Many of the problems in my previous post were instances of a general question:
Given three positive numbers V, S, and E, does there exist a cuboid with volume V in cubic inches, surface area S in square inches, and edge sum E in inches?
When I considered this question, I found that such a cuboid exists if and only if
E^2 S^2 - 2 S^3 - 16 E^3 V + 36 E S V - 108 V^2 >= 0.
The inequality was easy to obtain as soon as I realized that the desired cuboid exists if and only if the following polynomial in p has three positive roots:
p^3 - E p^2 + (S/2) p -V.
(The left-hand side of the inequality is the discriminant of the polynomial.) Proof sketches are appended at the end.
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The VSE inequality can be expressed in terms of two variables, the scaled surface area s = S/(2E^2) and scaled volume v = V/(E^3). In terms of these parameters, the space of all possible cuboids looks like so:
Each light-blue or dark-blue point corresponds to a cuboid (except for the dashed line along the bottom, which is s = 0). The two curved boundaries are where the discriminant vanishes, indicating a repeated root. Along the left boundary, there is a double root and its value is smaller than the other root; these are the rod-shaped cuboids. Along the right boundary, there is a double root and its value is larger than the other root; these are the tile-shaped cuboids. The right and left boundaries meet at a point right at the top; this cuboid is both rod-shaped and tile-shaped, i.e., it is a cube.
I put a little tour of the space up on YouTube here.
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Well, it turns out that a cuboid can't have V, S, and E numerically equal when measured in the same system of units. Seems a shame, really. Well, there's a desperate example in one of my notebooks with edge lengths 2, i, and -i.
But to keep it real, let's open things up to other shapes. Define the diameter of a 3D shape to be the greatest possible distance between two of its points. Can you find a shape for which the volume, surface area, and diameter are all numerically equal when measured in the same system of units? (A few examples are on my whiteboard right now.)
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It is nice to have an if-and-only-if condition, but the inequality is bit complicated. A few simpler, "only-if" inequalities (necessary conditions) are easy consequences of the inequality relating arithmetic and geometric means: Given n nonnegative numbers c_1, ..., c_n,
(1/n)(c_1+c_2 + ...+c_n) >= (c_1 c_2 ... c_n)^(1/n),
with equality if and only if all the c_i are equal. (This inequality played a role in one of our previous puzzles.)
If we choose the c_i to be the edge lengths x, y, and z, then we find
E >= 3V^(1/3),
with equality if and only if the cuboid is a cube.
This inequality verifies our intuition that a cuboid can't enclose a given volume using arbitrarily little edge length. For example, there is no cuboid with a volume of a million cubic inches and an edge sum of 1 inch.
The inequality can also be written
(E/3)^3 >= V
again with equality if and only if the cuboid is a cube.
The two forms of the inequality correspond to two optimal properties of the cube: Of all cuboids with a given volume, the cube has the least edge sum; and of all cuboids with a given edge sum, the cube has greatest volume.
(Proof of the second property: Let C be a cuboid with edge sum E. Suppose K is any cuboid with edge sum E that is not a cube. Denote the volumes by V_C and V_K. Applying the inequality to K, we have (E/3)^3 > V_K. But of course, (E/3)^3 = V_C. Therefore V_C > V_K.)
Now take the c_i to be xy, yz, zx. This gives
S >= 6V^(2/3)
again with equality if and only if the cuboid is a cube.
This again establishes something intuitively obvious, which is that a cuboid can't enclose a given volume using arbitrarily little surface area. For example, there is no cuboid with a volume of 1 cubic inch and a surface area of a millionth of an inch.
The inequality can also be written
(S/6)^(3/2) >= V
again with equality if and only if the cuboid is a cube.
The two forms again correspond to two optimal properties of the cube: Of all cuboids with a given volume, the cube has the least surface area; and of all cuboids with a given surface area, the cube has greatest volume.
The last two inequalities imply a third:
ES >= 18V.
This can also be found by applying the inequality of arithmetic and geometric means, taking the c_i to be the nine expressions xyz, xyz, xyz, x^2y, xy^2, x^2z, xz^2, y^2z, and yz^2.
For a cuboid with measures V, S, and E to exist, it is necessary but not sufficient that V, S, and E satisfy these simpler inequalities. For example, the numbers E = 8, S = 78, and V = 9 satisfy all of the "only-if" inequalities, although in fact there is no cuboid with these measures.
Another "only-if" inequality follows from applying the Cauchy-Schwarz inequality to the vectors (x, y, z) and (z, x, y). The result is (1/2)S <= x^2 + y^2 + z^2. But x^2 + y^2 + z^2 = E^2 - S, so we have
S <= (2/3)E^2
again with equality if and only if the cuboid is a cube.
This inequality can be used to show two more properties of the cube: Of all cuboids with a given surface area, the cube has the least edge sum; and of all cuboids with a given edge sum, the cube has greatest surface area.
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By the way, we still need the if-and-only-if VSE inequality. None of the "only-if" inequalities is strong enough to rule out the existence of a cuboid with a volume of 80 cubic inches, a surface area of 122 square inches, and an edge sum of 16 inches (though in fact no such cuboid exists).
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Proof sketch:
(1) Given that a cuboid exists with measures V, S, and E, to show that the polynomial p^3 - E p^2 + (S/2) p -V has three positive roots. Idea: the positive roots are none other than the edge lengths.
Denote the cuboid's edge lengths by (positive) x, y, and z. Then since x + y + z = E, 2(xy + yz + zx) = S, and xyz = V, the polynomial is p^3 - (x + y + z)p^2 + (xy + yz + zx)p - xyz. This however factors as (p-x)(p-y)(p-z), hence the polynomial has three positive roots.
(2) Given that the polynomial p^3 - E p^2 + (S/2) p -V has three positive roots, to show that a cuboid with measures V, S, and E exists. Idea: the edge lengths are none other than the positive roots.
Denote the positive roots by r1, r2, and r3. Since the polynomial is a monic cubic, the coefficients can be expressed in terms of the roots as symmetric functions, specifically as E = r1+r2+r3, S/2 = r1r2+r1r3+r2r3, and V = r1r2r3. But these functions are also the formulas for the measures of a cuboid. So create a cuboid with edge lengths r1, r2, and r3. The cuboid then has the desired measures.
(3) To arrive finally at the desired inequality.
It is obvious by inspection that every real root of p^3 - E p^2 + (S/2) p -V is positive. (Try substituting a negative number or zero for p.) Hence by (1) and (2), a cuboid exists having the specified measures if and only if the polynomial has three real roots. It now follows that the discriminant of the polynomial provides a necessary and sufficient condition for the existence of a cuboid with measures V, S, and E.
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p^n - m_1 p^(n-1) + m_2 p^(n-2) - ... + m_n
has n real roots.
Even supposing this principle does generalize, however, the discriminant for higher polynomials doesn't tell the real/complex nature of the roots in the simple way that it does for n = 2 and n = 3. So for n >= 4, I don't think the discriminant will give an immediately usable condition on the coefficients m_i.
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Usual disclaimers...I assume everything here that is correct is well-known. There may be some interesting related readings out there - if anybody knows of good ones, please put them in the comments!
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