I have one picture of us at the wedding this past weekend. I look like a basketball in a black dress.

The boy crashed out on the chair tonight.

Oh thank God! I've spent the better part of the past two nights wrestling with integrating the equation for a parabola (mostly wrestling with the arithmetic -- going from -b^3+2b^2sqrt(b^2-4ac)-b^3+4abc+.... to something slightly more compact just to get 1/6th of the equation written) and it has, despite all expectations, come out right (that is to say, I either had no stray dropped negatives in there, or I had enough that I could cancel them out!). It's taken up about five sheets of notebook paper... now to type it up semi-concisely. But if any of you need some kind of proof about the relationship between the area of a parabolic segment and the triangle inscribed therein, I'm your girl!

I have one picture of us at the wedding this past weekend. I look like a basketball in a black dress.

You look adorable!

Oh frell. I just realized that if I was going to assume it crossed the x-axis, I could just as well have assumed one endpoint was at 0, and then I would only have had to calculate from (-b+sqrt(b^2-4ac))/2a to 0, rather than from that to (-b-sqrt(b^2-4ac))/2a, but no. Instead I did twice the work, because I am a mow-ron.

(Sorry folks. Ignore the math and it'll go away. Just needed to vent.)

With the black dress and black tux we blend together like some freakish conjoined twins.

Gorgeous couple, cutiehead sleeping toddler

Annabel is mowing her way through a slice of pepperoni pizza. I never know what to expect when I give her foods these days.

The "pining for the fjords" sketch is on Monty Python on BBC America.

I hope I can get through the conference this weekend without my head exploding. In a possible bout of insanity, I volunteered to do the same job next year....

"This is an ex-parrot!"

How do you inscribe a triangle within a parabola? One point in the peak, one at each x-axis crossing? (assuming it's a parabola with exactly two real solutions)

Heh. Math geekery.

Oh, sorry, did I say that? It's meant to be a parabolic segment and its inscribed triangle, so I've said, basically, wherever the line is which cuts off the segment, move the segment so that line is at the x-axis and the segment is above it (so yes, the points of the triangle will be at the x crossings and at the vertex).

T:P is 3:4 right? That's a fun problem.

My method is below, all whitefonted for those that hate math (infidels!): Yeah, all parabolic segments can be defined by an upside-down (negative a) parabola with one of the solutions at (0,0) (so c=0). Then our parabola is of the form y = ax^2 + bx, with a < 0. Our key triangle points are the solutions to y=0: (0,0) and (-b/a, 0) and the peak, which is directly between them and turns out to be (-b / 2a, -b^2 / 4a) (can be found just by plugging x = -b/2a, the midpoint, or using derivative first to be sure). T = 1/2 * b * h, then, is b^3 / 8a^2

To find the area of the parabola segment, just need to integrate ax^2 + bx from 0->-b/a, which isn't too bad, and gives b^3 / 6a^2, so T:P = 3:4

I think this may have been a problem on the AP calc exam my senior year, actually. I've def' seen it somewhere. Figuring out the representation trick you did is definitely the crucial step.