Difference between revisions of "2015 AMC 10B Problems/Problem 25"
m (→Solution 2) |
m (→Solution 3 (Basically the exact same as Solution 1)) |
||
Line 62: | Line 62: | ||
==Solution 3 (Basically the exact same as Solution 1)== | ==Solution 3 (Basically the exact same as Solution 1)== | ||
− | The surface area is <math>2(ab+bc+ca)</math>, and the volume is <math>abc</math>, so equating the two yields | + | 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> | <cmath>2(ab+bc+ca)=abc.</cmath> | ||
Line 85: | Line 85: | ||
Minor Edit by Snow52 | Minor Edit by Snow52 | ||
+ | |||
+ | -Minor edit by Bobbob | ||
==See Also== | ==See Also== |
Revision as of 22:44, 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.