|
|
| (28 intermediate revisions by 9 users not shown) |
| Line 1: |
Line 1: |
| − | ==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?
| |
| − | | |
| − | <math>\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26</math>
| |
| − | | |
| − | ==Solution 1==
| |
| − | The surface area is <math>2(ab+bc+ca)</math>, the volume is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>.
| |
| − | | |
| − | Divide both sides by <math>2abc</math>, we have: <cmath>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.</cmath>
| |
| − | | |
| − | 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>.
| |
| − | | |
| − | Also note that <math>c\ge b\ge a>0</math>, we have <math>\frac{1}{a}\ge \frac{1}{b}\ge \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>.
| |
| − | | |
| − | So we have <math>a=3, 4, 5</math> or <math>6</math>.
| |
| − | | |
| − | 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}\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>
| |
| − | | |
| − | 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.
| |
| − | | |
| − | 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.
| |
| − | | |
| − | 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>.
| |
| − | | |
| − | 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>.
| |
| − | | |
| − | Thus, our answer is <math>\boxed{\textbf{(B)}\;10}</math>
| |
| − | | |
| − | ==Simplification of Solution 1==
| |
| − | The surface area is <math>2(ab+bc+ca)</math>, the volume is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>.
| |
| − | | |
| − | Divide both sides by <math>2abc</math>, we have: <cmath>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.</cmath>
| |
| − | 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>.
| |
| − | | |
| − | Also note that <math>c\ge b\ge a>0</math>, we have <math>\frac{1}{a}\ge \frac{1}{b}\ge \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>.
| |
| − | | |
| − | So we have <math>a=3, 4, 5</math> or <math>6</math>.
| |
| − | | |
| − | | |
| − | 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>.
| |
| − | | |
| − | Notice <math>immediately</math> that <math>b, c > q</math> This is our key step.
| |
| − | 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 \leq e</math>.
| |
| − | | |
| − | 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, 4, 6</math>.
| |
| − | | |
| − | - minor edit by Williamgolly, minor edit by Tiblis
| |
| − | | |
| − | ==Solution 2==
| |
| − | 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>.
| |
| − | | |
| − | 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.
| |
| − | | |
| − | ==Solution 3==
| |
| − | The surface area is <math>2(ab+bc+ca)</math>, and the volume is <math>abc</math>, so equating the two yields
| |
| − | | |
| − | <cmath>2(ab+bc+ca)=abc.</cmath>
| |
| − | Divide both sides by <math>2abc</math> to obtain<cmath>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.</cmath>
| |
| − | 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>.
| |
| − | | |
| − | 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>.
| |
| − | | |
| − | So we have <math>a=3, 4, 5</math> or <math>6</math>.
| |
| − | | |
| − | 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>.
| |
| − | | |
| − | 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.
| |
| − | | |
| − | 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.
| |
| − | | |
| − | 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>.
| |
| − | | |
| − | 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>.
| |
| − | | |
| − | Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions.
| |
| − | | |
| − | ==See Also==
| |
| − | {{AMC10 box|year=2015|ab=B|after=Last Question|num-b=24}}
| |
| − | {{MAA Notice}}
| |
| − | | |
| − | [[Category: Introductory Geometry Problems]] | |
