Convex Polygons

[ewmayer]

Originally Posted by Wacky

But, no trees were harmed by your experiment :)

Except possibly the one I banged my head against after realizing my elementary error - but I expect "its bark was worse than my beat".

[Orgasmic Troll]

Originally Posted by ewmayer

Except possibly the one I banged my head against after realizing my elementary error - but I expect "its bark was worse than my beat".

At what level of atrocity do puns become a bannable offense? If this wasn't over the line, it should be within spitting distance.

[Wacky]

Back to the original "conjecture"

Originally Posted by davar55

Show that a convex polygon completely containing another convex polygon necessarily has a greater perimeter.

In the trivial case where the polygons are identical, davar's conjecture fails.
However, I don't think that we should allow ourselves to dismiss it on such a technicality.

I also think that we can easily show that we can drop the first "convex". Any polygon completely containing a convex polygon cannot have a smaller perimeter than that of the polygon that it contains.

This is because any concave polygon necessarily has a larger perimeter than a convex one which shares the same "exterior" vertices.


I note that "Travis" made a reference to "over the line". Perhaps he is related to that revered Defender of the Alamo, William Barret Travis, who drew a line in 1836.

[cheesehead]

Suggested proof method: Divide the region between the two polygons into subregions such that a properly-constructed computation of the total area of the subregions between the two polygons, using the convexity properties wherever useful, shows that a nonzero area between the polygons implies that the outer perimeter exceeds the inner perimeter.

That is, find a way of expressing the area of the region between the polygons as a function of the lengths of all (inner and outer) perimeter segments such that a nonzero area between the polygons implies that the outer perimeter exceeds the inner perimeter. The area function must contain other terms such as vertext angles, but construct it in such a way that those angle-terms all become irrelevant to showing that a nonzero area implies that the outer perimeter exceeds the inner.

[Orgasmic Troll]

Originally Posted by Wacky

I note that "Travis" made a reference to "over the line". Perhaps he is related to that revered Defender of the Alamo, William Barret Travis, who drew a line in 1836.

Nope, Travis is my first name, and despite my surname starting with a T, my name is not Travis Travis, that would be a travesty indeed.

[Orgasmic Troll]

Given any convex polygon P, if a vertex is moved away from the origin to form a convex polygon P', P' has a larger perimeter than P

Thus if P is contained in Q, there exists a deformation of P into Q by a sequence of convex polygons with monotonically increasing perimeters.

(vertices can be added onto P as necessary)

[mfgoode]

Originally Posted by davar55

Show that a convex polygon completely containing another
convex polygon necessarily has a greater perimeter.

I am very happy you have made a comeback. We need independent and original thinkers like you to keep us pondering. Before someone says whose the 'we' i speak for my self and use the royal plural.

Well I can see your very first problem has tied all the big wigs in a knot with their impressive proofs.

Well you are not quite right in shall we say your conjecture. It is true in most polygons but not in all and therefore cannot be generalised.

There is a certain curve called Koch's curve or snowflake known after his name which encloses a finite area but is infinite in length. If you construct it there will come a time when its perimeter will be greater than any that encloses it so you dont have to go so far as to strictly not be a convex polygon.

Yes if you had to say the same about areas then that would be more plausible in two dim.

en.wikipedia.org/wiki/Koch_snowflake - 20k

Mally

[Orgasmic Troll]

Originally Posted by mfgoode

I am very happy you have made a comeback. We need independent and original thinkers like you to keep us pondering. Before someone says whose the 'we' i speak for my self and use the royal plural.

Well I can see your very first problem has tied all the big wigs in a knot with their impressive proofs.

Well you are not quite right in shall we say your conjecture. It is true in most polygons but not in all and therefore cannot be generalised.

There is a certain curve called Koch's curve or snowflake known after his name which encloses a finite area but is infinite in length. If you construct it there will come a time when its perimeter will be greater than any that encloses it so you dont have to go so far as to strictly not be a convex polygon.

Yes if you had to say the same about areas then that would be more plausible in two dim.

en.wikipedia.org/wiki/Koch_snowflake - 20k

Mally

CONVEX polygon, mally. Go look it up before you post anymore nonsense.

sheesh

[cheesehead]

Originally Posted by TravisT

< snip >

Ah, how the rust has accumulated since I was young and mentally agile! Too much thinking in numerical mode and not enough visualizing in dynamic geometrical mode nowadays.

Well, also there were reasons I suspect I wouldn't have made it as a professional mathematician anyway -- too often missing the simple stuff even back then.

[Zeta-Flux]

Travis,

What do you mean by the origin? Are you talking about an arbitrary point in the interior, or are you assuming the polygon has the origin in its interior? In any case, I don't think you have sufficiently justified the answer (for example, what happens if Q has LESS vertices?).

I think your argument can be modified a little however. (I would start by moving the outer polygon so that it has at least 2 points in common with the inner polygon. Then "reduce" the outer polygon to the inner.)

[mfgoode]

Originally Posted by TravisT

CONVEX polygon, mally. Go look it up before you post anymore nonsense.

sheesh

I dont recognise the end 'sheesh' Travis as I know your surname begins with T as your first name. Should I take it as a mideast term 'sheesh kebab' lamb sandwich ?
Well if you are of the erroneous opinion that my understanding of convex is wrong and you have a better idea of it pray kindly enlighten me dear sheesh.
In the meantime please take the trouble of looking up what a Koch curve is
I'm sure you have never heard of it in your formal education.

Mally