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From: William Elliot on 7 Mar 2010 02:33
On Fri, 5 Mar 2010, Brian Chandler wrote:

> Richard Tobin wrote:

>> Given a square ABCD, centre O, with the triangle ABO removed, is it
>> possible to dissect the remainder AOBCD into four congruent pieces?
>
> Um, I'm no good at ASCII art, and not much good at any art, but will
> this napkin sketch do?
> http://imaginatorium.org/private/puzzle.png
>
Guessing how the crumpled, stained and smudged paper may read,
it appears the same method could be used to dissect the remainder
into 2^n (n = 0,1,2,...) congruent pieces.

> (How does one really attack proving things about puzzles like this?
> I wish I knew.)
>
By insight and intuition as you did and, in this case,
by "going outside of the box" of a connected piece.

(Richard. I do not reply to posts that don't contain the original
problem. Reconstructing them takes too much unneeded time and effort.

----
From: Brian Chandler on 7 Mar 2010 09:06
William Elliot wrote:
> On Fri, 5 Mar 2010, Brian Chandler wrote:
> > Richard Tobin wrote:
>
> >> Given a square ABCD, centre O, with the triangle ABO removed, is it
> >> possible to dissect the remainder AOBCD into four congruent pieces?
> >
> > Um, I'm no good at ASCII art, and not much good at any art, but will
> > this napkin sketch do?
> > http://imaginatorium.org/private/puzzle.png
> >
> Guessing how the crumpled, stained and smudged paper may read,
> it appears the same method could be used to dissect the remainder
> into 2^n (n = 0,1,2,...) congruent pieces.

It does? I don't immediately see how... In particular, I can only see
one way to divide Richard's 'M' shape into two congruent bits, by
drawing a line down the middle. In fact, it's easy to find a way of
dissecting a resulting half (call it 'h', of which it is roughly the
envelope) into four 'h' shapes of linearly half size, so it _is_ clear
that a dissection into 2^(2n+1) pieces is possible. But I can't
immediately find a way to divide the 'M' into 16 pieces of half-linear-
size like my pieces on the napkin, and I _certainly_ can't see it's
obvious that this process could be repeated indefinitely. I can see
it's "obvious" that the 'M' cannot be dissected into any number of
smaller 'M's, because whatever corner stuck into the peak of the 'M'
there would be another corner alongside in the wrong place.

But this is all dreadfully ad hoc, which is what my question was
about. I suppose we can take comfort(?) from David Cantrell's posts on
circle packing -- I presume he's some sort of expert on this, from the
number of results he produces, yet it all seems inelegant, just
computational crunching. Presumably this is all we can manage?

The original 'M' problem just "looked" soluble, because it divides
into 6 triangles in the obvious way, so each piece has to be 1 1/2 of
these... But consider, for example, a hexagon, made of 6 equilateral
triangles in the obvious way; remove one triangle. Can the remaining
five-triangle shape be dissected into four congruent shapes? I guess
not, but I can't see how to _begin_ making any sound subclaims on the
way to proving it.

Brian Chandler
From: William Elliot on 8 Mar 2010 00:53
On Sun, 7 Mar 2010, Brian Chandler wrote:

> William Elliot wrote:
>> On Fri, 5 Mar 2010, Brian Chandler wrote:
>>> Richard Tobin wrote:
>>
>>>> Given a square ABCD, centre O, with the triangle ABO removed, is it
>>>> possible to dissect the remainder AOBCD into four congruent pieces?
>>>
>>> Um, I'm no good at ASCII art, and not much good at any art, but will
>>> this napkin sketch do?
>>>
>> Guessing how the crumpled, stained and smudged paper may read,
>> it appears the same method could be used to dissect the remainder
>> into 2^n (n = 0,1,2,...) congruent pieces.
>
> It does? I don't immediately see how... In particular, I can only see
> one way to divide Richard's 'M' shape into two congruent bits, by
> drawing a line down the middle. In fact, it's easy to find a way of
> dissecting a resulting half (call it 'h', of which it is roughly the
> envelope) into four 'h' shapes of linearly half size, so it _is_ clear
> that a dissection into 2^(2n+1) pieces is possible. But I can't
> immediately find a way to divide the 'M' into 16 pieces of half-linear-
> size like my pieces on the napkin, and I _certainly_ can't see it's
> obvious that this process could be repeated indefinitely. I can see
> it's "obvious" that the 'M' cannot be dissected into any number of
> smaller 'M's, because whatever corner stuck into the peak of the 'M'
> there would be another corner alongside in the wrong place.

Use monspace font and adjust for vertical stretch.
Here, without explantion, is ascii art for the cases n = 0,1,2
or 1,2 and 4 congruenet pieces. For n = 3, ie 8 congruenet
peices, divide all the triangles again by droping a perpendicular
to the hypotneuse from the opposite vertex. Continue for each
successor n.

