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 ==
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== 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>.
  
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== Solution 2 ==
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Similarly as above, we solve for <math>p,</math> but we express the denominator differently:
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<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>
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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.
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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>
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<cmath>\implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(n+2)}{5(n+2)}=\dfrac{n-10}{10(n+2)},</cmath>
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where we simply let <math>n=11</math> to achieve <math>p=\boxed{130}.</math>
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~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

Problem

An $m\times n\times p$ rectangular box has half the volume of an $(m + 2)\times(n + 2)\times(p + 2)$ rectangular box, where $m, n,$ and $p$ are integers, and $m\le n\le p.$ What is the largest possible value of $p$?

Solution 1

\[2mnp = (m+2)(n+2)(p+2)\]

Let’s solve for $p$:

\[(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)\] \[[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)\] \[p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}\]

Clearly, we want to minimize the denominator, so we test $(m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9$. The possible pairs of factors of $9$ are $(1,9)(3,3)$. These give $m = 3, n = 11$ and $m = 5, n = 5$ respectively. Substituting into the numerator, we see that the first pair gives $130$, while the second pair gives $98$. We now check that $130$ is optimal, setting $a=m-2$, $b=n-2$ in order to simplify calculations. Since \[0 \le (a-1)(b-1) \implies a+b \le ab+1\] We have \[p = \frac{2(a+4)(b+4)}{ab-8} = \frac{2ab+8(a+b)+32}{ab-8} \le \frac{2ab+8(ab+1)+32}{ab-8} = 10 + \frac{120}{ab-8} \le 130\] Where we see $(m,n)=(3,11)$ gives us our maximum value of $\boxed{130}$.

  • Note that $0 \le (a-1)(b-1)$ assumes $m,n \ge 3$, but this is clear as $\frac{2m}{m+2} = \frac{(n+2)(p+2)}{np} > 1$ and similarly for $n$.

Solution 2

Similarly as above, we solve for $p,$ but we express the denominator differently:

\[p=\dfrac{2(m+2)(n+2)}{(m+2)(n+2)-4(m+n+2)} \implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(m+n+2)}{(m+2)(n+2)}.\] Hence, it suffices to maximize $\dfrac{m+n+2}{(m+2)(n+2)},$ under the conditions that $p$ is a positive integer.

Then since $\dfrac{m+n+2}{(m+2)(n+2)}>\dfrac{1}{2}$ for $m=1,2,$ we fix $m=3.$ \[\implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(n+2)}{5(n+2)}=\dfrac{n-10}{10(n+2)},\] where we simply let $n=11$ to achieve $p=\boxed{130}.$

~Generic_Username

See also

1998 AIME (ProblemsAnswer KeyResources)
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

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