2015 AMC 10B Problems/Problem 25
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 needSince , 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.
Solution 3
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.
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
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