Dec. 10th, 2011

stoutfellow: Joker (Joker)
So, the Big D is really pissed over Fuchsia's defection, and he's taking it out on the guys at DevilTech. (I don't think we've ever heard of them before, but I could be mistaken.)

Pass over the fact that he refers to Fyoosh as a "henchman". PC language isn't his forte. The thing is, he creates the devil-girls himself; DevilTech has nothing to do with it. If he made a bad selection (and it's increasingly looking as though the orange girl, whatever her name is, is another), it's his fault, not theirs.

Definitely unfair. But then, he's the Devil. Fairness isn't his forte either.

("The Devil - the prowde spirit - cannot endure to be mocked", as Thomas More tells us.)
stoutfellow: Joker (Joker)
The onset of cold weather (low of 10F this morning, and it'll go lower) has exposed several shortcomings in my wardrobe: not enough long-sleeved shirts, fraying long johns, a slight shortage of dress pants.... I just got back from a visit to Kohl's, repairing those deficiencies. They are (of course) having a sale, and I got something like 50% off on everything. I should have done this earlier, but time hasn't been on my side.

I gave the last lectures of the semester last week. In my geometry class, I'd run out of material from the text, so I finished off with a chat about my current research. (Of course, this won't be on the test....) I also was on the committee for a Masters' presentation on Thursday (tough material, and well-presented); I'm on another one next Wednesday, and my Senior Project student will be making her presentation on Thursday. She wants to do a dry run for me on Wednesday.... I'll be giving my calculus final on Monday, and the take-homes from the other two classes are due Tuesday, which is also the day of the department Christmas party. Lots to do, and not much time: the family Christmas party is Saturday, so I'll be flying out on Friday. I've found a volunteer to dogsit, as well.

I could use a break. The only troublesome thing is that my research is moving well, and being away from my software may be a hindrance. On the other hand, not having Mathematica to calculate for me, I'll have no choice but to just think, which is really what I need to do at this point. We'll see.
stoutfellow: Joker (Joker)
One for the funny coincidence file:

My current research involves identifying possibly-interesting classes of polygons. (You know: like rectangles, or trapezoids, or whatever.) I've devised a systematic procedure for doing so. Unfortunately, the procedure is algebraic, so I get a description of the class of polygons in terms of an equation. Extracting geometric content from that equation is a difficult task (and one I need to improve on before I can get anything publishable from this stuff).

So. One class of quadrilaterals that looks as though it should be very interesting has been refusing to reveal its secrets to me, and I've been worrying at it for a week or so now, without making any visible progress. It involves geometrically describing a certain line (or, actually, a pair of perpendicular lines), which points in... a certain, algebraically-described direction.

This morning, as I was fixing dinner, an idea struck me. I've seen that direction before. I published a paper about three years ago, in an area which isn't really my cup of tea; I had stumbled across something, and it turned out to be publishable. Well, that very paper mentions that pair of lines! I just needed to recognize that.

I'm not quite out of the woods, though. Here's the deal. Associated with every quadrilateral is a parallelogram, which I call its "first component"; the difference between the parallelogram and the quadrilateral is a vector, the "second component". Now, if you have a parallelogram, there's an ellipse which is tangent to all four sides at their midpoints. The directions of the axes of that ellipse are the directions I'm looking for; and the geometric description is that the second component points along one of the axes of that ellipse. Short of actually constructing the first and second components and then the ellipse, though, I don't see how to relate that to the original quadrilateral - but that should be a much easier task than I was originally faced with.

(It's got to be an interesting class. I've been able to verify that it includes all parallelograms, kites, isosceles trapezoids, antiparallelograms, and "flats" - quadrilaterals whose vertices all lie along the same line.)

Serendip ho!

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