Difference between revisions of "1998 AIME Problems/Problem 14"
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An <math>m\times n\times p</math> rectangular box has half the volume of an <math>(m + 2)\times(n + 2)\times(p + 2)</math> rectangular box, where <math>m, n,</math> and <math>p</math> are integers, and <math>m\le n\le p.</math> What is the largest possible value of <math>p</math>? | An <math>m\times n\times p</math> rectangular box has half the volume of an <math>(m + 2)\times(n + 2)\times(p + 2)</math> rectangular box, where <math>m, n,</math> and <math>p</math> are integers, and <math>m\le n\le p.</math> What is the largest possible value of <math>p</math>? | ||
− | == Solution == | + | == Solution 1 == |
<cmath>2mnp = (m+2)(n+2)(p+2)</cmath> | <cmath>2mnp = (m+2)(n+2)(p+2)</cmath> | ||
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*Note that <math>0 \le (a-1)(b-1)</math> assumes <math>m,n \ge 3</math>, but this is clear as <math>\frac{2m}{m+2} = \frac{(n+2)(p+2)}{np} > 1</math> and similarly for <math>n</math>. | *Note that <math>0 \le (a-1)(b-1)</math> assumes <math>m,n \ge 3</math>, but this is clear as <math>\frac{2m}{m+2} = \frac{(n+2)(p+2)}{np} > 1</math> and similarly for <math>n</math>. | ||
+ | == Solution 2 == | ||
+ | |||
+ | Similarly as above, we solve for <math>p,</math> but we express the denominator differently: | ||
+ | |||
+ | <cmath>p=\dfrac{2(m+2)(n+2)}{(m+2)(n+2)-4(m+n+2)} \implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(m+n+2)}{(m+2)(n+2)}.</cmath> | ||
+ | Hence, it suffices to maximize <math>\dfrac{m+n+2}{(m+2)(n+2)},</math> under the conditions that <math>p</math> is a positive integer. | ||
+ | |||
+ | Then since <math>\dfrac{m+n+2}{(m+2)(n+2)}>\dfrac{1}{2}</math> for <math>m=1,2,</math> we fix <math>m=3.</math> | ||
+ | <cmath>\implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(n+2)}{5(n+2)}=\dfrac{n-10}{10(n+2)},</cmath> | ||
+ | where we simply let <math>n=11</math> to achieve <math>p=\boxed{130}.</math> | ||
+ | |||
+ | ~Generic_Username | ||
== See also == | == See also == | ||
{{AIME box|year=1998|num-b=13|num-a=15}} | {{AIME box|year=1998|num-b=13|num-a=15}} |
Revision as of 17:30, 18 April 2016
Contents
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
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.