Problem:
Let B be the set of rectangular boxes with surface area 54 and volume 23. Let r be the radius of the smallest sphere that can contain each of the rectangular boxes that are elements of B. The value of r2 can be written as qp, where p and q are relatively prime positive integers. Find p+q.
Solution:
Consider a rectangular box with dimensions a×b×c from B. From the given conditions ab+bc+ca=27 and abc=23. The diagonal of the box is a2+b2+c2. The maximum value of a2+b2+c2 among all possible values is the diameter of the smallest sphere that can contain each box from B. This is equivalent to maximizing s=a+b+c, because (a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)=a2+b2+c2+54.
Consider f(x)=(x−a)(x−b)(x−c)=x3−sx2+27x−23. The problem is then equivalent to finding the maximum possible s for which this cubic has three positive real roots (allowing multiplicity). Because uv+vw+wu≤u2+v2+w2 for all real numbers u,v, and w, it follows that (u+v+w)2≥3(uv+vw+wu) and thus
(a+b+c)2≥3(ab+bc+ca)and(ab+bc+ca)2≥3abc(a+b+c),
which together imply that 9≤s≤23243. When s=9 or s=23243, equality must hold in the above inequalities, implying both a=b=c=3 and abc=23. Thus the cubic does not have three real roots (allowing multiplicity) for s at or near 9 or 23243.
As s increases across the interval (9,23243), the graph of the cubic curve moves down. There are two limiting cases with three positive real roots. When s is the greatest possible, the local maximum will be a double root (which will be less than the third root). When s is the least possible, the local minimum will be a double root (which will be greater than the third root). If the three roots are a,b,b, then b2+2ab=27 and ab2=23. Hence b2+b46=27, and b is a root of b3−27b+46=(b−2)(b2+2b−23), so b=2 and b=−1±26.
The double root −1−26 is not positive, so consider two boundary cases with roots
2,2,423and26−1,26−1,(26−1)223.
The former triple of roots yields the greatest value for s because the value of the double root is less than the third root. Thus the maximum of a2+b2+c2 occurs when a=423 and b=c=2. In that case r2=41⋅(a2+b2+c2)=64657. The requested sum is 657+64=721.
OR
A box in B has dimensions (x,y,z), where x,y,z are positive real numbers satisfying
xyz2xy+2yz+2zx=23and=54.(1)(2)
As in the first solution, it is sufficient to find (x,y,z) that maximizes x+y+z. Because 2xy+2yz+2zx=54, the dimensions cannot all be less than 3, so assume x≥3. The value of x determines both yz and y+z:
yzy+z=x23=Pand=x27−x223=S.(3)(4)
This system has positive solutions for y and z if P and S are both positive and satisfy S2−4P≥0. When x≥3, both P and S are positive, so x is the dimension of a box in B if and only if
(x27−x223)2−4(x23)≥0,
which is equivalent to
g(x)=−92x3+272x2−(2⋅23⋅27)x+232≥0.
Note that g(21)>0,g(1)<0,g(3)>0, and g(10)<0. It follows that this cubic has three real roots, all in the interval (21,10). The Rational Root Theorem suggests that 1 and 423 might be roots in the interval (21,10), and indeed, 423 is a root. It follows that g(x)≥0 when x∈[3,423], and these are the values of x that correspond to boxes in B.
To find the value of x in [3,423] that maximizes
f(x)=x+y+z=x+x27−x223,
note that
f(423)−f(x)=x21⋅(x−423)(x−2)2,
which is 0 at 423 and positive elsewhere in [3,423]. In this case y=z=2 and