Difference between revisions of "2015 AMC 10B Problems/Problem 25"
m (→Solution 3 (Basically the exact same as Solution 1)) |
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<cmath>2(ab+bc+ca)=abc.</cmath> | <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> | + | 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>. | 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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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. | ||
− | + | -minor edit by Snow52 | |
− | - | + | -minor edit by Bobbob |
==See Also== | ==See Also== |
Revision as of 22:45, 5 December 2020
Contents
[hide]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 1
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 1
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 .
- minor edit by Williamgolly, minor edit by Tiblis
Solution 2
We need:Since , we get . Thus . From the second equation we see that . Thus .
If , we need . We get five roots: If , we need . We get three roots: . If , we need , which is the same as . We get only one root: (corresponding to ) . If , we need . Then . We get one root: . Thus, there are solutions.
-minor edit by Bobbob
Solution 3 (Basically the exact same as Solution 1)
The surface area is , and the volume is , so equating the two yields:
Divide both sides by to obtain: First consider the bound of the variable . Since we have , or .
Also note that , hence . 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 , we get , so . We find the solutions , , , , , for a total of solutions.
When , we get , so . We find the solutions , , , for a total of solutions.
When , we get , so . The only solution in this case is .
When , is forced to be , and thus .
Thus, there are solutions.
-minor edit by Snow52
-minor edit by Bobbob
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.