Difference between revisions of "1998 AIME Problems/Problem 14"
Mathcool2009 (talk | contribs) (→Solution) |
(→Solution) |
||
Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
− | < | + | <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>. | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
== See also == | == See also == |
Revision as of 17:32, 14 September 2015
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 :
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 We have Where we see gives us our maximum value of .
- Note that assumes , but this is clear as and similarly for .
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.