2015 AMC 12B Problems/Problem 23

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

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.

Solution 2

Find that 2ab+2bc+2ac = abc

I'm not going to go through the bashing process of this solution, I will just give the outline

Split it up into cases:

Case 1: a=b=c You should get a=6

Case 2: a=b You should get a^2(2-c)+4ac = 0. The easiest way to bash this is to set c equal to an integer, starting from 3 (so that a^2 is negative). You should get a=12, 8, 6, 5, but cut off 6 as we have already counted that in Case 1. Thus, three cases here. However, we can also do b=c and a=c on top of this - thus, we get 3*3 = 9 cases.

Case 3: Can we have a/b/c independent of each other? Simply put, no. It's just not possible for there to be no overlap. You can verify this by setting persay a equal to some integer, plugging it in to the equation, and then using SFFT, but you will be to no avail.

A practical approach here is to just look at the answer choices. If a/b/c were independent of each other, and a/b/c have nothing differentiating them, then for each independent ordered triple {a,b,c}, we should have 6 solutions. That would mean that the solution has to be 10+6n, where n is some integer. Obviously the only answer choice here that works is B, as the the other answer choices we cannot plug in an integral n to get them. Thus, either by POE or complicated induction, we find the answer to be 10.

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2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2015 AMC 12B Problems/Problem 23

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Solution

Solution 2

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