Prove that the ratio of the surface ares of the golden cuboid to that of the sphere is Phi : Pi?

Question

A golden cuboid is defined as a rectangular prism whose length, width and height are in the ratio of phi : 1 : 1/phi.
Prove that the ratio of the surface areas of the golden cuboid to that of the sphere that circumscribes it is Phi : Pi

Answer 1

Add up the area of the six sides of the cuboid to compute its surface area. There are two sides that are Φa x a, two that are (a/Φ) x a, and two that are (a/Φ) x (aΦ), where "a" is some constant. So the area is 2Φa^2 + 2(a^2)/Φ + 2a^2, or (2a^2)(Φ + (1/Φ) + 1). Remember that Φ^2 - Φ - 1 = 0 by definition of the golden ratio, so from this you can derive 1/Φ = Φ - 1 and the surface area becomes (2a^2)(Φ + (Φ-1) + 1) = (4a^2)(Φ).

The long diameter of the cuboid will be the diameter of the sphere too. So if "d" is the diameter, then the surface area of the sphere is 4π (d/2)^2.

This diagonal is also the hypotenuse of a right triangle, with the other legs being the short height (a/Φ) of the cuboid, and the diagonal face of the largest side, namely the aΦ x a side. So if d is the length of the diagonal, we have:
(a/Φ)^2 + (√((aΦ)^2 + a^2))^2 = d^2

This is:
(a/Φ)^2 + (aΦ)^2 + a^2 = d^2
(a^2) [ (1/Φ)^2 + Φ^2 + 1 ] = d^2
(a^2) [ (Φ-1)^2 + Φ^2 + 1 ] = d^2
(a^2) [ (Φ^2 - 2Φ + 1) + Φ^2 + 1 ] = d^2
(a^2) [ 2Φ^2 - 2Φ + 2 ] = d^2
(2a^2) [ Φ^2 - Φ + 1 ] = d^2
(2a^2) [ (Φ^2 - Φ - 1) + 1 + 1 ] = d^2
(2a^2) [ (0) + 1 + 1 ] = d^2
4a^2 = d^2
d = 2a.
d/2= a

So the ratio of the cuboid's surface area to that of the sphere is:
(4a^2)(Φ) / 4π (a)^2 =
Φ / π

Answer 2

You're bringing about an idea as old as mathematics. The problem is for every example of orderliness, I can find you one that blows the human mind and puts into question the very validity of mathematics. (Not their utility, but their ability to represent the world). Nobody argues that mathematics is indeed very useful, but whether it is a true representation of the world or not is still a subject of debate. I mean, that 3 rocks plus 3 rocks is six rocks is generally accepted as banal, but what the heck does it mean, when we say that the speed of an object, captured as a moment in time, is the derivative of its acceleration? What if the acceleration is not constant? Then its velocity becomes the derivative of the derivative of a function of its changing acceleration. Anyway... The Pythagoreans, which are some of the people who took this kind of harmony most seriously bumped upon some problems which left them baffled and worried. They expected mathematics to show the deep harmony of the universe as well. The Pythagorean Theorem, of course, is well known: a^2+b^2=c^2. Thus, a triangle with legs of 4 and 3 units has a hypotenuse of 5. But here's a problem that particularly bothered the Pythagoreans. In fact, they were forbidden to discuss this with outsiders, lest the pristine beauty they ascribed to number become the object of doubt: Imagine now a square triangle with two equal sides, or the diagonal crossing a square (it's the same basic problem). Here, a and b are equal, therefore, 2a^2=c^2. Solving such a problem requires the introduction of irrationals in our numbers system, because there are no two rational numbers that will satisfy both a and c (this was proven by people who are much better than I am at math). If a is a rational, than c is not. if c is a rational, than a is not. What this means, is that if you take the line of natural numbers, and you draw a line of equal length to a given distance on it (i.e. you go from 0 to 3 and draw a line of distance 3 at a right angle), and join these two lines with a straight diagonal, your diagonal's length can not be rational. Worse yet, you can now flatten this line and measure it against the line of natural numbers, which must mean that the continuity of the rationals is an illusion! The line is full of holes! The irrationals are there, escaping our best efforts at giving them a precise and final value. (unless you're satisfied with sqrt(2) as a value) - The Pythagoreans were not). Most mathematicians today are formalists (i.e. they take a pragmatic approach), and they can live with that, but mathematical platonists have always found this sort of thing very bothersome.

Answer 3

The surface of the sphere is π(1/φ² + 1 + φ²). The surface of the cuboid is 2(1/φ + 1 + φ). If φ = (1/2)(1+√5), then the ratio comes out to φ/π.