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2015 AMC 12B Problems/Problem 23

Revision as of 21:34, 3 February 2020 by Lcz (talk | contribs) (→‎Solution 2)

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

We need \[abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).\] Since $ab, ac \le bc$, 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 $bc = 4(b+c) \Rightarrow (b-4)(c-4)=16$. We get three roots $\{(4,5,20), (4,6,12), (4,8,8)\}$.
  • If $a=5$ we need $3bc = 10(b+c)$, which is the same as $9bc=30(b+c)\Rightarrow (3b-10)(3c-10)=100$. We get only one root (corresponding to $100=5\cdot 20$) $(5,5,10)$.
  • If $a=6$ we need $4bc = 12(b+c)$. Then $(b-3)(c-3)=9$. 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.

Note

This is also problem 25 of the AMC 10B Problem 25, but they don't connect. Someone should fix this.

See Also

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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