Difference between revisions of "1998 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
| − | An <math>m\times n\times p</math> rectangular box has half the volume of an <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 == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 19:38, 4 July 2013
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
Let’s solve for :
![$[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)$](http://latex.artofproblemsolving.com/e/2/c/e2c32ed27109f6e9f84372c378e822f0bf7dc712.png)
For the denominator, we will use a factoring trick (colloquially known as SFFT), which states that 
.
Clearly, we want to minimize the denominator, so 
. 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 can quickly test for the denominator assuming other values to convince ourselves that 
is the best possible value for the denominator. Hence, the solution is .
Proof that setting the denominator 
to is optimal: Suppose 
, and suppose for the sake of contradiction that there exist 
such that 
for some 
and such that
![\[\frac{2(m + 2)(n + 2)}{(m - 2)(n - 2) - 8} < \frac{2(m' + 2)(n' + 2)}{(m' - 2)(n' - 2) - 8}.\]](http://latex.artofproblemsolving.com/6/4/d/64d254cc1f46f8446528027ae6c0d5af41a87530.png)
This implies that
![\[d(m + 2)(n + 2) < (m' + 2)(n' + 2),\]](http://latex.artofproblemsolving.com/7/1/c/71c55f1dc5d9c7ff2170826c9e8ee88d8c5a4c4b.png)
and
![\[d((m - 2)(n - 2) + 4(m + n)) < (m' - 2)(n' - 2) + 4(m' + n').\]](http://latex.artofproblemsolving.com/f/1/c/f1cfd7ed8f44aef39dd003938edad94962def2cf.png)
Substituting gives
![\[d(9 + 4(m + n)) < 8 + d + 4(m' + n'),\]](http://latex.artofproblemsolving.com/d/7/a/d7abf8ede181597dd4a9ed6333934c54819547da.png)
which we rewrite as
![\[d(8 + 4(m + n)) < 24 + 4((m' - 2) + (n' - 2)).\]](http://latex.artofproblemsolving.com/1/2/9/129c0206a67fc55f1151490f0b339a91b0c3e787.png)
Next, note that for 
to be positive, we must have 
and 
be positive, so
![\[(m' - 2) + (n' - 2) \leq (m' - 2)(n' - 2) + 1 = 9 + d.\]](http://latex.artofproblemsolving.com/2/c/0/2c06938893c3b73e512b275727f98ad1bb705561.png)
So
![\[d(8 + 4(m + n)) < 8 + 4(9 + d)\]](http://latex.artofproblemsolving.com/0/2/3/0236df0d66c50bffec72de3c597d11026aeeb8a5.png)
![\[d(4 + 4(m + n)) < 44\]](http://latex.artofproblemsolving.com/5/d/6/5d66dd6cbb51fce24f388cba36bcc01c1066f92b.png)
![\[d(1 + m + n) < 11\]](http://latex.artofproblemsolving.com/0/3/2/0328864442864929e311e929b2c00d7356b8bfe4.png)
Next, we must have that 
and 
are positive, so 
and 
. Also, 
by how we defined 
. So
![\[2(1 + 3 + 3) < 11,\]](http://latex.artofproblemsolving.com/f/9/5/f953b56aab66465153582d621d79c390a422755f.png)
a contradiction. We already showed above that there are some values of 
and 
such that that work, so this proves that one of these pairs of values of and must yield the maximal value of .
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. 
