now go through cases for <math>a</math> and you end up with the same result. However, now you don't have to guess solutions. For example, when <math>a=3</math>, then <math>de = 36</math> and <math>d=1, 2, 3, 4, 6</math>.
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 +
 
 +
 
 +
 
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== Video Solution ==
 +
https://www.youtube.com/watch?v=JFUpe32aWnw&t=1941s
  
 
==See Also==
 
==See Also==
 +
{{AMC10 box|year=2015|ab=B|after=Last Question|num-b=24}}
 
{{AMC12 box|year=2015|ab=B|num-a=24|num-b=22}}
 
{{AMC12 box|year=2015|ab=B|num-a=24|num-b=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:42, 7 October 2024

The following problem is from both the 2015 AMC 10B #25 and 2015 AMC 12B #23, so both problems redirect to this page.

Problem

A rectangular box measures $a \times b \times c$, where $a$, $b$, and $c$ are integers and $1\leq a \leq b \leq c$. The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?

$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26$

Solution 1

We need \[abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).\] Since $a\le b$ and $a,b,c$ are all positive$,ac \le bc$. From the first equation we get $abc \le 6bc$. Thus $a\le 6$. From the second equation we see that $a > 2$. Thus $a\in \{3, 4, 5, 6\}$.

  • If $a=3$ we need $bc = 6(b+c) \Rightarrow (b-6)(c-6)=36$. We get five roots $\{(3, 7, 42), (3, 8, 24), (3,9,18), (3, 10, 15), (3,12,12)\}$.
  • If $a=4$ we need $2bc = 8(b+c) \Rightarrow bc = 4(b+c) \Rightarrow (b-4)(c-4)=16$. We get three roots $\{(4,5,20), (4,6,12), (4,8,8)\}$.
  • If $a=5$ we need $3bc = 10(b+c) \Rightarrow 9bc=30(b+c) \Rightarrow (3b-10)(3c-10)=100$. We get one root $\{(5,5,10)\}$.
  • If $a=6$ we need $4bc = 12(b+c) \Rightarrow bc = 3(b+c) \Rightarrow (b-3)(c-3)=9$. We get one root $\{(6,6,6)\}$.

Thus, there are $5+3+1+1 = \boxed{\textbf{(B)}\; 10}$ solutions.

Solution 2

The surface area is $2(ab+bc+ca)$, and the volume is $abc$, so equating the two yields

\[2(ab+bc+ca)=abc.\]

Divide both sides by $2abc$ to obtain \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.\]

First consider the bound of the variable $a$. Since $\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},$ we have $a>2$, or $a\geqslant3$.

Also note that $c \geq b \geq a > 0$, hence $\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}$. Thus, $\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}$, so $a \leq 6$.

So we have $a=3, 4, 5$ or $6$.

Before the casework, let's consider the possible range for $b$ if $\frac{1}{b}+\frac{1}{c}=k>0$. From $\frac{1}{b}<k$, we have $b>\frac{1}{k}$. From $\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k$, we have $b \leq \frac{2}{k}$. Thus $\frac{1}{k}<b \leq \frac{2}{k}$.

When $a=3$, we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{6}$, so $b=7, 8, 9, 10, 11, 12$. We find the solutions $(a, b, c)=(3, 7, 42)$, $(3, 8, 24)$, $(3, 9, 18)$, $(3, 10, 15)$, $(3, 12, 12)$, for a total of $5$ solutions.

When $a=4$, we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{4}$, so $b=5, 6, 7, 8$. We find the solutions $(a, b, c)=(4, 5, 20)$, $(4, 6, 12)$, $(4, 8, 8)$, for a total of $3$ solutions.

When $a=5$, we get $\frac{1}{b}+\frac{1}{c}=\frac{3}{10}$, so $b=5, 6$. The only solution in this case is $(a, b, c)=(5, 5, 10)$.

When $a=6$, $b$ is forced to be $6$, and thus $(a, b, c)=(6, 6, 6)$.

Thus, there are $5+3+1+1 = \boxed{\textbf{(B)}\; 10}$ solutions.

Simplification of Solution 2

The surface area is $2(ab+bc+ca)$, the volume is $abc$, so $2(ab+bc+ca)=abc$.

Divide both sides by $2abc$, we have: \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.\] First consider the bound of the variable $a$. Since $\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},$ we have $a>2$, or $a\ge 3$.

Also note that $c\ge b\ge a>0$, we have $\frac{1}{a}\ge \frac{1}{b}\ge \frac{1}{c}$. Thus, $\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{3}{a}$, so $a\le 6$.

So we have $a=3, 4, 5$ or $6$.


We can say $\frac{1}{b}+\frac{1}{c}=\frac{1}{q}$, where $\frac{1}{q} = \frac{1}{2}-\frac{1}{a}$.

Notice $\emph{\text{immediately}}$ that $b, c > q$. This is our key step. Then we can say $b=q+d$, $c=q+e$. If we clear the fraction about b and c (do the math), our immediate result is that $de = q^2$. Realize also that $d \leq e$.

Now go through cases for $a$ and you end up with the same result. However, now you don't have to guess solutions. For example, when $a=3$, then $de = 36$ and $d=1, 2, 3, 4, 6$.



Video Solution

https://www.youtube.com/watch?v=JFUpe32aWnw&t=1941s

See Also

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Question
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
All AMC 10 Problems and Solutions
2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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
All AMC 12 Problems and Solutions

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