8 major triads possible with 12 pitches?

On Fri, 8 May 1998, Paul H. Erlich wrote:
> Paul Hahn wrote,
> > 25/18 25/24 25/16
>
> > 10/9 5/3 5/4 15/8
>
> > 16/9 4/3 1/1 3/2 9/8
>
> [ . . . ] this may be the most just major triads
> one can achieve with 12 notes. Anyone care to prove this, or find a
> counterexample?

Well, I can prove it, but it's not particularly elegant. The rough
outlines of the proof go like this: eight just major triads have eight
1/1s, eight 3/2s, and eight 5/4s, or 24 members in all. To get this out
of fewer actual pitches many pitches must serve as different members of
multiple triads. No pitch can be a member of more than three different
major triads, and if you examine the geometry of scales in the
triangular lattice there must be at least three pitches in a scale that
each are members of only a single triad. That leaves at most 9 pitches
to function as the other 21 members. This requires at least three
pitches to serve triple duty (21 = 3 * 3 + 2 * 6), but the most
economical way (pitch-wise) to do that is this:

. .
. * .
. * * .
. . .

and that already uses up your twelve pitches and still only gets you
seven major triads. Q.E.D.

--pH http://library.wustl.edu/~manynote
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