I do it like erislover and caracal does it, but I take notice of where both ends are in the “tube”. Tighten the tube up until the ends are at the greatest diameter of the tube (the end on the inside of the tube is at one end of the circle and the outside edge is at the other end of the circle…when both edges are at the diameter, flatten tube. You’ll end up almost perfect folds. I know I didn’t explain very well, but if you try it you’ll see what I mean.
Teebone
i put the envelope flush to the top of the letter
put a dot on the paper just back from the bottom of the envelope
and fold
second fold is to match first
its never 1/3 exactly
so far all my letters have reached the recipients but they have not sent me a reply??
Roadfood has it right. Use your envelope.
Ah, I forgot to state an assumption – or, actually, a restriction I was placing on the conditions. (Ain’t that always the case with mathematics?) You can certainly third a piece of paper with origami. What I’ve never been able to do, and have never seen done (and have thereby, perhaps wrongly, assumed to be impossible) is third a piece of paper directly, without imposing a bunch of non-third folds on the paper – i.e. without putting weird superfluous folds across your letter. I’ve always associated this with the angle trisection problem – the straightedge being the edge of the paper and the angle being the angle from any reference point on the paper’s surface to the corners of the edge you’re trying to trisect. However, I’m no mathematician and only an amateur origamist, so I’ll defer to the experts in either field.
On reflection, I suppose this is an even more restrictive set of conditions than the standard angle-trisection problem, because the straightedge in this setup is “locked” in place relative to the angle. So if the straightedge-compass angle-trisection is impossible, this setup should be even impossibler. Once again, I’ll defer to a true mathematician.
I vote with Roadfood and rodent (hmmm, do those two go together?). Which leads me to ask the OP and others:
Why do you want to fold it exactly in thirds? It should have two folds and fit into the envelope, and ideally the folds should be exactly perpendicular to the edges, but why exactly thirds?
BTW I fold the bottom up then the top down, then insert the paper into the envelope with the folded-down top edge facing the back of the envelope.
I fold one end over without creasing it until I am sure that I will end up with two sections of equal width showing. When sure, I crease the fold, insert the other end into the fold and make the second crease. I was off by less than 1/32" when I tried it a minute ago.
bryanmcc, I’m having a hard time seeing exactly what constuctions you intend to be allowed in your system.
Also, I should point out this:
These are two entirely different things. Consider, for example, an isosceles right triangle. If we trisect one of the 45[sup]o[/sup] angles, we end up with a 15[sup]o[/sup] angle, of course. Furthermore, if we assume that angle trisection is equivalent to trisecting the edge opposite that angle (in any given triangle), we would expect that tan(15[sup]o[/sup]) = 1/3, which is clearly not the case.
OK, thanks.
You so crazy.
Look, I’m not the sharpest tool in the shed and so it’s probably me being slightly dim, but I’ve tried this with a few different squares of paper and it doesn’t seem to work.
In front of me is a square that’s 21 centimetres a side. I’ve followed the instructions, and the resulting intersection is 6.7cm up the left edge. Now, I grant you that this is close to the required 7cm point, but it’s no closer than I could probably get by approximation.
Possibly intriguing footnote: there have been a few references in this thread to visual approximation, which some people find easier than others. Among other things, I’m a magician (some UK Dopers have even seen me do a little card magic) and one thing many magicians need to practise is cutting a deck exactly in half. In other words, cutting 26 cards out of 52. Not 25, not 27, but exactly 26. The weird thing is that this is possible. The weirder thing is how easy it becomes after a little practice. Those who can do it simply reach a point where they can actually tell the difference between 25 cards and 26. Strange but true.
I know that a lot of people access the SDMB from work. Now I have a mental picture of people all over the country/world sitting at their desks playing with paper, trying to get it to fold just so, trying to follow various suggestions and recommendations from their fellow dopers. Workers of the world, FOLD! You have nothing to lose but your jobs!
CC, I lost interest in my question hours ago. Now I’m just making origami boulders. 
O gees I just caught something. You’re not supposed to fold the letter in exact thirds. The top edge should stick out a little, to make it easier to open. If you screw up, and do it exactly in thirds, hey, it’s OK, say law vee.
It concerns me that people are placing such a great emphasis on having the perfect fold, especially in light of some of the odd methods being suggested here 
My method is simple because the substantial majority of all things I put in envelopes are letters. I have created a standard letter template that I use with all my letters, so I know that the bottom of the text on the second subject line is right at the top 1/3 mark.
I just fold the bottom third up to that mark and make the bottom crease, then fold the top third down and make the top crease.
As for non-letters, I just eye it and do quite well. Sometimes, guys, you just have to let it go and not worry about a less than perfect fold. So please just take it easy out there before you all bleed to death from papercuts while practicing all these new techniques.
Give it to me. Being able to fold a piece of paper into even thirds just by eyeballing it is my stupid human trick.
ianzin, it’s probably either do to an inaccuracy in your folds, or maybe you’re not using a perfect square (or maybe you misunderstood what I said).
Here’s an outline of the proof (which may be a little difficult to follow since I can’t provide a drawing):
Assume the square is one unit length per side. Show that the fold I described will hit the right edge of the square at a point 5/8 units from the bottom right corner.
After the fold has been made, consider the triangle formed by the (right half of the) top edge and right edge of the square (the right edge of the square provides two sides of the triangle, since it’s been folded). This is a right triangle whose sides have the proportions 3:4:5.
By similarity, the triangle formed by the (left half of the) top edge, bottom edge, and left edge, is also a 3:4:5 right triangle. In particular, for this triangle, the ratio of the top edge to the left edge is 3/4. Since the top edge is 1/2 unit long, the left edge is 2/3 units. This shows that the intersection of the left edge with the bottom edge is 1/3 unit from the bottom left corner.