Problem:
An m×n×p rectangular box has half the volume of an (m+2)×(n+2)×(p+2) rectangular box, where m,n, and p are integers, and m≤n≤p. What is the largest possible value of p?
Solution:
First notice that 2mnp=(m+2)(n+2)(p+2) implies m+22m=np(n+2)(p+2)>1, which shows that m≥3. Next rewrite the equation as
2mnp=mnp+2(mn+np+mp)+4(m+n+p)+8mnp−2(mn+np+mp)+4(m+n+p)−8=8(m+n+p)(m−2)(n−2)(p−2)=8(m+n+p)
which suggests replacing m−2,n−2, and p−2 by the positive integers a,b, and c, respectively. Notice that 1≤a. The problem is now to find the largest c that satisfies the equation abc=8(a+b+c+6), which can be rewritten 8c=ab−8a+b+6. Because (a−1)(b−1) is nonnegative, it follows that a+b≤ab+1, hence that
8c=ab−8a+b+6≤ab−8ab+1+6=ab−8ab−8+15=1+ab−815
This shows that c can be no larger than 8⋅16=128, and that c attains this value if ab=9 and a+b=ab+1=10. Thus c=128 when a=1 and b=9, and m=3, n=11, and p=130 are the dimensions of a possible box.
The problems on this page are the property of the MAA's American Mathematics Competitions