Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 AMC 10B Problems/Problem 25"

m (→‎Solution 2)
m (→‎Solution 2)
Line 56: Line 56:
 
If <math>a=4</math> we need <math>bc = 4(b+c) \Rightarrow (b-4)(c-4)=16</math>. We get three roots: <math>\{(4,5,20), (4,6,12), (4,8,8)\}</math>.
 
If <math>a=4</math> we need <math>bc = 4(b+c) \Rightarrow (b-4)(c-4)=16</math>. We get three roots: <math>\{(4,5,20), (4,6,12), (4,8,8)\}</math>.
 
If <math>a=5</math> we need <math>3bc = 10(b+c)</math>, which is the same as <math>9bc=30(b+c)\Rightarrow (3b-10)(3c-10)=100</math>. We get only one root (corresponding to <math>100=5\cdot 20</math>) <math>(5,5,10)</math>.
 
If <math>a=5</math> we need <math>3bc = 10(b+c)</math>, which is the same as <math>9bc=30(b+c)\Rightarrow (3b-10)(3c-10)=100</math>. We get only one root (corresponding to <math>100=5\cdot 20</math>) <math>(5,5,10)</math>.
−
If <math>a=6</math> we need <math>4bc = 12(b+c)</math>. Then <math>(b-3)(c-3)=9</math>. We get one root <math>(6,6,6)</math>.
+
If <math>a=6</math> we need <math>4bc = 12(b+c)</math>. Then <math>(b-3)(c-3)=9</math>. We get one root: <math>(6,6,6)</math>.
 
Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions.
 
Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions.
 +
-Minor edits by Bobbob
  
 
==Solution 3 (Basically the exact same as Solution 1)==
 
==Solution 3 (Basically the exact same as Solution 1)==

Revision as of 22:36, 4 December 2020

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

- minor edit by Williamgolly, minor edit by Tiblis

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 edits 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

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_25&oldid=139033"