PSA for Clive and Calamrin - Surface area of two nested convex shapes

[jack]

If you have two convex 2D shapes edit: one entirely inside the other, why must the smaller have a smaller circumference?

Last time I saw this mentioned, I don't know why I didn't see the solution. Now, the answer suddenly seems to jump out at me. Take a section of the inner shape length delta. Draw normals to the line at either end of it. Then the outer shape must cross between these lines, and this section of its curve must be at least as long as a straight line approximating the inner curve.

Then integrate. The sum of the inner lines tends to the inner circumference, the outer lines are at least as long.

(Technical faff. You need to assume all this converges, which it should because its a convex shape. It's a slight cheat because if the inner line is curved, it's slightly longer than the straight line we're comparing the outer to -- I *think* this could be fixed by observing that being curved makes the nromals bend apart. It would need to be to show the outer is *strictly* greater. Corners are special cases, make sure the partition of the circumference includes them, and take normals to the lines on either side, thus missing out a bit of the outer shape, which would need to be remembered to show strictly greater.)

I think the same thing should work fine for the 3d case, except you'd need a surface integral, which is a bit of a faff to actually calculate.

OK, and I use the maths tag for real for the first time.

Comments

[gerald_duck]

Hmm. My original problem was expressed in terms of containment of one shape inside the other. You say "smaller", by which I assume you mean containment not, for example, a comparison of areas (…in the 2D case; volumes in the 3D case).

I have a concern, here. Your proof seems to rely on the obviousness of "and this section of its curve must be at least as long as a straight line approximating the inner curve". Wlog, I choose a section of the inner shape that is its entirety, and in that case this is suddenly non-obvious.

Or, at least, it's as obvious as it ever was, but has no more obvious a proof than ever. (-8

In fact, I think we did some good handwaving on Friday evening that I will describe in my journal this evening.

[cartesiandaemon.livejournal.com]

Smaller: yes, oops. Edited, thanks.

Ack. I see the problem. But not quite there: the line segment in that case would be zero length, which is shorter, but casts doubt on it being a good approximation to the curve.

But I think it is ok, because as you have more sections, the total length of straight segments tends to the length of the inner circumference (I think this is a definition), and the outer circumference always being longer[1] than each total length means it must be longer limit, ie. the inner circumference

[1] Non-strictly.

[simont]

Your proof seems to rely on the obviousness of "and this section of its curve must be at least as long as a straight line approximating the inner curve". Wlog, I choose a section of the inner shape that is its entirety, and in that case this is suddenly non-obvious.

Doesn't matter; the proof relies on this condition continuing to hold as the perimeter of the inner shape becomes more and more subdivided and the sections become smaller and smaller. So although the case you describe doesn't satisfy the condition, as you subdivide further and further it will begin to, and will continue to thereafter as you approach infinity.

That something is not true for N=0 does not prevent a successful proof that it's always true for N >= (say) 10.

[gerald_duck]

Hmm.

On re-reading, I realise that when Jack speaks of a straight line approximating a section of the inner curve, he has in mind specifically the one joining the endpoints of that section, not some line of best fit.

I'm now convinced. I think. (-8

[cartesiandaemon.livejournal.com]

Sorry, I should have included the diagram. HTML is not the best medium for this.

I'm now convinced. I think. (-8

I wavered myself. I think my way does work though, though the polygonal approximation step isn't *tidy*.

And come to think of it, I don't *know* this works in 3d.

What sort of method were you proposing?

[gerald_duck]

OK; I was intending to get around to procrastinating, but you've forced my hand. :-p

I've now written it up, here.

a geometric argument

[ewx.livejournal.com]

Let P and Q be distinct points on the inner curve If we draw normals to the curve at P and Q then these normals must diverge on their way out, converge on their way out, or be parallel.

For Jack's argument to work, for every P it must be the case that you can find a maximum distance along the curve between PQ such that they are parallel or diverge on their way out for any Q within that distance.

Suppose there is no such maximum distance. That is to say, no matter how close P and Q are, the normals converge on their way out at point C. PQC is a triangle. Consider the line PQ to be the base and C the top. Now the angles at P and Q must be strictly less than 90 degrees, because otherwise the angles of the triangle would add up to more than 180 (the angle at C cannot be 0 because then the normals would be parallel).

But the tangents to the inner curve at P and Q are at right angles to PC and QC (respectively). So they must point downwards with respect to the triangle, as illustrated:

So at least the ends of line PQ must lie outside the inner curve. But the inner curve is convex, so this cannot be the case, by definition of "convex".

What's it aclled?

[(Anonymous)]

Hello


G'night