If you take a rectangle and construct, inside it, a square on the shorter of the two sides, the portion of the rectangle that isn't inside the square is, itself, a rectangle. In general, this second rectangle is a different shape than the first; but it can be the same shape, albeit smaller, if the rectangle's shape is exactly right. A rectangle for which this holds true is claimed to be more æsthetically satisfying than other rectangular shapes.

Rectangles are so simple that the shape of one is encoded entirely, given that it is a rectangle, in the ratio between the lengths of the sides. If the longer side is k times as long as the shorter side (so k is at least one) then the sides of the rectangle obtained by snipping off a square on the shorter side, as above, are 1 and k−1 times the original rectangle's shorter side. For the resulting rectangle to be the same size as the original, k/1 must be the same ratio as either 1/(k−1) or (k−1)/1; however, k−1 and k are not equal, so the second candidate can't work. This then requires k/1 = 1/(k−1), i.e. k.(k−1) = 1, i.e.

- 0 = k.k −k −1 = (k −1/2)**2 −1/4 −1, which implies
- 5 = (2.k −1)**2

whence 2.k −1 is one of the square roots of five, so

- k = (1 ± √5) / 2

both values for which are necessarily irrational (i.e. not the ratio of any two whole numbers), since √5 is. Since √5 > 1, the – case of ± yields a negative solution; the only positive solution is

- k = (1 + √5) / 2

This last is known as **the golden ratio**. Its value
is roughly 1.618

So, I am bound to ask myself, can we do an analogous thing in higher dimensions ? Geometrically, that would mean cutting a cube out of the end of a cuboid. But this won't yield a cuboid unless the original cuboid has a square cross-section perpendicular to its longest side. So, instead, consider the problem as one in the world of ratios; can we get k>h>1 for which the ratios among k, h and 1 match those among k−1, h−1 and 1, in some order ? Clearly this requires that the latter triad's 1 not try to take the place of the former triad's 1; which, in turn, requires that h−1 < 1. We are then left with two possibilities:

- k−1 > 1 > h−1
If these are in the same ratios as k>h>1, then we have k−1 = k/h and 1/(h−1) = h. The latter implies that h is the golden ratio; the former can be re-arranged to give k as 1/(1−1/h) = h/(h−1) = h.h = 1+h (exploiting the above properties of the golden ratio).

- 1 > k−1 > h−1
If these are in the same ratios as k>h>1, then we have k−1 = h/k and h−1 = 1/k, whence 1+1/k = h = k.(k−1), so k+1 = k.k.(k−1) and 0 = 1 +k +k.k −k.k.k.

We have a cubic polynomial equation to solve, so apply the standard technique. First scale the equation by −27 then substitute y = 3.k−1 to obtain

- 0 = 27.(k.k.k −k.k −k −1)
- = (3.k).(3.k).(3.k) −3.(3.k).(3.k) −9.(3.k) −27
- = (y+1).(y+1).(y+1) −3.(y+1).(y+1) −9.(y+1) −27
- = y.y.y +3.y.y +3.y +1 −3.y.y −6.y −3 −9.y −9 −27
- = y.y.y −12.y −38

which is y.y.y −3.E.y −2.F with E = 4 and F = 19, yielding F.F −E.E.E = 361 −64 = 297 > 0. This is consistent with our only having one (real) solution for y; it will be p+q for some p, q whose cubes are 19±√297 = 19±3.√33. I use

`python`as my calculator:>>> 33.**.5 5.7445626465380286 >>> 3 * _ 17.233687939614086 >>> 19 + _, 19 − _ (36.233687939614086, 1.7663120603859142) >>> map(lambda x: x**(1./3), _) [3.3090564799660944, 1.2088037856763885] >>> _[0] + _[1] 4.5178602656424829

That gives y's value; then k = (y+1)/3 = 1.839286755214161 and h = 1+1/k = 1.5436890126920764.

