Convex Polygons

[Orgasmic Troll]

Originally Posted by 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.)

Translation is an isometry, so "origin" vs. "arbitrary interior point" is moot.

Vertices on P need not be pushed to vertices on Q, just to the boundary of Q.

It's not a complete proof, there are conditions on how far a vertex can be pushed out such that P' is convex, but these are minor considerations. However, moving Q could be tricky. Once you've moved it, how do you know that P is still contained in Q?

[mfgoode]

Originally Posted by TravisT

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)

Yes Travis! what is the origin and where is the vertex and in relation to what.?

You have cloaked a simple geometrical descritption into high brow topological language. It certainly does not impress me I can assure you.

First brush up on elementary geometry the way we elders have been disciplined in.

If you know any thing on pyramids esp. the Pyramid of Cheops you will find the largest stones have been put at the base and the vertex is among the smallest of the rest. BTW it is no longer there symbolising the unending knowledge possible reaching to the highest pinnancle one can imagine.

[QUOTE = Travis] 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)[/QUOTE]

Thats a partial Koch !

Mally

[Orgasmic Troll]

Originally Posted by mfgoode

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

your understanding of convex is wrong. (edit: On a second reading, your notion of convex may be correct. I assumed you were providing the koch snowflake as a contradiction to Davar's "conjecture." If this is the case, you are definitely wrong, but after reading the statement "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," there are two alternate possibilities: you have posted a worthless nonsequiter or utter gibberish. As I find these alternatives to be more damning, I leave my post unchanged from my initial assumption)

I have heard of a koch curve and a snowflake, enough to know the difference between the two.

Since you won't do your homework, here are three equivalent definitions of a (simple) convex polygon (in Euclidean space):

i) the intersection of a finite number of half planes
ii) a polygon P such that if x and y are in P, then the line segment containing x and y is entirely contained in P
iii) a polygon P where every interior angle is less than pi

I hope that you will put the effort forth on your own to see why the koch snowflake fails to satisfy the definition.

BTW, "sheesh" is an expression of exasperation.

[mfgoode]

Originally Posted by Xyzzy

Live in the moment. Forget the past.

(I think the Dog Whisperer says something like this!)

How very true Xyzzy.

Here one on the same lines- a poem found in 'Easy Eddie's" pocket when he was riddled by bullets in a lonely Chicago Street. He was the lawyer for Al Capone and then mended his ways and turned from crime to decency.

The poem read:
The clock of life is wound but once
And no man has the power,
To tell just when the hands will stop
At late or early hour.
Now is the only time you own,
(So) Live, love, toil with a will,
Place no faith in time,
For the clock may soon be still.

I hope you like it as a motto of mine.

Mally

[philmoore]

This is not a proof, only the mere beginning of an attempt to make Travis' idea work:

Let the inner polygon have the consecutive vertices P[sub]1[/sub], P[sub]2[/sub], ... , P[sub]n[/sub]. Pick a point C in the interior of this polygon and draw rays from C through each vertex, as in Ernst's idea, which then intersects the outer polygon in the respective points Q[sub]1[/sub], Q[sub]2[/sub], ... , Q[sub]n[/sub]. The polygon defined by these Q-vertices has a perimeter less than or equal to that of the outer polygon and also contains the inner polygon, so it suffices to prove that the perimeter of the Q-polygon is greater than or equal to that of the inner, P-polygon.

Now, using C as the center, or origin, consider the transformation which takes each P[sub]k[/sub] to a point P'[sub]k[/sub] which is either s times as far from the origin C or is the point Q[sub]k[/sub], whichever is closer to C. As the parameter s starts at 1 and increases to the maximum of all the distance ratios CQ[sub]k[/sub]/CP[sub]k[/sub], we will deform the original polygon into the polygon defined by the Q-vertices. It is not true, as Richard (Wacky) has explained above, that the length of each side individually necessarily increases in length, but it seems to me that we should be able to use the convexity here to prove that the total distance does increase.

Perhaps the idea of Zeta-Flux, to instead deform the outer polygon to the inner, can help here. Perhaps, rather than moving all the vertices together, the Q-vertices can be moved inward one-by-one in an order that guarantees that the perimeter only decreases.

Travis' idea has a certain appealing elegance, but I am finding that some of the details of implementation are rather subtle!

[davieddy]

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.

Mally

I don't know the history of this (presumably) persona non grata
but an endorsement from the intellectually challenged Mally
is the worst possible encouragement I could imagine.

[Zeta-Flux]

Originally Posted by TravisT

Translation is an isometry, so "origin" vs. "arbitrary interior point" is moot.

Vertices on P need not be pushed to vertices on Q, just to the boundary of Q.

It's not a complete proof, there are conditions on how far a vertex can be pushed out such that P' is convex, but these are minor considerations. However, moving Q could be tricky. Once you've moved it, how do you know that P is still contained in Q?

Fix any direction vector (or any line). One can translate Q along this vector (or line) the minimum distance until Q intersects P. This guarantees one point of intersection. Rotation about this point of intersection will guarantee a second (if it doesn't already exist).

[Orgasmic Troll]

Originally Posted by Zeta-Flux

Fix any direction vector (or any line). One can translate Q along this vector (or line) the minimum distance until Q intersects P. This guarantees one point of intersection. Rotation about this point of intersection will guarantee a second (if it doesn't already exist).

You're assuming you can rotate and keep P inside Q. Also, you haven't explained to me why having points of intersection is a good thing anyway

[Orgasmic Troll]

Okay, maybe this will be clearer

Look at three consecutive edges of P, call them e1, e2 and e3

we can add a triangle that has e2 as one of its edges as long as the opposite point is contained in the region outside of P bounded by the lines containing e1, e2, and e3 (this region is either a triangle or a an infinite trunctad "cone") and this new polygon (P with the triangle added) will still be convex and have greater perimeter than P by the triangle inequality.

[Zeta-Flux]

Travis,

It is obvious you can rotate until Q bumps up against P at a second point. You just rotate the minimum amount until this happens. (If Q already intersects P at two points, or along a line, no rotation is necessary.)

I was going to leave it to the rest of you to figure out why this is beneficial. The reason is as follows.

Label all of the vertices of Q (along with all intersection points between P and Q--ignoring line segments in common). Then, given any two consecutive edges which are not colinear, you can force them to be colinear, by moving the middle vertex. This always results in a convex polygon with smaller perimeter. There are two cases.

1) Doing this straightening doesn't make Q hit P. In this case, the straightening reduces the number of vertices.

2) During the straightening process, Q hits P. Then the straightening has to stop, and there is a new intersection point between Q and P. One must show that only finitely many new vertices can be added in this way (left to the reader). Eventually, the edges of Q and the edges of P become common edges.

Thus, after a finite number of steps, the outer polygon is reduced to the smaller polygon, and the perimeter always lost length.

Cheers,
Zeta-Flux

P.S. A better way than translating and rotating, is just linearly reducing the entire perimeter of Q until it cannot be reduced further without going inside P (i.e. take the smallests polygon which is similar to Q, but also still able to contain P).

[mfgoode]

Originally Posted by davieddy

I don't know the history of this (presumably) persona non grata
but an endorsement from the intellectually challenged Mally
is the worst possible encouragement I could imagine.

Well Davi you can see for your self. This problem is an excellent one given by Davar 55.

Its a general rule that original thinkers are never welcome anywhere.

Instead of using the poison pen try cracking this problem out for a change.

I have pondered it long and deep and tried to work it out but came to a conclusion that the inner perimeter turns out to be greater than the outer one ! so Im waiting for enlightenment from you.

Mally