source file: m1412.txt
Date: Mon, 11 May 1998 10:42:44 -0500 (CDT)

Subject: 8 major triads possible with 12 pitches?

From: Paul Hahn <Paul-Hahn@library.wustl.edu>

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 <manynote@lib-rary.wustl.edu> http://library.wustl.edu/~manynote
    O
   /\        "Churchill? Can he run a hundred balls?"
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