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| − | ==Problem==
| + | #redirect [[2015 AMC 12B Problems/Problem 23]] |
| − | A rectangular box measures <math>a \times b \times c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>1\leq a \leq b \leq c</math>. The volume and the surface area of the box are numerically equal. How many ordered triples <math>(a,b,c)</math> are possible?
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| − | <math>\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26</math>
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| − | ==Solution==
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| − | The surface area is <math>2(ab+bc+ca)</math>, the volume is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>.
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| − | Divide both sides by <math>2abc</math>, we have: <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\ge 3</math>.
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| − | Also note that <math>c\ge b\ge a>0</math>, we have <math>\frac{1}{a}>\frac{1}{b}>\frac{1}{c}</math>.
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| − | Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{3}{a}</math>, so <math>a\le 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>.
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| − | From <math>\frac{1}{b}<k</math>, we have <math>b>\frac{1}{k}</math>. From <math>\frac{2}{b}\ge \frac{1}{b}+\frac{1}{c}=k</math>, we have <math>b\le \frac{2}{k}</math>. Thus <math>\frac{1}{k}<b\le \frac{2}{k}</math>
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| − | When <math>a=3</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{6}</math>, so <math>b=7, 8, \cdots, 12</math>. The solutions we find are <math>(a, b, c)=(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12)</math>, for a total of <math>5</math> solutions.
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| − | When <math>a=4</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{4}</math>, so <math>b=5, 6, 7, 8</math>. The solutions we find are <math>(a, b, c)=(4, 5, 20), (4, 6, 12), (4, 8, 8)</math>, for a total of <math>3</math> solutions.
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| − | When <math>a=5</math>, <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, our answer is <math>\boxed{\textbf{(B)}\;10}</math>
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| − | ==Simplification of Solution==
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| − | The surface area is <math>2(ab+bc+ca)</math>, the volume is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>.
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| − | Divide both sides by <math>2abc</math>, we have: <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\ge 3</math>.
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| − | Also note that <math>c\ge b\ge a>0</math>, we have <math>\frac{1}{a}>\frac{1}{b}>\frac{1}{c}</math>. Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{3}{a}</math>, so <math>a\le 6</math>.
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| − | So we have <math>a=3, 4, 5</math> or <math>6</math>.
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| − | We can say <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{q}</math>, where <math>\frac{1}{q} = \frac{1}{2}-\frac{1}{a}</math>.
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| − | Notice <math>immediately</math> that <math>b, c > q</math>! This is our key step.
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| − | Then we can say <math>b=q+d</math>, <math>c=q+e</math>. If we clear the fraction about b and c (do the math), our immediate result is that <math>de = q^2</math>. Realize also that <math>d<e</math>.
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| − | Now go through cases for <math>a</math> and you end up with the same result. However, now you don't have to guess solutions. For example, when <math>a=3</math>, then <math>de = 36</math> and <math>d=1, 2, 3, 6</math>.
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| − | ==See Also==
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| − | {{AMC10 box|year=2015|ab=B|after=Last Question|num-b=24}}
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| − | {{MAA Notice}}
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| − | [[Category: Introductory Algebra Problems]] | |
