Planes

Has anyone taken a micro-ATX desktop on a planeflight? In their carry on luggage?

I was thinking of making a bulid with this case: newegg.com/Product/Product.aspx?Item=N82E16811205011&nm_mc=AFC-C8Junction&cm_mmc=AFC-C8Junction-PCPartPicker, LLC-_-na-_-na-_-na&cm_sp=&AID=10446076&PID=3938566&SID=

what do you think? Will TSA refuse to allow me to board with it?

Other urls found in this thread:

duckware.com/tech/worldshardesteasygeometryproblem.html
myredditvideos.com/
twitter.com/SFWRedditVideos

No they probably won't, but they will check trough it. If you don't mind the extra ass rape then go ahead.

Also check the power outlets on the plane, they don't tend to be the same as the ones in houses. I couldn't even charge my laptop properly at one plane.

Absolutely 30 degrees

I wasn't going to plug it in, I just wanted a way to ship my computer without actual shipping costs + dangers, you know what I mean? If my computer breaks or is lost in a transit I can't really afford another one

a = 50

That picture's degrees are so off it hurts. If you calculate the top left triangle with 60 and 70, you'd get only 50 degrees for the 3rd angle.

That angle is clearly over 90, and that 70 is clearly closer to 45.

I used to have a computer around that size back in 2005ish, would bring it through airports four times a year until it was replaced around 2008. I never had any problems until sixteen year old me was asked "Is this a computer?" by Dublin airport security. Being an insolent little brat I answered "No, it's a washing machine." I was going through security four hours before my flight, they let me out of the interview room just as the flight started boarding.

How the fuck is the 50 angle bigger then the 130?

The artist didn't into math.

Figures are intentionally not drawn properly so you can't just measure it.

It's obviously 60 degrees. ABED is a cyclic quadrilateral.

Take the CPU / GPU / HDD/SSD with you as on-carry item. Case/mobo/psu/etc through luggage.

It's 20 degrees. Draw it out to scale and then measure.

It's 10 degrees

I used a protractor and got 36°

Bruh I would not trust luggage handlers with my PC

If you have to put foam around the case or something

Anyone claiming a solution is either assuming symmetry or scale when they shouldn't, or just have faulty reasoning. It's a troll problem, just accept it as such.

Here, I even fired up MATLAB to prove it. The working should be straightforward enough for anyone to grasp.

god I fucking hate these kinda tasks because I always fail to find at least one linearly independent equation

calling the angle below a x, and the other two correspondingly b,y one gets
a b x y |
1 1 0 0 | 130
1 0 1 0 | 150
0 1 0 1 | 140

There must be one equation where the coefficients of a & y don't add up the same as the ones from b & x. The rest would be trivial matrix inversion.

Of course, you picked linearly dependent equations (see above). I could not be arsed to thin about whether there there EXISTS an equation that can be inferred, though

>you picked linearly dependent equations

I didn't pick shit, buddy. That's the information that the problem gives you, and I plugged it in to demonstrate that it's insoluble.

>(see above)

That's three equations and four unknowns, so obviously worthless, no working required. I'll give you the benefit of the doubt and assume you just forgot to type the fourth one out.

>I could not be arsed to think about whether there there EXISTS an equation that can be inferred, though

If there is, I'm not seeing it. It should be fairly plain to see though, since there isn't much information there.

The only other equations I can see there is to sum angles around bigger polygons (eg. ABC, ACE, CEFD where F is the intersection of BD, AE), but that gives you a matrix row of (1 1 1 1), ie. wow it's fucking nothing.

Would love to be proven wrong though, it'd be the first case of a "You should be able to solve this" being legit solved.

>I didn't pick shit, buddy
yes you did. you chose the 4 equations which are immediately obvious ASSUMING there does not exist some potentially cryptic relation between the angles that would need to be constructed (circles and straight lines, the usual.)
You missed my point. I provided the subset of linearly independent equations I found and proposed properties a 4th equation would require to make the system solvable.

From a clean, purely mathematical point of view we're not done.
We'd have to prove that there exists more than one geometrically valid solution.
"I don't see another equation" is not a valid argument, you gotta prove there is none.

>I plugged it in to demonstrate that it's insoluble
Thanks for showing that the problem doesn't dissolve in water I guess?

of course, not that I want or care to, I'm just saying that we can only assume it's not solvable, showing it would need an annoying amount of work or a good eye for that stuff.

