# 2015 AMC 12B Problems/Problem 23

The following problem is from both the 2015 AMC 12B #23 and 2015 AMC 10B #25, 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

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}, 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}.

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 AMC 10B Problem 25, but the pages are separate. Someone should fix this. EDIT: fixed it, but someone help with the link

EDIT #2: fixed all