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KL
science forum beginner

Joined: 05 Feb 2005
Posts: 12

Posted: Fri Jun 30, 2006 12:13 pm    Post subject: Decomposition of convex polygons

Hi all,

I'm trying to find a way to build an arbitrary convex polygon out of
isoceles triangles (for use in a 3d computer construction), and I've
just about convinced myself this is impossible. Can it be done? What if
the triangles are 'nearly' isoceles (say within 50%)? And of course,
*how* could this be done? The polygon is specified only by the lenght
of all of the sides.

The limitation to isoceles (or nearly such) is built into the computer
system I am working with.

This seems to me to be a relatively interesting geometrical question
which is simply beyond my abilities anymore. I appreciate any help
y'all can give. TIA.

- KL
Peter Webb
science forum Guru Wannabe

Joined: 05 May 2005
Posts: 192

Posted: Fri Jun 30, 2006 3:05 pm    Post subject: Re: Decomposition of convex polygons

"KL" <k.lambda@gmail.com> wrote in message
 Quote: Hi all, I'm trying to find a way to build an arbitrary convex polygon out of isoceles triangles (for use in a 3d computer construction), and I've just about convinced myself this is impossible. Can it be done? What if the triangles are 'nearly' isoceles (say within 50%)? And of course, *how* could this be done? The polygon is specified only by the lenght of all of the sides. The limitation to isoceles (or nearly such) is built into the computer system I am working with. This seems to me to be a relatively interesting geometrical question which is simply beyond my abilities anymore. I appreciate any help y'all can give. TIA. - KL

If I understand you properly (and I probably don't) then it seems very
easily possible.

Pick any point on the surface P and draw a "circle" one unit away. Fill this
with as many isoceles triangles with the pointy end on P and the base
connecting two points on the "circle" - say 6 triangles, so each is
equilateral if it is a flat point. Pick any point on the circle that you
have already used and choose this as your new centre P, and repeat ... you
will tile your surface with icoceles triangles, and "almost" equilateral if
you want. The smaller the size of the unit you use, the finer the modelling,
and the same is true for the number of triangles you fill each circle with,
but I don't know how to optimise the answer.
KL
science forum beginner

Joined: 05 Feb 2005
Posts: 12

Posted: Fri Jun 30, 2006 6:01 pm    Post subject: Re: Decomposition of convex polygons

Peter Webb wrote:
 Quote: "KL" wrote in message news:1151669627.649574.303670@75g2000cwc.googlegroups.com... Hi all, I'm trying to find a way to build an arbitrary convex polygon out of isoceles triangles If I understand you properly (and I probably don't) then it seems very easily possible. Pick any point on the surface P and draw a "circle" one unit away. Fill this with as many isoceles triangles with the pointy end on P and the base connecting two points on the "circle" - say 6 triangles, so each is equilateral if it is a flat point. Pick any point on the circle that you have already used and choose this as your new centre P, and repeat ...

Sorry. Say I want to contruct an irregular pentagon out of isoceles
tris- I don't see how your method would do it.

perhaps I should have been more concrete initially...

- KL
jasen
science forum beginner

Joined: 28 Jun 2006
Posts: 16

Posted: Fri Jun 30, 2006 10:07 pm    Post subject: Re: Decomposition of convex polygons

On 2006-06-30, KL <k.lambda@gmail.com> wrote:
 Quote: Hi all, I'm trying to find a way to build an arbitrary convex polygon out of isoceles triangles (for use in a 3d computer construction), and I've just about convinced myself this is impossible. Can it be done? What if the triangles are 'nearly' isoceles (say within 50%)? And of course, *how* could this be done? The polygon is specified only by the lenght of all of the sides.

any polygon can be constructed of arbitrary triangles, so your problem
reduces to constructing an arbitrary triangle from isosceles triangles,

any triangle can be constructed from two right-angle triangles,

(divide it from the longest side to the other corner at a right-angle to the
longest side)

and any right triangle can be constructed from two isosceles
triangles, divide it from the centre of longest side (hypoteneuse) to the
right-angle.

is that proof enough ?

