Yarr!

[sorcyress]

Fuck man, I love math. Maths. Whatever. It's just so...perfect! And the numbers line up and do exactly as they're supposed to, and then Bam! Finished math test, without ever having stopped to think.

And I'm *good* at it. I love being good at something that people tend to have trouble with, it's just soothing to know that I'm not a complete waste of Earth. And of all the things in the world to be good at, I love that it's simple sums, just adding and subtracting and multiplying the numbers in my head until they do exactly as I say.

Math is soothing, is what I'm trying to say. I enjoy it. The classes I've had so far this year that I've liked the most have been the ones where I go in and take a maths exam.

In short, eight AM statistics exam, after a night spent up late watching Coupling? Still didn't stand a chance. I kicked that exams ass so hard that it had to go import some cheap, lead painted ass from China, just so I could kick that too.

~Sor
MOOP!

Comments

[heptadecagram.livejournal.com]

[heptadecagram.livejournal.com]

(Yes, this can be done without using trig identities, only geometry)

[shield-toad111.livejournal.com]

Fun. I feel all validated that vet school hasn't taken away my math abilities. (I start to doubt myself when our entire class seems confounded by basic algebra.)

[shield-toad111.livejournal.com]

Or not... curses.

[thorog.livejournal.com]

Considering I tutor maths, I should not be having this much trouble. I can get down to 0º≤x≤130º...everything seems satisfied with any angle in this range...

[heptadecagram.livejournal.com]

There is a unique solution.

Mwa-ha! And also ha!

[marcmagus]

I'm willing to go so far as to say 0° < x < 130°. Neither x nor <BDE can be 0 and have the other constraints hold. Hmmm...I just had a flash of insight on some tighter bounds for x, but proving it would take a bit of work [I think it's right, though] (and I don't want to spoil anything for anybody)

[blaisepascal]

I've got it narrowed down to a set of 4 linear equations in 4 unknowns which (unfortunately) don't seem to nonsingular. But it appears to be a tighter constraint than 0° < x < 130°

I have:
<CDE + <DEC = 160&deg; <DEC + <AED = 150&deg; <AED + <EDB = 130&deg; <EDB + <CDE = 140&deg; At this point, I realized that there must be some relationship, probably involving triangle ACB being isosceles, that I'm unaware of. The next rabbit hole I'm tempted to run down involved bisecting <DBE to get point F, with FE parallel to BA, and see if that gives me any more promising relationships. It does... It gives me an upper bound x < 70&deg;

[marcmagus]

Interesting. I got my upper bound of x < 70° in a completely different way, having to do with the fact that AD < BE. So I agree with your finding.

Running headlong down the trail of extrapolating information about the relative lengths of different line segments from known angles has yielded x < 65°, but I'm a lot less confident I didn't mess anything up here.

I haven't come up with anything which gives me a lower bound on x other than x > 0° yet. An insight on that could be heartening.

I also played with bisecting <ACB (or raising an altitude to intersect C). Unfortunately, it doesn't seem to generate anything I can use.

[marcmagus]

Damn you. Now I'm late for work, and all I have is a not particularly impressive upper bound on x.

Maybe what I'm missing will come to me in the car...

[xalolo.livejournal.com]

Alright, where's the mack truck?

[timotab.livejournal.com]

xkcd for the win

[xalolo.livejournal.com]

Aha! Got it.

But you're not online to check with.

[heptadecagram.livejournal.com]

You could probably state the answer, as the "fun" part is figuring out what it is and how to get there.

[mrbelm.livejournal.com]

Got it. x = 35 degrees

[heptadecagram.livejournal.com]

Nope.

[ksatyr.livejournal.com]

20

[heptadecagram.livejournal.com]

Correct! Prove your work!

[ksatyr.livejournal.com]

Draw a line from C to intersect AB at right angles. Everything else important becomes isosceles and all angles can then be calculated.

[heptadecagram.livejournal.com]

Bingo!

[kdsorceress.livejournal.com]

Liam sweetie, I am to ask you if a hypothetical line drawn from C to intersect at right angles with AB would intersect the intersection between DB and AE.

Because apparently it wouldn't, and therefore that solution is wrong? Something like that.

Thanks

~Sor

[markbark.livejournal.com]

Did you see today's XKCD? (http://imgs.xkcd.com/comics/nerd_sniping.png)

Could YOU be a victim?

--MAB

[scandiamaxie.livejournal.com]

I suck at higher levels of math...I have to shake, rattle, and roll the numbers, letters, and intergers to get the wrong solution then flip the solutions with previous mentioned numbers and stone them to death to get the correct solution!

thats how i work! Pretty violent processes I have!