The generalisation to dimension n would then call for a decreasing list (:q|n) with q(n−1) equal to 1 and (: (q(n−2)−1).q(i) ←i :n−1) consisting of the same entries as (: q(i)−1 ←i :n−2) with a 1 inserted somewhere in the latter list. The sequence from this insertion point onwards will simply be the sequence from some (possibly smaller) n's solution with 1 inserted at the start, analogous to the second case above; any earlier entries in q will simply be obtained by backwards extrapolation, as in the first case above. Thus the only interesting cases are the ones where the 1 is inserted at the start; thus

- q(0).(q(n−2) −1) = 1 and
- q(i+1).(q(n−2) −1) = q(i) −1

for each i in n−1, with the case i = n−2 being fatuous as q(i+1) = 1. We can use the first identity to write q(n−2) −1 as 1/q(0), so the other can be re-written as q(i+1) = q(0).(q(i) −1) for each i in n−2. Starting at i = 0 and working upwards, we thus express each q(i) as a polynomial in q(0), of degree i+1, ending with q(n−2) = 1 +1/q(0) equal to some polynomial of degree n−1 in q(0). Multiplying through by q(0) and rearranging, we get a simple polynomial equation in q(0), of degree n. Best of all, the polynomials for q(i) in terms of q(0) don't depend on n; if we take k = q(0), we get:

- q(0) = k
- q(1) = k.(k −1)
- q(2) = k.(k.k −k −1)
- q(3) = k.(k.k.k −k.k −k −1)
- q(4) = k.(k.k.k.k −k.k.k −k.k −k −1)

and, for general i > 0, q(i) = k.(power(i, k) −(power(i, k) −1)/(k −1)), as may be confirmed by observing that it holds true for the case i = 1 and, inductively, gives

- q(1+i) = k.(q(i) −1)
- = k.(k.(power(i, k) −(power(i, k) −1)/(k −1)) −1)
- = k.(power(i+1, k) −k.(power(i, k) −1)/(k −1) −1)
- = k.(power(i+1, k) −(power(i+1, k) −k)/(k −1) −(k −1)/(k −1))
- = k.(power(i+1, k) −(power(i+1, k) −k +k −1)/(k −1))
- = k.(power(i+1, k) −(power(i+1, k) −1)/(k −1))

which is the corresponding form for i+1. We thus infer

- 1 +1/k = q(n−2) = k.(power(n−2, k) −(power(n−2, k) −1)/(k −1)), whence

Multiplying through first by k, then by k−1, this becomes:

- k.k −1
- = k.k.(power(n−2, k).(k −1) −power(n−2, k) +1)
- = power(n+1, k) −2.power(n, k) +k.k

whence 0 = 1 +power(n+1, k) −2.power(n, k), which we can equally write as 2 = k +1/power(n, k). Given how we obtained this, it is no surprise that k = 1 is a solution (we recently multiplied by k−1), but we know this isn't a relevant solution (it would give q(i) = 0 for all i > 0 and we require q(i) > 1 for all i < n−1). Rearranging and dividing through by k−1, we obtain:

- power(n, k) = sum(: power(i, k) ←i |n)

in which the two sides grow monotonically in k, when k is positive, for any n > 1, as does the derivative of each, with the left side ultimately growing faster than the right. At k = 1, the left side is 1 and the right is n > 1; at k = 2, the left side exceeds the right side by 1; and our problem requires k > 1. Thus, for each n > 1, the solution is unique, with 1 < k < 2. We can sensibly extend to n = 1, with the equation reducing to k = 1 as the fatuous solution.

Let P(n) = (: power(n, x) −sum(: power(i, x) ←i |n) ←x :) be the polynomials we're solving; let ({reals}: K :{naturals}) give its solutions, such that P(n, K(n)) = 0, with 1≤K(n)<2 for each positive natural n. Observe that P(1+n) = P(n) +power(1+n) −2.power(n), evaluated at K(n) < 2, gives P(1+n, K(n)) = 0 +(K(n) −2).power(n, K(n)) < 0 with P(1+n) monotonically increasing between 1 and 2, hence K(1+n) is necessarily greater than K(n), so K is (strictly) monotonically increasing – each of our generalised golden ratios is bigger than the previous.

Written by Eddy.