# 1998 AIME Problems/Problem 14

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## 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 $p=\boxed{130}.$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 