Difference between revisions of "2015 AMC 10B Problems/Problem 25"
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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=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=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=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>. | 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. | Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions. | ||
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-Minor edits by Bobbob | -Minor edits by Bobbob | ||
Revision as of 22:37, 4 December 2020
Contents
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.
-Minor edits 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
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.