Difference between revisions of "1998 AIME Problems/Problem 14"

(Solution)
(Solution)
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== Solution ==
 
== Solution ==
<div style="text-align:center;"><math>2mnp = (m+2)(n+2)(p+2)</math></div>
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<cmath>2mnp = (m+2)(n+2)(p+2)</cmath>
  
 
Let’s solve for <math>p</math>:
 
Let’s solve for <math>p</math>:
  
<div style="text-align:center;"><math>(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)</math><br />
+
<cmath>(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)</cmath>
<math>[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)</math><br />
+
<cmath>[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)</cmath>
<math>p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4}</math></div>
+
<cmath>p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}</cmath>
  
For the denominator, we will use a factoring trick (colloquially known as [[Simon's Favorite Factoring Trick|SFFT]]), which states that <math>xy + ax + ay + a^2 = (x+a)(y+a)</math>.  
+
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>.
  
<div style="text-align:center;"><math>p = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}</math></div>
+
*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>.
 
 
Clearly, we want to minimize the denominator, so <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 can quickly test for the denominator assuming other values to convince ourselves that <math>1</math> is the best possible value for the denominator. Hence, the solution is <math>130</math>.
 
 
 
 
 
Proof that setting the denominator <math>(m - 2)(n - 2) - 8</math> to <math>1</math> is optimal: Suppose <math>(m - 2)(n - 2) = 9</math>, and suppose for the sake of contradiction that there exist <math>m', n'</math> such that <math>(m' - 2)(n' - 2) = 8 + d</math> for some <math>d > 1</math> and such that
 
<cmath>\frac{2(m + 2)(n + 2)}{(m - 2)(n - 2) - 8} < \frac{2(m' + 2)(n' + 2)}{(m' - 2)(n' - 2) - 8}.</cmath>
 
This implies that
 
<cmath>d(m + 2)(n + 2) < (m' + 2)(n' + 2),</cmath>
 
and
 
<cmath>d((m - 2)(n - 2) + 4(m + n)) < (m' - 2)(n' - 2) + 4(m' + n').</cmath>
 
Substituting gives
 
<cmath>d(9 + 4(m + n)) < 8 + d + 4(m' + n'),</cmath>
 
which we rewrite as
 
<cmath>d(8 + 4(m + n)) < 24 + 4((m' - 2) + (n' - 2)).</cmath>
 
Next, note that for <math>p'</math> to be positive, we must have <math>m' - 2</math> and <math>n' - 2</math> be positive, so
 
<cmath>(m' - 2) + (n' - 2) \leq (m' - 2)(n' - 2) + 1 = 9 + d.</cmath>
 
So
 
<cmath>d(8 + 4(m + n)) < 8  + 4(9 + d)</cmath> //shouldn't this be 24 + 4(9+d)? then the entire proof is fallacious (it results in 2(1 + 3 + 3) < 15 which is true)
 
<cmath>d(4 + 4(m + n)) < 44</cmath>
 
<cmath>d(1 + m + n) < 11</cmath>
 
Next, we must have that <math>m - 2</math> and <math>n - 2</math> are positive, so <math>3 \leq m</math> and <math>3 \leq n</math>. Also, <math>2 \leq d</math> by how we defined <math>d</math>. So
 
<cmath>2(1 + 3 + 3) < 11,</cmath>
 
a contradiction. We already showed above that there are some values of <math>m</math> and <math>n</math> such that <math>(m - 2)(n - 2) = 9</math> that work, so this proves that one of these pairs of values of <math>m</math> and <math>n</math> must yield the maximal value of <math>p</math>.
 
  
 
== See also ==
 
== See also ==

Revision as of 17:32, 14 September 2015

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

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

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 $n$.

See also

1998 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AIME Problems and Solutions

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