Difference between revisions of "1998 AIME Problems/Problem 14"
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== Solution 3 == | == Solution 3 == | ||
Observe that | Observe that | ||
− | <cmath>2 = \left ( 1 + \frac{2}{m} \right ) \left ( 1 + \frac{2}{n} \right ) \left (1 + \frac{2}{p} \right ) \leq \left ( 1 + \frac{ | + | <cmath>2 = \left ( 1 + \frac{2}{m} \right ) \left ( 1 + \frac{2}{n} \right ) \left (1 + \frac{2}{p} \right ) \leq \left ( 1 + \frac{2}{m} \right )^3</cmath> thus <math>m < 7</math>. |
Now, we can use casework on <math>m</math> and Simon's Favorite Factoring Trick to check that <math>m = 5,2,1</math> have no solution and for <math>m = 3,4,6</math>, we have the corresponding values of <math>p</math>: <math>130,54,16</math>. | Now, we can use casework on <math>m</math> and Simon's Favorite Factoring Trick to check that <math>m = 5,2,1</math> have no solution and for <math>m = 3,4,6</math>, we have the corresponding values of <math>p</math>: <math>130,54,16</math>. | ||
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~amplreneo | ~amplreneo | ||
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== See also == | == See also == | ||
{{AIME box|year=1998|num-b=13|num-a=15}} | {{AIME box|year=1998|num-b=13|num-a=15}} |
Latest revision as of 14:17, 3 May 2020
Problem
An rectangular box has half the volume of an rectangular box, where and are integers, and What is the largest possible value of ?
Solution 1
Let’s solve for :
Clearly, we want to minimize the denominator, so we test . The possible pairs of factors of are . These give and respectively. Substituting into the numerator, we see that the first pair gives , while the second pair gives . We now check that is optimal, setting , in order to simplify calculations. Since We have Where we see gives us our maximum value of .
- Note that assumes , but this is clear as and similarly for .
Solution 2
Similarly as above, we solve for but we express the denominator differently:
Hence, it suffices to maximize under the conditions that is a positive integer.
Then since for we fix where we simply let to achieve
~Generic_Username
Solution 3
Observe that thus .
Now, we can use casework on and Simon's Favorite Factoring Trick to check that have no solution and for , we have the corresponding values of : .
Thus, the maximum value is .
~amplreneo
See also
1998 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.