2015 AMC 12B Problems/Problem 23: Difference between revisions

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 47: / Line 47: @@
 ==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}}
 {{MAA Notice}}

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