--

Bye.
Jasen
Patrick Hamlyn
science forum beginner

Joined: 03 May 2005
Posts: 45

Posted: Sat Jul 01, 2006 4:22 am    Post subject: Re: Decomposition of convex polygons

jasen <jasen@free.net.nz> wrote:

 Quote: On 2006-06-30, KL wrote: Hi all, I'm trying to find a way to build an arbitrary convex polygon out of isoceles triangles (for use in a 3d computer construction), and I've just about convinced myself this is impossible. Can it be done? What if the triangles are 'nearly' isoceles (say within 50%)? And of course, *how* could this be done? The polygon is specified only by the lenght of all of the sides. any polygon can be constructed of arbitrary triangles, so your problem reduces to constructing an arbitrary triangle from isosceles triangles, any triangle can be constructed from two right-angle triangles, (divide it from the longest side to the other corner at a right-angle to the longest side) and any right triangle can be constructed from two isosceles triangles, divide it from the centre of longest side (hypoteneuse) to the right-angle. is that proof enough ?

That's nice... is there any way of proceeding in similar fashion to prove that
any polygon can be constructed from equilateral triangles?
jasen
science forum beginner

Joined: 28 Jun 2006
Posts: 16

Posted: Sun Jul 02, 2006 2:26 am    Post subject: Re: Decomposition of convex polygons

On 2006-07-01, Patrick Hamlyn <path@multipro.N_OcomSP_AM.au> wrote:
 Quote: jasen wrote: On 2006-06-30, KL wrote: Hi all, I'm trying to find a way to build an arbitrary convex polygon out of isoceles triangles (for use in a 3d computer construction), and I've just about convinced myself this is impossible. Can it be done? What if the triangles are 'nearly' isoceles (say within 50%)? And of course, *how* could this be done? The polygon is specified only by the lenght of all of the sides. any polygon can be constructed of arbitrary triangles, so your problem reduces to constructing an arbitrary triangle from isosceles triangles, any triangle can be constructed from two right-angle triangles, (divide it from the longest side to the other corner at a right-angle to the longest side) and any right triangle can be constructed from two isosceles triangles, divide it from the centre of longest side (hypoteneuse) to the right-angle. is that proof enough ? That's nice... is there any way of proceeding in similar fashion to prove that any polygon can be constructed from equilateral triangles?

any polygon, or any convex polygon?

there's a simple recipe to do any convex polgon,
pick any point on the perimeter of the polygon (or even inside it)
and divide the polygon into triangles along lines to every other vertex,
then treat the triangles individually.

I think it can be shown that polygons with concaves can also be done (by
dividing them into convex polygons and then into triangles.

treatment for polygons where lines cross (eg 4 points describing a "bow-tie"
or 5 describing a pentagram) depends on what it means when the edges cross.

Bye.
Jasen
Peter Webb
science forum Guru Wannabe

Joined: 05 May 2005
Posts: 192

Posted: Mon Jul 03, 2006 5:47 pm    Post subject: Re: Decomposition of convex polygons

"Patrick Hamlyn" <path@multipro.N_OcomSP_AM.au> wrote in message
news:03uba25cgpdhjs3hhq6qmatain3ttutln9@4ax.com...
 Quote: jasen wrote: On 2006-06-30, KL wrote: Hi all, I'm trying to find a way to build an arbitrary convex polygon out of isoceles triangles (for use in a 3d computer construction), and I've just about convinced myself this is impossible. Can it be done? What if the triangles are 'nearly' isoceles (say within 50%)? And of course, *how* could this be done? The polygon is specified only by the lenght of all of the sides. any polygon can be constructed of arbitrary triangles, so your problem reduces to constructing an arbitrary triangle from isosceles triangles, any triangle can be constructed from two right-angle triangles, (divide it from the longest side to the other corner at a right-angle to the longest side) and any right triangle can be constructed from two isosceles triangles, divide it from the centre of longest side (hypoteneuse) to the right-angle. is that proof enough ? That's nice... is there any way of proceeding in similar fashion to prove that any polygon can be constructed from equilateral triangles?

No. Every vertex has n equilateral triangles meeting, so must be of angle
n*pi/6.

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