Triangle to square
salonowiec
simple problem. Is it possible for irregular triangle?
Robert Israel
> Cutting equilateral triangle so that pieces create a square is a known and
> simple problem. Is it possible for irregular triangle?
>
Look up the Bolyai-Gerwien (or Wallace-Bolyai-Gerwien) theorem.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Philippe 92
> "salonowiec" <debrza...@poczta.onet.pl> writes:
>
>> Cutting equilateral triangle so that pieces create a square is a known and
>> simple problem. Is it possible for irregular triangle?
>>
>
> Look up the Bolyai-Gerwien (or Wallace-Bolyai-Gerwien) theorem.
Yep.
This theorem states than any given polygon can be dissected in a
finite number of pieces, so that pieces create any other given polygon.
For practical dissection, better search for "geometric dissections"
than theoretical Bolyai-Gerwien theorem.
The construction used to dissect equilateral triangles into a square
works perfectly with other irregular triangles, under some limit
conditions, in similar four pieces.
See http://mathafou.free.fr/pbg_en/sol110a.html for details and limits
for which the same dissectiion works.
Regards.
--
Philippe C., mail : chephip, with domain free.fr
site : http://mathafou.free.fr/ (mathematical recreations)
salonowiec
Uzytkownik "Philippe 92" <nos...@free.invalid> napisal w wiadomosci
news:mn.7c017db2e...@free.fr...
Thank tou, that's it... The above said Bolyai-Gerwien theorem is powerful
but gives no practical way of irregular triangle dissection...
Brian Chandler
> Uzytkownik "Philippe 92" <nos...@free.invalid> napisal w wiadomosci
> news:mn.7c017db2e...@free.fr...
> > Robert Israel wrote :
> >> "salonowiec" <debrza...@poczta.onet.pl> writes:
> >>
> >>> Cutting equilateral triangle so that pieces create a square is a known
> >>> and
> >>> simple problem. Is it possible for irregular triangle?
> Thank tou, that's it... The above said Bolyai-Gerwien theorem is powerful
> but gives no practical way of irregular triangle dissection...
You've had a general result, and a particular dissection (but it
doesn't seem to work in all cases). Here's a very simple three stage
process for any triangle.
(Let the area of the triangle be Q, and let s^2 = Q. So s is side of
square required.)
(1) For any triangle, slicing off half the height gives a triangle
which can be fitted back to make a parallelogram. (Cut between
midpoints of two sides)
(2) Any parallelogram can be rearranged to a parallelogram with one
side equal to s. Make a cut between two opposite sides of length s.
Need to show that this is possible: the product of the heights of the
parallelogram cannot exceed Q; therefore at least one height is not
more than s. But at least one diagonal is not less than s. Therefore
at some angle between perpendicularly across the parallelogram and
diagonally there is a cut of length s. These two pieces can be
rearranged to make a parallelogram with one side length s.
(3) Since one side is s, and area is Q, height must also be s. Cut
perpendicularly from the blunt corner at one end of this side. If the
blunt angle is not more than 135deg, you now have two pieces that make
a square. If it is more, you need to keep slicing, in a way that is
obvious once you draw it on paper, until you have a square.
In the one case we have only made 3 cuts, so cannot possibly have more
than 8 pieces. But in the second ("repeated slicing") case we may have
any number of pieces. We console ourselves by noting that there cannot
be a general maximum number of pieces. Why? Consider a triangle with
sides 2, 1000000, 1000001: we must make at least 999 cuts to get about
1000 narrow strips.
A quick trial with an "ordinaryish" triangle resulted in six pieces.
If you cut a triangle from a sheet of A3 paper, using a long side and
third vertex on the opposite side, the area is 1/2^4 m^2, so the side
of the square is exactly 25 cm, which ought to be the basis for some
sort of parlour trick.
HTH
Brian Chandler
Philippe 92
> salonowiec wrote:
>> Uzytkownik "Philippe 92" <nos...@free.invalid> napisal w wiadomosci
>> news:mn.7c017db2e...@free.fr...
>>> Robert Israel wrote :
>>>> "salonowiec" <debrza...@poczta.onet.pl> writes:
>>>>
>>>>> Cutting equilateral triangle so that pieces create a square is a known
>>>>> and
>>>>> simple problem. Is it possible for irregular triangle?
>
>> Thank tou, that's it... The above said Bolyai-Gerwien theorem is powerful
>> but gives no practical way of irregular triangle dissection...
>
> You've had a general result, and a particular dissection (but it
> doesn't seem to work in all cases). Here's a very simple three stage
> process for any triangle.
>
>
Yes, this is a way of proving the BWG theorem...
But ! It doesn't give the *minimum* number of pieces !
> A quick trial with an "ordinaryish" triangle resulted in six pieces.
>
But "ordinaryish" triangles require just 4 pieces !!
> If you cut a triangle from a sheet of A3 paper, using a long side and
> third vertex on the opposite side, the area is 1/2^4 m^2, so the side
> of the square is exactly 25 cm, which ought to be the basis for some
> sort of parlour trick.
Especially all those triangles cut from an A3 in that way.
Hence your puzzle :
On an A4 (A3 is too large for my compass and straightedge ;-)
mark a point C on the large side, opposite to large side AB, and cut
off triangle ABC.
Then construct a 4-pieces only dissection of triangle ABC into a square
**all drawings being inside the triangle itself** (sides and vertices
included).
spoiler to solution
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See figure at http://cjoint.com/?0cqbd5ZNcmv (available 21 days)
Construct midpoints M, N and O of resp. AC, BC and AB.
(easy, not detailed)
Construct altitude CH and K = intersect of CH with MN
(hence HK = half altitude)
Copy AI = HK
An arc of circle with center O and radius OA (diameter = AB)
intersects the perpendicular to AB from I in point P
AP is equal to the side of the equivallent square.
(For AP^2 = AI.AB = AB.CH/2)
Then construct D with ND = AP and E with DE = MN
Draw the perpendiculars MF and EG to ND
Cut along ND, MF, EG.