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> |
Let’s solve for <math>p</math>: | Let’s solve for <math>p</math>: | ||
| − | < | + | <cmath>(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)</cmath> |
| − | < | + | <cmath>[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)</cmath> |
| − | < | + | <cmath>p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}</cmath> |
| − | + | Clearly, we want to minimize the denominator, so we test <math>(m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9</math>. The possible pairs of factors of <math>9</math> are <math>(1,9)(3,3)</math>. These give <math>m = 3, n = 11</math> and <math>m = 5, n = 5</math> respectively. Substituting into the numerator, we see that the first pair gives <math>130</math>, while the second pair gives <math>98</math>. We now check that <math>130</math> is optimal, setting <math>a=m-2</math>, <math>b=n-2</math> in order to simplify calculations. Since | |
| + | <cmath>0 \le (a-1)(b-1) \implies a+b \le ab+1</cmath> | ||
| + | We have | ||
| + | <cmath>p = \frac{2(a+4)(b+4)}{ab-8} = \frac{2ab+8(a+b)+32}{ab-8} \le \frac{2ab+8(ab+1)+32}{ab-8} = 10 + \frac{120}{ab-8} \le 130</cmath> | ||
| + | Where we see <math>(m,n)=(3,11)</math> gives us our maximum value of <math>\boxed{130}</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+5)}{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> |
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| − | < | + | ~Generic_Username |
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== See also == | == See also == | ||
Latest revision as of 20:05, 29 May 2023
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
![\[2mnp = (m+2)(n+2)(p+2)\]](http://latex.artofproblemsolving.com/d/4/8/d488161a60c3b671e6a099deae2b1bd8743f01e9.png)
Let’s solve for :
![\[(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)\]](http://latex.artofproblemsolving.com/7/b/1/7b143cfbb2985678c4a7d6424e032e016fa2a879.png)
![\[[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)\]](http://latex.artofproblemsolving.com/8/d/8/8d85d749a6a153c00912c02d1cc6bd841fb8f0f2.png)
![\[p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}\]](http://latex.artofproblemsolving.com/4/c/a/4ca431090d2a4c440e06936242d05d118a9b4789.png)
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
![\[0 \le (a-1)(b-1) \implies a+b \le ab+1\]](http://latex.artofproblemsolving.com/f/c/b/fcbb8bd5842ec9e3e12154c710e3221711c4ec51.png)
We have
![\[p = \frac{2(a+4)(b+4)}{ab-8} = \frac{2ab+8(a+b)+32}{ab-8} \le \frac{2ab+8(ab+1)+32}{ab-8} = 10 + \frac{120}{ab-8} \le 130\]](http://latex.artofproblemsolving.com/b/4/b/b4b5f08e82b91dea4d2a91f26251aa9db9b5a164.png)
Where we see 
gives us our maximum value of 
.
- Note that
assumes 
, but this is clear as 
and similarly for 
.
assumes 
, but this is clear as 
and similarly for 
.Solution 2
Similarly as above, we solve for 
but we express the denominator differently:
![\[p=\dfrac{2(m+2)(n+2)}{(m+2)(n+2)-4(m+n+2)} \implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(m+n+2)}{(m+2)(n+2)}.\]](http://latex.artofproblemsolving.com/7/a/9/7a9a475822c4e657de3c3eb43c11a824f4f13b11.png)
Hence, it suffices to maximize 
under the conditions that is a positive integer.
Then since 
for 
we fix 
![\[\implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(n+5)}{5(n+2)}=\dfrac{n-10}{10(n+2)},\]](http://latex.artofproblemsolving.com/b/3/1/b3151acf0e20a8435251529044c4713537ebe32b.png)
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. 
