The Silver and Bronze Means

The silver and bronze means are not well-known as such.  They arise from the extension to 3-D of the triangular matrix.  Triangular matrices generate convergent ratios when applied recursively to a base vector.

The silver and bronze means are closely associated with the regular heptagon and the seven-pointed star.

AB=s; DE=m; FG=l.

Imagine that you could unfold this star across the base AB so that C would now be on the other side, pointing down.  Then DEC would be a triangle similar to ABC.  Fold once more across DE and now C is at the top again and triangle FGC has a base equal to the current long arm and a side equal to the sum of the three.

The following sequence converges to the limit s:m:l where s (=AB) is the side of the heptagon, m (=DE) is the distance between alternate vertices and l (=FG) is the distance between vertices 3 apart.

Using the letters s, m, l to denote the smaller, medium and larger terms, s/l converges to about 4/9, s/m to about 5/9 and m/l to about 4/5.

^s/_m + ^s/_l = 1.

We propose to call s/m the silver mean and s/l the bronze mean.

The obvious geometric intepretation of the 3-D ratio is a box with dimensions s, m, l.

Recall that the geometry of the golden mean leads to an infinite series of squares (and a spiral) as we step through the Fibonacci sequence.  What can be the properties of the magic box?  Well, here are two:

the long diagonal from corner (0, 0, 0) to corner (s, m, l) has length l+m-s

the volume sml = ( (l-m) × (m-s) × s ) + ( (s+m-l) × s × (l-s) ) + 3s ³

- that's two smaller magic boxes and three cubes.

The Matrix Aspect

We can generate the 3-D series by recursively transforming a base vector by the 3-D triangular matrix:

We could multiply the resultant vector by the matrix in a recursive manner or we could multiply the matrix by itself first and then the vector to achieve the same result.  This generates a series of matrices, successive powers of the first.  As for the golden mean, the identity matrix in the centre is preceded/succeeded by mutually inverse matrices, stepping 1 power up/down.

Unlike the 2-D triangular matrix, the 3-D triangular matrix can be composed as the product of two other matrices:

These two matrices are members of the core of a field of matrices which conserve the s:m:l ratio.  The core members of the field comprise the solutions of the 3-D equations (there are four).  There are 6 matrices in the group, corresponding to transformations from s to m, s to l, m to l and their inverses.  As in the case of the golden mean, negative solutions are included: so 12 solutios need 4 cubic equations.

The 3-D Equations

These are cubic equations and as in the 2-D case, each ratio appears as a solution, both negative and positive.  Since there are 6 ratios in the core (s/l, s/m, m/l and inverses), this makes 4 equations.  To derive them one approach is:-

s/m = m/(l+ m) = 1 - l/(l + m) = 1 - s/l;

s/m . m/l = s/l; m/l = (s/l) / (1 - s/l);

(s + m + l) / (s + m + l) = 1 = s/(s + m + l) + m/(s + m + l) + l/(s + m + l);

=(^s/_l))² + ^s/l/_{(1 - s/l)} . ^s/_l + ^s/_l;

substituting r for s/l we obtain: r² + r²/(1 - r) + r = 1;

multiplying through by (1 - r): r²(1 - r) + r ² + r(1 - r) = 1 - r;

multiplying by -1: r³ - r² - 2r + 1 = 0

This has the solutions s/l (of course), m/s and -l/m. The other three equations are:

r³ - 2r² - r + 1 = 0;

r³+ r² - 2r - 1 = 0;

r³ + 2r² - r - 1 = 0;

Another approach would be to write down all the anticipated solutions in some order and make an educated guess how to group them into four groups of three.  Fortunately this works in the 3-D case.

(ordered by increasing size) -l/s, -m/s, -l/m, -m/l, -s/m, -s/l, s/l, s/m, m/l, l/m, m/s, l/s

Group 1: -l/s, -s/m, m/l

Group 2: -m/s, -s/l, l/m

Group 3: -l/m, s/l, m/s

Group 4: -m/l, s/m, l/s

Graphical Interpretation

The four cubics are the same shape, all being translations of the equation r³ - ⁷/₃ r = 0

The width of the cubic at AB is a maximum: as you move up or down the end approaching the turning point loses more width than is gained by the end moving outward.  AB is fractionally longer than (l+m-s)/s in Figure 7.

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