------------------
\ | /|
\ | / |
\ | / |
\ | / |
\ | /\ |
\ | / \ |
\ | / \ |
\|/ \|
X--------|
/|\ /|
/ | \ / |
/ | \ / |
/ | \/ |
/ | \ |
/ | \ |
/ | \ |
/ | \|
~~~~~~~~~~~~~~~~~~

> But this is all dreadfully ad hoc, which is what my question was
> about. I suppose we can take comfort(?) from David Cantrell's posts on
> circle packing -- I presume he's some sort of expert on this, from the
> number of results he produces, yet it all seems inelegant, just
> computational crunching. Presumably this is all we can manage?
>
> The original 'M' problem just "looked" soluble, because it divides
> into 6 triangles in the obvious way, so each piece has to be 1 1/2 of
> these... But consider, for example, a hexagon, made of 6 equilateral
> triangles in the obvious way; remove one triangle. Can the remaining
> five-triangle shape be dissected into four congruent shapes? I guess
> not, but I can't see how to _begin_ making any sound subclaims on the
> way to proving it.
>
From: Brian Chandler on 8 Mar 2010 04:25
William Elliot wrote:
> On Sun, 7 Mar 2010, Brian Chandler wrote:
>
> > William Elliot wrote:
> >> On Fri, 5 Mar 2010, Brian Chandler wrote:
> >>> Richard Tobin wrote:
> >>
> >>>> Given a square ABCD, centre O, with the triangle ABO removed, is it
> >>>> possible to dissect the remainder AOBCD into four congruent pieces?
> >>>
> >>> Um, I'm no good at ASCII art, and not much good at any art, but will
> >>> this napkin sketch do?
> >>>
> >> Guessing how the crumpled, stained and smudged paper may read,
> >> it appears the same method could be used to dissect the remainder
> >> into 2^n (n = 0,1,2,...) congruent pieces.
> >
> > It does? I don't immediately see how...

> Here, without explantion, is ascii art for the cases n = 0,1,2
> or 1,2 and 4 congruenet pieces. For n = 3, ie 8 congruenet
> peices, divide all the triangles again by droping a perpendicular
> to the hypotneuse from the opposite vertex. Continue for each
> successor n.

Perhaps I'm being dim, but I still can't see how your diagram divides
the 'M' (well, yours is a capital sigma...) into four congruent
pieces. Here it is again, with the triangles labelled:


------------------
\ | /|
\ | / |
\ a | b / |
\ | / y |
\ | /\ |
\ | / \ |
\ | / x \ |
\|/ \|
X--------|
/|\ /|
/ | \ z / |
/ | \ / |
/ | \/ |
/ | \ w |
/ c | d \ |
/ | \ |
/ | \|
~~~~~~~~~~~~~~~~~~

Can you name the four pieces as made from these 8 triangles? I can't
see how the *whole* piece can possibly be the same shape.

Brian Chandler
From: William Elliot on 8 Mar 2010 06:00
On Mon, 8 Mar 2010, Brian Chandler wrote:
> William Elliot wrote:
>> On Sun, 7 Mar 2010, Brian Chandler wrote:
>>
>>> William Elliot wrote:
>>>> On Fri, 5 Mar 2010, Brian Chandler wrote:
>>>>> Richard Tobin wrote:
>>>>
>>>>>> Given a square ABCD, centre O, with the triangle ABO removed, is it
>>>>>> possible to dissect the remainder AOBCD into four congruent pieces?
>>>>>
>>>>> Um, I'm no good at ASCII art, and not much good at any art, but will
>>>>> this napkin sketch do?
>>>>>
>>>> Guessing how the crumpled, stained and smudged paper may read,
>>>> it appears the same method could be used to dissect the remainder
>>>> into 2^n (n = 0,1,2,...) congruent pieces.
>>>
>>> It does? I don't immediately see how...
>
>> Here, without explantion, is ascii art for the cases n = 0,1,2
>> or 1,2 and 4 congruenet pieces. For n = 3, ie 8 congruenet
>> peices, divide all the triangles again by droping a perpendicular
>> to the hypotneuse from the opposite vertex. Continue for each
>> successor n.
>
> Perhaps I'm being dim, but I still can't see how your diagram divides
> the 'M' (well, yours is a capital sigma...) into four congruent
> pieces. Here it is again, with the triangles labelled:

> ------------------
> \ | /|
> \ | / |
> \ a | b / |
> \ | / y |
> \ | /\ |
> \ | / \ |
> \ | / x \ |
> \|/ \|
> X--------|
> /|\ /|
> / | \ z / |
> / | \ / |
> / | \/ |
> / | \ w |
> / c | d \ |
> / | \ |
> / | \|
> ~~~~~~~~~~~~~~~~~~
>
> Can you name the four pieces as made from these 8 triangles? I can't
> see how the *whole* piece can possibly be the same shape.
>
Ok, I get your point. The best I can do is two pieces
of a large and a small triangle that are disconnected
and two pieces of a large and a small triangle sharing
a point.

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