Conference rusure::math

    Yes.  Just as you can have 2-dimensional objects with finite area but
    infinite perimeter.
    
    Many fractals have this property (such as the Mandelbrot set), but
    there are simpler objects with this property also.
    
    Consider the object bounded by:
    
    	The line segment X=1 for Y=(0 to 1)
    
    	The ray Y=0 for X>=1
    
    and The curve Y=X^2 for X>=1
    
    This object has finite area yet has an infinite perimeter.
    
    Similar 3D objects are easy to construct.
    
    --  Barry

    Re: .1
    
    I think you meant Y=X^(-2).
    
    Re: .0
    
    The solid of revolution created by revolving the region in .1 around
    the x-axis will have finite volume and infinite surface area.
    You should be able to find the detail of this in any calculus texts
    in the chapter on Improper Integrals. This is very well-known.
    This even have some Greek's name attached to it, if I remember
    correctly.

    Re .2:
    
    	Oops, right you are.
    
    Re .3:
    
    	I think that that is what we said in .1, and .2, ... ?
    
    --  Barry

The 2D object that was just described has an infinite size.  That is, there's
no finite-sized box we could "put" it in.

I'll describe an object whose area AND size are both fixed, yet it has an
infinite perimeter:

	Start with an equilateral triangle, each size length 1.

	Divide each side into equal thirds, and let a smaller equilateral
	triangle grow out of each , one side of which is a common edge with
	the middle third of a side of the larger triangle.

	So now the figure looks like a Star of David (jewish star).

	Again, divide each segment into thirds and let triangles grow out.
	Now the outer edge of the figure has 48 edges.

	Keeping doing this forever, and then...

	o.k. Let's prove that this has finite area but infinite perimeter.

	If the area of large triangle is A, we then add three little triangles,
	which each are A/9, so we've added 3*(A/9).

	The next set of smaller ones we add four on each side, in other words
	twelve total , and each is 1/9 area of the previous, so we're adding
	12*(1/9)*(A/9).

	Next, we'll add sixteen on each side, 48 total, each with 1/9 area
	of previous, so we're adding 48*(1/9)*(1/9)*(A/9).

	In fact, each time we add the next layer, we add one triangle per
	segment, but it introduces FOUR smaller segments for next time.
	So each time we'll have FOUR times more triangles, each of which is
	1/9 the previous area.

	Writing all that as a series, it's

	TOT_AREA = A + 3*(A/9) + 3*4*(1/9)*(A/9) + 3*4*4*(1/9)*(1/9)*(A/9)...

	Each term has a factor of 4*(1/9) multiplied in, which is a constant
	less than 1, so we have a finite geometric series.  (If this isn't
	clear, write to me, I have a neat way of explaining why geometric
	series work this way!).  The formula in general is s=a/(1-r) and in
	our case a is A, and r is 4/9, so

	TOT_AREA = 9A/5, or about twice the original triangle's area.


	Now to prove that perimeter is infinite.

	If we examine one side of triangle, the first layer buckles the
	middle third, so that perimeter changes from 3/3 to 4/3.

	In general, each resultant segment, which has some length, call it
	S, gets buckled by the next layer to have a new length of 4S/3.

	So each layer causes the entire perimeter to multiply by 4/3 so
	the perimeter is infinite.

	There's one more important thing to prove about this object.  That is,
	we must prove that the protruding sides never collide with each
	other, in which case the overlaps would need to be SUBTRACTED
	from some of the tallies.

	Such a subtraction would obviously keep the area finite, but what
	about the perimeter?

	Can anyone think of an easy way to prove we won't have a collision
	in this object ?

/Eric

    Re .5:
    
    Thats a fratcal, as I mentioned in .1, Eric.  It does meet your
    requirements of finte size & area with infinite perimeter, but as you
    saw there are pretty complicated objects.
    
    It so happens that there are fairly simple (non-fractal) objects which
    also have these properties.
    
    --  Barry

    A simpler example is a comb, where you repeatedly replace each tooth by
    two thinner teeth.
    
    It serves as a 2- or 3-dimensional example. However combing your hair
    in Flatland requires a much simpler implement, since your hair doesn't
    tangle.
    
    Dick (and it's not just Flatlanders who have little use for a comb :-)
    

    Actually, depending on how you define the spacing and width of the
    teeth, .7 either meets my definition of a fractal, or else fails to
    satisfy the common notion of an object (ie. parts of the perimeter are
    either indeterminate, or else enclose no area), or both.
    
    --  Barry

    For those of you of the non-Euclidean persuasion:

	Fix a circle of unit radius, C, and for N > 2, choose any N points 
    distinct on its circumference. Form the curved N-sided polygon by 
    connecting  each of the N pairs  of adjacent points which 
    are not antipodal with the unique circular arc which is orthogonal 
    to C; for an antipodal pair (at most one), connect them with 
    a straight line segment. 


	The constructed N-gon minus its 
    vertices can be intepreted as region of the Poincare disc 
    conformal model (i.e., angles but not distance are preserved)
    of hyperbolic/Lobachevskian geometry.

    The hyperbolic area of the N-gon region is just (N-2)*PI. 
    Each bounding arc is a complete hyperbolic line (geodesic,
    locally length minimizing) of infinite length. 

    
	For N = 3, all such constructed triangles have area PI and 
    are in fact congruent.
    


    -Franklin

    And for those of you with an Euclidian, Non-Fractal bent:
    
    Consider a logarithmic spiral starting at (1,0) and converging to
    (0,0).
    
    Now add a copy of it rotated 180 degrees about the origin (ie. a
    logarithmic spiral starting at (-1,0) ).
    
    Now close the 2 curves with a semi-circle from (1,0) to (-1,0).
    
    This object has Infinite perimeter, yet finite area AND the finite
    'size' that Eric requested in .5.  It is also non-fractal and
    completely Euclidian.
    
    --  Barry