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Difference between revisions of "2015 AMC 12B Problems/Problem 23"
m (→‎Solution: consistency)
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We need <cmath>abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).</cmath>
 
We need <cmath>abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).</cmath>
−
Since <math>ab, ac \le bc</math>, we get <math>abc \le 6bc</math>. Thus <math>a\le 6</math>. From the second equation we see that <math>a > 2</math>. Thus <math>a\in \{3, 4, 5, 6\}</math>.  
+
Since <math>ab, ac \le bc</math>, from the first equation we get <math>abc \le 6bc</math>. Thus <math>a\le 6</math>. From the second equation we see that <math>a > 2</math>. Thus <math>a\in \{3, 4, 5, 6\}</math>.
 +
 
 +
*If <math>a=3</math> we need <math>bc = 6(b+c) \Rightarrow (b-6)(c-6)=36</math>. We get '''five''' roots <math>\{(3, 7, 42), (3, 8, 24), (3,9,18), (3, 10, 15), (3,12,12)\}</math>.
 +
 
 +
*If <math>a=4</math> we need <math>2bc = 8(b+c) \Rightarrow 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) \Rightarrow 9bc=30(b+c) \Rightarrow (3b-10)(3c-10)=100</math>. We get '''one''' root <math>\{(5,5,10)\}</math>.
 +
 +
*If <math>a=6</math> we need <math>4bc = 12(b+c) \Rightarrow bc = 3(b+c) \Rightarrow (b-3)(c-3)=9</math>. We get ''' one''' root <math>\{(6,6,6)\}</math>.
  
−
*If <math>a=3</math> we need <math>bc = 6(b+c) \Rightarrow (b-6)(c-6)=36</math>. We get '''five''' roots <math>\{(3, 7, 42), (3, 8, 24), (3,9,18), (3, 10, 15), (3,12,12)\}.</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=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.
  

Revision as of 23:23, 30 October 2021

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 $ab, ac \le bc$, from the first equation 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 $2bc = 8(b+c) \Rightarrow bc = 4(b+c) \Rightarrow (b-4)(c-4)=16$. We get three roots $\{(4,5,20), (4,6,12), (4,8,8)\}$.
  • If $a=5$ we need $3bc = 10(b+c) \Rightarrow 9bc=30(b+c) \Rightarrow (3b-10)(3c-10)=100$. We get one root $\{(5,5,10)\}$.
  • If $a=6$ we need $4bc = 12(b+c) \Rightarrow bc = 3(b+c) \Rightarrow (b-3)(c-3)=9$. 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

EDIT #2: fixed all

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
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

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