Convex Polygons

[davar55]

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

[99.94]

davar55 is back?

[akruppa]

On parole.

Alex

[Xyzzy]

Live in the moment. Forget the past.

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

[ewmayer]

Originally Posted by akruppa

On parole.

Alex

I though we agreed on Animal Hous-style double-secret probation?

[philmoore]

I think he is redeeming himself with a great puzzle.

[petrw1]

Originally Posted by davar55

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

As a teacher of my high school friend used to say: "The answer is intuitive to even the most casual observer."

Is that a good enough answer?

[ewmayer]

Originally Posted by petrw1

As a teacher of my high school friend used to say: "The answer is intuitive to even the most casual observer."

Is that a good enough answer?

No, because intuition is not the same as mathematical proof.

Anyway, surely many different ways to prove this, so the question comes down to which proofs are the "most simple and elegant," i.e. of searching for an Erdos-style "book" proof.

A simple one I thought of, which also strikes me as reasonably elegant - it would need a few more dotting-of-the-i's to make into a formal proof, but I think those needed steps are clear from the outline:

Pick a point lying strictly inside the inner polygon. This is necessarily also inside the outer polygon. Then, a ray drawn from this "origin" point will necessarily intersect both polygons. Draw as many such rays as are needed so that one ray intersects each vertex of each of the 2 polygons (some rays may intersect one vertex of each polygon, but the important thing is, no vertex is left ray-less.) Now, any pair of adjacent rays segments off a straight-line segment of each polygon. Since the rays are all distinct and the 2 polygons do not cross (although they may have coincident segments), the length of the line segment of the outer polygon defined by any 2 adjacent rays is always >= the length of the line segment of the inner polygon defined by the same 2 rays, and will in fact be strictly > unless the 2 line segments exactly coincide, in which case the lengths will be equal. Now just add up the lengths of all the resulting line segments, and you get

(length of perimeter of outer polygon) >= (length of perimeter of inner polygon),

with equality obtaining only if the 2 polygons are identical. QED

[Wacky]

Originally Posted by ewmayer

Since the rays are all distinct and the 2 polygons do not cross (although they may have coincident segments), the length of the line segment of the outer polygon defined by any 2 adjacent rays is always >= the length of the line segment of the inner polygon defined by the same 2 rays

This is exactly the approach that I thought of. However, I realized that the above statement is not correct.

Please construct the contradiction as follows:

Draw two adjacent rays. Call their intersection the "origin". Choose a point on each to be a vertex of the inner polygon. Let one of them be significantly closer to the origin than the other. Now, let the farther point also be a vertex of the outer polygon. Construct a side of that polygon such that it is perpendicular to the other ray.

Now, within this sector, the portion of the side of the outer polygon is shorter than the side of the inner polygon.

[ewmayer]

Originally Posted by Wacky

This is exactly the approach that I thought of. However, I realized that the above statement is not correct.

Ah - good catch, Richard - serves me right for drawing an imaginary sketch in my head. That's "worth the paper it's drawn on," apparently.

So using this approach, one would need to show every such occurrence where an inner-polygon segment is longer than the corresponding outer-polygon line segment due the former intersecting the 2 rays sufficiently more obliquely would be more than compensated for by the instances where such a thing does not occur, i.e. the inequality holds for the respective sums, even though individual terms contributing to the latter may not obey the same inequality. That makes things rather more interesting. I'm still reasonably sure that there is a simple proof, but it may not lie in the above approach.

[Wacky]

Originally Posted by ewmayer

Ah - good catch, Richard - serves me right for drawing an imaginary sketch in my head. That's "worth the paper it's drawn on," apparently.

But, no trees were harmed by your experiment :)