Conference rusure::math

If the right triangle is isosceles, the answer is 1, because otherwise the final
division must produce two equal-area triangles.

But this case was presumably meant to be excluded...

I may not correctly understand your condition; please clarify if needed. 
According to my current understanding there is no upper bound if you follow the
strategy of always dividing the subtriangle with the smallest area.  Is there
some constraint I overlooked that forbids this?

re .1
    
>If the right triangle is isosceles, the answer is 1, because otherwise the final
>division must produce two equal-area triangles.
>
>But this case was presumably meant to be excluded...
    
    The intention of "start with an arbitrary right-angled triangle" was
    that you could start with any triangle you like (provided that it is
    right-angled) in an attempt to maximise the number of divided triangles
    possible. That probably wasn't clear. "Start with a general
    right-angled triangle" might have been better.
    
>I may not correctly understand your condition; please clarify if needed. 
>According to my current understanding there is no upper bound if you follow the
>strategy of always dividing the subtriangle with the smallest area.  Is there
>some constraint I overlooked that forbids this?
    
    I don't believe additional constraints are needed but perhaps I have
    not communicated clearly. Did you actually try your suggestion? The
    problem is that you cannot get away with only dividing one subtriangle.
    Continuing the example in .0, if you, for example, started with the
    third divided triangle (rather than starting with the original triangle
    and repeating the sequence of divisions) then you would *at least* have
    to divide every sub-triangle since otherwise there would be duplicates
    between the third divided triangle and your proposed fourth divided
    triangle.

    Derek
    
    I'm confused as well. I thought that always dividing the _smaller_
    sub-triangle would let you go on for ever. I must have misread
    something.
    
    
    > ...................................................... In order to
    > avoid duplicating an area from the second triangle we had to divide
    > both of the sub-triangles from the second triangle ..........
    
    Why? You could have just divided one of them:

            ^
           /|  `
          / |     `
         /  |    /   `
        /   |   /       `
       /    |  /           `
      /     | /               `
     /      |/                   `
    +-------+-----------------------+
    
    or
            ^
           /|  `
          / |     `
         /  |        `
        /   |           `
       /    |              `
      /`    |                 `
     /    ` |                    `
    +-------+-----------------------+
    
    Dividing both introduces an equal-area difficulty.
    
    Dick

    If a < b, where a and b are the areas you get from dividing the area-1
    triangle, then after n steps of dividing the smaller triangle the areas
    are
      b, ba, ba^2, ..., ba^(n-1), a^n
    which are in descending size and hence different.

re .0
    
>With the constraint that no sub-triangle in any divided triangle has the same
>area as any other sub-triangle in the same or another divided triangle,
    
    I think people may be overlooking that when looking for duplicate
    areas, the entire set of divided triangles is included. Thus, no
    matter how many unique sub-triangle areas you have in one divided
    triangle, it and the original triangle only make 2 divided triangles
    (which is not more than the 3 that I exhibited in .0).

    Sorry, I'm still not sure I get it, but I have a new guess now.
    
    Is the problem to find a maximal set of divided triangles such that
    there are no duplicated areas anywhere in the set (where only areas
    of single, i.e. non-divided, triangles count)?
    
    That is, roughly speaking, to maximize the number of "root" triangles?
    (I had thought, using this analogy, you started with one root and
    tried to maximize the number of leaves.)

    re .6                                      
    
>    Is the problem to find a maximal set of divided triangles such that
>    there are no duplicated areas anywhere in the set (where only areas
>    of single, i.e. non-divided, triangles count)?
    
    Yes.
    
    More specifically, we need only prove that the least upper bound for
    the size of the set is N (maximising over all possible original
    triangles�). From the example in .0 we can see that N is at least 3.
    
>    That is, roughly speaking, to maximize the number of "root" triangles?
>    (I had thought, using this analogy, you started with one root and
>    tried to maximize the number of leaves.)
    
    You start with one root, the original triangle, and try to maximise the
    number of leaves. At least in principle, all leaves are direct
    descendants of the root i.e. each leaf is formed by applying 0 or more
    divide operations to the original triangle.
    
    �As far as I can see there are only two cases for the original
    triangle. If it is isosceles then the size of the set is 1. It is not
    possible to divide the triangle and meet the conditions in .0. If it is
    not isosceles then the size of the set is N (regardless of the
    triangle). It remains to be proven what the value of N is.

N may be the same for all non-isosceles cases, but I'd like your reaction to the
following comment.

If a and b are the areas you get when dividing an area-1 triangle, then there
are extra duplicate areas to worry about whenever a^r = b^s for some positive r
and s.  Thus, if a = 1/phi is the inverse of the golden ratio, we have b =
1/phi^2 = a^2, so subdividing the area-a triangle creates a duplicate of the
area-b one.

If a is transcendental, this can't occur, and the algebraic numbers have measure
zero, so we can assume the "typical" case excludes this.  But is your derivation
of the value of N affected by this?

    re .8
    
    Yes, you're right. My claim that N is the same for all non-isosceles
    triangles is not justified (although without further thought not
    necessarily false).
    
    You are probably making this harder than is needed. Any unexpected
    duplicates just potentially reduce N for that triangle but not N as
    defined (least upper bound over all initial triangles). I suggest you
    assume that if the area could be different for some a (i.e. by looking
    only at the expression for the area) then it is.

    OK, I think I have it.
    
    If a = 1/2 (the isosceles case), N = 1.
    If a = 1/phi or 1/phi^2 (the golden ratio case), N = 2.
    Otherwise N = 3.
    
    So my nitpicking turns out to matter!  The corresponding hint for
    the golden ratio case is to note that all areas are powers of 1/phi.
    
    I'll post my proof in my next reply.

Assume N >= 4.  Then by induction we will show that for every n >= 0, there are
at least 2n+4 triangles (either further divided, or not) whose area is
expressible as a degree-n term in a and b (i.e., a^r*b^s where r+s=n and r,s are
nonnegative integers).  This means we have an infinity of triangles, so N<=3.

For n=0, we have at least 4, so the base of the induction holds.

Otherwise, suppose this is true for n=m.  Of the 2m+4 triangles, at most m+1 are
not further subdivided (one each for a^n, a^(n-1)b, ..., b^n), so at least m+3
are further subdivided; this means there are at least 2(m+3) = 2(m+1) + 4
triangles whose area is expressible as a degree m+1 term.

Now to show N >= 3.  The 3 triangles given in .0 have subtriangles with areas 1,
a, b, a^2, ab, b^2, a^2b, and ab^2.  If a /= b, using obvious inequalities shows
that the only possible duplication of areas occurs if a = b^2 or a^2 = b, both
of which lead to the golden-ratio case.

The remaining nontrivial part is to show that N <= 2 in the golden ratio case. 
A proof similar to the one above, left to the reader, shows that if we start
with >= 3 triangles, for every even power P of 1/phi there must be at least 3
triangles of area P.

    re .11
    
    A delightful proof, more than justifying the difficulties I had in
    communicating the problem. (Notes is at its worst on geometry type
    problems.) The proof in .11 is much nicer than the one that I was aware
    of.