OKay, we came up with x=1/(1+sqrt(2)=sqrt(2)-1.

Trying A) x = [-2 +/- sqrt (4-4*1*-1)]/2 = -1 +/- sqrt(8)/2 = -1 + sqrt(2), so A works.

Not a pretty question.

I am enchanted by the math, but I do not know from whence it came.

so, I am at work, not doing work and no one else is really here and so it is quiet and peaceful and whatnot. That is nice. Natter is put put putting, though, yes? are people working? or not at work? or too bored to post?

I'm at work, not working, but was all caught up in doing math. And now I'm going home.

It was the math talk. It drove people away

I'm sulking because no one cares about the sheep urine bus.

Kidding.

I'm the only one here now. So no one will know if I go home early.

I have concocted a clever plan which relies upon this fact....

What I did to make it work, -t (like how I fold and spindle my sentence structure to avoid the hyphen at the beginning of the line?) is the following:

There's a bar vertically down the center which is 1 by x. There's another one horizontally which overlaps, so: x + x - x^2. Then you have the 4 triangle bits, each of which have a hypotenuse of x, so you could put them all together (hypotenuses out) to form a square of side x, so: 2x - x^2 + x^2 = 2x. So the only thing outside the octagon is four more little triangles which are exactly the same, so another x^2, and that fills out the original square, so: x^2 (outer triangles) + 2x (horizontal and vertical bars) = 1 (area of original square), so x^2 + 2x - 1 = 0.

More tangrams than geometry, but there you have it.

the urine bus disturbs me. moreso wondering why someone came up with the idea.

There's a bar vertically down the center which is 1 by x. There's another one horizontally which overlaps, so: x + x - x^2. Then you have the 4 triangle bits, each of which have a hypotenuse of x, so you could put them all together (hypotenuses out) to form a square of side x, so: 2x - x^2 + x^2 = 2x. So the only thing outside the octagon is four more little triangles which are exactly the same, so another x^2, and that fills out the original square, so: x^2 (outer triangles) + 2x (horizontal and vertical bars) = 1 (area of original square), so x^2 + 2x - 1 = 0.

That? Makes me want to run screaming. I don't know how you mathy types do it.

Sorry, everybody. I have about a month and a half to cover linear algebra, discrete math, trigonometry, and number theory, so that's pretty much my main focus (especially since I only have the rest of the month before I'm doing two clases as well).