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

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- minor edit by Williamgolly, minor edit by Tiblis
 
- minor edit by Williamgolly, minor edit by Tiblis
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Solution
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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>.
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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>
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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>.
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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>.
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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>.
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Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions.
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Solution 2
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The surface area is <math>2(ab+bc+ca)</math>, and the volume is <math>abc</math>, so equating the two yields
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<cmath>2(ab+bc+ca)=abc.</cmath>
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Divide both sides by <math>2abc</math> to obtain<cmath>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.</cmath>
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First consider the bound of the variable <math>a</math>. Since <math>\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},</math> we have <math>a>2</math>, or <math>a\geqslant3</math>.
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Also note that <math>c \geq b \geq a > 0</math>, hence <math>\frac{1}{a} \geq \frac{1}{b}  \geq \frac{1}{c}</math>. Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}</math>, so <math>a \leq 6</math>.
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So we have <math>a=3, 4, 5</math> or <math>6</math>.
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Before the casework, let's consider the possible range for <math>b</math> if <math>\frac{1}{b}+\frac{1}{c}=k>0</math>. From <math>\frac{1}{b}<k</math>, we have <math>b>\frac{1}{k}</math>. From <math>\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k</math>, we have <math>b \leq \frac{2}{k}</math>. Thus <math>\frac{1}{k}<b \leq \frac{2}{k}</math>.
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When <math>a=3</math>, we get <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{6}</math>, so <math>b=7, 8, 9, 10, 11, 12</math>. We find the solutions <math>(a, b, c)=(3, 7, 42)</math>, <math>(3, 8, 24)</math>, <math>(3, 9, 18)</math>, <math>(3, 10, 15)</math>, <math>(3, 12, 12)</math>, for a total of <math>5</math> solutions.
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When <math>a=4</math>, we get <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{4}</math>, so <math>b=5, 6, 7, 8</math>. We find the solutions <math>(a, b, c)=(4, 5, 20)</math>, <math>(4, 6, 12)</math>, <math>(4, 8, 8)</math>, for a total of <math>3</math> solutions.
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When <math>a=5</math>, we get <math>\frac{1}{b}+\frac{1}{c}=\frac{3}{10}</math>, so <math>b=5, 6</math>. The only solution in this case is <math>(a, b, c)=(5, 5, 10)</math>.
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When <math>a=6</math>, <math>b</math> is forced to be <math>6</math>, and thus <math>(a, b, c)=(6, 6, 6)</math>.
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Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions.
  
 
==See Also==
 
==See Also==

Revision as of 19:56, 5 July 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

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

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 $immediately$ 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 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.

See Also

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Question
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All AMC 10 Problems and Solutions

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