2015 AMC 10B Problems/Problem 25

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

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 $6$.

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 $\frac{2}{b}\ge \frac{1}{b}+\frac{1}{c}=k$, we have $b\le \frac{2}{k}$. Thus $\frac{1}{k}<b\le \frac{2}{k}$

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

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

When $a=5$, $\frac{1}{b}+\frac{1}{c}=\frac{3}{10}$, 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, our answer is $\boxed{\textbf{(B)}\;10}$

Simplification of Solution 1

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 $d=1, 2, 3, 4, 6$.

Solution 2

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.

-minor edit by Bobbob

Solution 3 (Basically the exact same as Solution 1)

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.

-minor edit by Snow52

-minor edit by Bobbob

Note

This is also AMC 12B Problem 23, but the pages are separate.

See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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