21 Mar 07
Originally posted by luskinProve it.
What's all the fuss about here? As Xanthos has conceded "Without the allowance of 0.5m the answer is trivial (as many sides as you can have as you start to approximate a solution)". So with or without that stipulation, the closest approximation to a circle is a regular convex polygon with as many sides as possible.
21 Mar 07
Originally posted by PBE6In your post you said constructed according to the stipulations, so I naturally thought that you were still trying to include the edges shortening rule when you don't need to. My original post was trying to say that.
You're not reading very well, either. That's exactly what I'm trying to do! Geez, just read leisurelysloth's post if you refuse to understand mine.
So far we have:
1. The polygon with largest area for a given perimeter is convex.
2. The polygon with largest area for a given perimeter is cyclic.
21 Mar 07
Originally posted by luskinIt's intuitively quite clear that this is the case. But that's not the same as proving it, which isn't trivial at all.
What's all the fuss about here? As Xanthos has conceded "Without the allowance of 0.5m the answer is trivial (as many sides as you can have as you start to approximate a solution)". So with or without that stipulation, the closest approximation to a circle is a regular convex polygon with as many sides as possible.
21 Mar 07
Originally posted by mtthwYes I know. I'm not trying to prove it. My point was that XanthosNZ(who posed this question in the first place remember) seems to imply that, without the 0.5metre reduction factor, the proof would be trivial. I don't see how that makes any difference to the 'regular polygon conjecture'.
It's intuitively quite clear that this is the case. But that's not the same as proving it, which isn't trivial at all.
21 Mar 07
4 edits
The area of a cyclic polygon is calculated by drawing a point in the middle of the containing circle and joining the vertices of the triangle to the dots, we get n triangles each with two sides of length r (where r is the radius of the circle) and edge length L(i) where i labels the edge. We want to maximize the area with the constraint that sum L(i) = constant = C. We therefore need to find extrema of the function:
F = sum_i (1/2 * L(i) * sqrt(r^2 - L(i)^2) ) + lamda * (sum_i ( L(i) ) - C)
dF / dL(i) = 1/2 * sqrt(r^2 - L(i)^2) *( (r^2 - 2*L(i)^2) / (r^2 - L(i)^2)) + lambda = 0
This equation depends only on r, the radius of the circle passing through all the polygons vertices and lambda. This shows that all the L(i) are the same.
So we have that the cyclic polygon with largest area is equilateral. All cyclic polygons that are equilateral are regular. So the conjecture is proven.
21 Mar 07
1 edit
Of course I haven't shown that cyclic polygons have the largest area, I can only show that if for a given set of polygons with some fixed set of edge lengths at least one is cyclic (then if you keep the bits of arc from the circle that goes through the cyclic one's vertices any non-cyclic polygon with the same edges is contained in a shape made from the bits of arc and that shape is smaller than a circle with the same perimeter and so the non-cyclic polygons are smaller than the cyclic ones - for the same set of edge lengths).
21 Mar 07
Originally posted by DeepThoughtVery confusing. What I understand from your response is that a cyclic polygon can be reshaped to be non-cyclic, but that non-cyclic polygon will not take up as much room within the same circle. However, this is meaningless because if you stretch a cyclic polygon into a shape approaching a line (keeping the side lengths and number of vertices/bends constant of course), you will find that shape no longer fits inside the same circle. Of course, the area will be much smaller than a regular polygon, but the circles have (and prove) nothing to do with this.
Of course I haven't shown that cyclic polygons have the largest area, I can only show that if for a given set of polygons with some fixed set of edge lengths at least one is cyclic (then if you keep the bits of arc from the circle that goes through the cyclic one's vertices any non-cyclic polygon with the same edges is contained in a shape made from the ...[text shortened]... the non-cyclic polygons are smaller than the cyclic ones - for the same set of edge lengths).
21 Mar 07
2 edits
Originally posted by PBE6No, you've misunderstood what I'm doing. Draw a cyclic polygon on a piece of card, also draw the circumscribing circle, this circle includes each of the vertices of the polygon on its circumference. Cut out the polygon and also cut along the circumference of the circle, so you get n bits of card which are segments of the circle. The straight parts of the segments can be fitted together to form all polygons with the same initial edge lengths as the cyclic polygon, with the arcs pointing outwards. The area of the polygon so formed is the area of the enclosing shape bounded by the arcs minus the area of the segments. Either the enclosing shape is a circle, in which case the area of the polygon is equal to the area of the cyclic one or it is not, in which case the area of the new shape must be smaller than that of the cyclic polygon because the area of the segments is fixed and the perimeter of the bounding shape is the same length as the circles circumference but it is not a circle - and we are allowed to assume that the circle is the shape with maximum area for a given circumference.
