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.
whence 2.k -1 is one of the square roots of five, so
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
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:
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).
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
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 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 (:k|n) with k(n-1) equal to 1 and (: (k(n-2)-1).k(i) ←i :n-1) consisting of the same entries as (: k(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 k will simply be obtained by backwards extrapolation, as in the first case above.
Written by Eddy.