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==Solution== | ==Solution== | ||
− | The surface area is <math>2(ab+bc+ca)</math>, the | + | 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 | + | 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\ | + | 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\ | + | 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}\ | + | 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>. | 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>. | 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}\ | + | 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=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=6</math>, <math>b</math> is forced to be <math>6</math>, and thus <math>(a, b, c)=(6, 6, 6)</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, | + | Thus, our answer is <math>\boxed{\textbf{(B)}\;10}</math> |
+ | |||
+ | ==Simplification of Solution== | ||
+ | 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}>\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>. | ||
+ | |||
+ | 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<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, 6</math>. | ||
==See Also== | ==See Also== | ||
− | {{AMC10 box|year=2015|ab=B|after=Last | + | {{AMC10 box|year=2015|ab=B|after=Last Question|num-b=24}} |
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Introductory Algebra Problems]] |
Revision as of 14:03, 2 February 2019
Problem
A rectangular box measures , where , , and are integers and . The volume and the surface area of the box are numerically equal. How many ordered triples are possible?
Solution
The surface area is , the volume is , so .
Divide both sides by , we have:
First consider the bound of the variable . Since we have , or .
Also note that , we have . Thus, , so .
So we have or .
Before the casework, let's consider the possible range for if .
From , we have . From , we have . Thus
When , , so . The solutions we find are , for a total of solutions.
When , , so . The solutions we find are , for a total of solutions.
When , , so . The only solution in this case is .
When , is forced to be , and thus .
Thus, our answer is
Simplification of Solution
The surface area is , the volume is , so .
Divide both sides by , we have: First consider the bound of the variable . Since we have , or .
Also note that , we have . Thus, , so .
So we have or .
We can say , where .
Notice that ! This is our key step. Then we can say , . If we clear the fraction about b and c (do the math), our immediate result is that . Realize also that .
Now go through cases for and you end up with the same result. However, now you don't have to guess solutions. For example, when , then and .
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 |
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