Very confusing. What I understand from your response is that a cyclic polygon can be reshaped to be non-cyclic, but that non-cyclic polygon will not take up as much room within the same circle. However, this is meaningless because if you stretch a cyclic polygon into a shape approaching a line (keeping the side lengths and number of vertices/bends constant of ...[text shortened]... e much smaller than a regular polygon, but the circles have (and prove) nothing to do with this.
Provided for each polygon with some set of fixed edge lengths there exists at least one cyclic polygon with the same edge lengths then we have proved the conjecture. I haven't a clue how you'd go about proving that though.
21 Mar 07
Originally posted by DeepThoughtAha! Still confusing, but much clearer than before. As I now understand:
No, you've misunderstood what I'm doing. Draw a cyclic polygon on a piece of card, also draw the circumscribing circle, this circle includes each of the vertices of the polygon on its circumference. Cut out the polygon and also cut along the circumference of the circle, so you get n bits of card which are segments of the circle. The straight parts of ...[text shortened]... hen we have proved the conjecture. I haven't a clue how you'd go about proving that though.
1) For every cyclic polygon, circumscribe a circle and link the constructed arc segments to their respective polygon sides.
2) The area of the polygon is equal to the area of the circumscribed circle minus the arc segments (which have fixed area).
3) The sum of the exterior perimeters of the arc segments is equal to the circumference of the circle.
4) As the vertices are moved, the area changes but the exterior perimeter of the figure does not.
5) Using the result that a circle encompasses the greatest area for a given perimeter, greater than any arbitrary curve with the same perimeter, the configuration of the polygon with attached segments that creates the original circumscribed circle contains the greatest area.
This proves the result for a cyclic polygon. However, it does not apply to non-cyclic polygons and it is not always possible to make a cyclic polygon out of a non-cyclic one. So the proof is still elusive.
21 Mar 07
6 edits
Originally posted by PBE6We need to show that given some set of edge lengths {L(i) | i=0, ... , n-1} you can always construct at least one polygon which is cyclic. This is definitely true for triangles (all triangles are cyclic). It might be possible on that basis to construct an iterative proof like the one you mentionned earlier.
Hmmm, maybe you're right...I was thinking about a square with one of the corners pulled very far out, but if you are allowed to re-arrange the order of the sides then it seems possible. Anyone got a proof that it is?
Edit: just thought of this: Take n vertices and n-1 edges, put them onto the circumference of a circle and scale the circle up or down, keeping all the edge lengths fixed so that the vertices move relative to their angular position on the circle, until you can fit in the last edge. This shows that for a set of edge lengths it is always possible to find a cyclic polygon and is the last bit we need to complete the proof, unless there's another loophole.
21 Mar 07
Originally posted by luskinI just don't see how your posts add anything to this thread. I'm not trying to prove it, I'm just saying.
Yes I know. I'm not trying to prove it. My point was that XanthosNZ(who posed this question in the first place remember) seems to imply that, without the 0.5metre reduction factor, the proof would be trivial. I don't see how that makes any difference to the 'regular polygon conjecture'.
21 Mar 07
Originally posted by PBE6conjecture: Any arbitrary set of n sides, which can form an n-gon, can also form a cyclic n-gon and the circle will be identical regardless of the order in which the sides are arranged.
Hmmm, maybe you're right...I was thinking about a square with one of the corners pulled very far out, but if you are allowed to re-arrange the order of the sides then it seems possible. Anyone got a proof that it is?
21 Mar 07
Originally posted by leisurelyslothJust checked Wikipedia under "cyclic polygons"...apparently any triangle, quadrilateral or simple regular polygon is cyclic, but it's mute on polygons with 5 sides or more (I'm assuming this is because it's not always true for them, but I have no proof).
conjecture: Any arbitrary set of n sides, which can form an n-gon, can also form a cyclic n-gon and the circle will be identical regardless of the order in which the sides are arranged.