Obviously it's not formally proven. To truly prove that would probably require machine verification, or else someone could always suggest the possibility of information that wasn't considered, as you are doing now (and rightly so).

That's part of the insidiousness of these types of problems. They require incredible rigor to prove their insidiousness, and by that point everyone who knows anything already agrees with you, and 99% of people are too stupid to follow along anyway.

What I'm saying is: here is the information that the problem clearly presents, and that information isn't enough to solve it. There's _probably_ no more information to be found, and if there is, it's very well hidden. If anyone can find it, fantastic. Otherwise, consider it impossible.

In any case, the statement "You should be able to solve this" is wrong.

"Soluble" has more than one accepted usage, kid. If you had just taken five seconds to google "define soluble" you could have avoided making an ass of yourself.

>"Soluble" has more than one accepted usage, kid.
Do you need me to awkwardly explain the joke and ruin any slight humor it may have delivered or are you good?
>you could have avoided making an ass of yourself
Sup kettle

>I-I was just pretending to be retarded!

Silly me for assuming a serious tone in what was a fairly serious discussion up until this point. Not to mention how incredibly common it is for people to not know the alternative definition of soluble.

Your post came off as a correction, not a joke. Not my fault that you can't make it clear when something is a joke. Use a caret-nosed smiley or something next time, that would have made it pretty obvious.

Not that I've found a solution yet, but your analysis is wrong. The figure clearly exists and there is noting contradictory about the angle given. I'm trying to find a system of equations that is linearly independent. The solution should be around 20'

>your analysis is wrong.

It's a pretty fucking straightforward analysis, it should be trivial to point out where I'm wrong, yet you aren't going to?

>The figure clearly exists and there is noting contradictory about the angle given.

Yes, and? I'll assume you just misread my analysis, go over it again please (or possibly revise your linear algebra).

>I'm trying to find a system of equations that is linearly independent.

That's the whole point of the discussion thus far - if a LI system exists, it's extremely well hidden.

If you want to make actual progress, find a geometry software package that can analyse it for you. This is 2016 after all.

>If you want to make actual progress, find a geometry software package that can analyse it for you. This is 2016 after all.
No wonder you can't solve it. You rely on computers to think for you.

let's see..
A triangle is uniquely defined up to similarity if for example all 3 angles are known.
let's call the intersection of the lines AE and BD F
as we can see for example in the triangles (ABD) and (ABE) are fully defined, as well as the triangles (ABF),(AFD)&(BEF) by extension.
What remains is a quadrilateral (ECDF) of which we know it's 4 angles. Is that shape up to similarity uniquely defined?
Because if we cut it into 2 triangles we'd need information worth 6 angles in total.

Here's your (You)

Something tells me the key to the problem is drawing a line that isn't shown connecting two of the points. A unique solution definitely exists, and I see no reason why you couldn't deduce this from angles alone.

>Something tells me

I'm gonna stop you right there faggot.

The whole point of these types of problems is they try to fool you into thinking they're possible when they are actually not. You just fell for it hook line and sinker.

If you're so sure it's possible, stop blabbering and show your work.

This, in college I worked ramp pt at a downline station.

When you have 24 minutes to turn a regional jet, some of those bags get thrown pretty hard.

I want to fuck a kuro

If you're trying to take that as carry on and your flight includes a small regional jet they might make you gate check it anyway.

Far as I know there's no regulation on desktop computers though, they make you take the laptop out of bags during screening so if you're not carrying it in a bag just as long as you have it out at the checkpoint you should be fine.

I've added the only extra geometry I could think to. I think this will get us there.

Actually, this is a dumb picture, but its getting at what I mean. You have to extend the figure to solve this problem.

Can you split AD and create a right triangle leading to the angle (50) at the top, find that measurement and then use it to deduce alpha?

I guessed 40, and the math all worked out.
So the answer is 40.

Maybe you guys should use trial and error for once instead of relying on your ivy-league math methods and expensive computer programs.

I guessed 999, and the math all worked out.
So the answer is 999.

Maybe you guys should use trial and error for once instead of relying on your ivy-league math methods and expensive computer programs.

I got 60.
C=20 because A and B are known.
Since DBE is known and equal to C, then BD on CD, DE is then height.
BDE=180-20-90=70
EOD=AOB=180-70-60=50
a=180-70-50=60.

...

>DE is then height
That's only true if you assume it's at a right angle from the base, which it isn't

Fug, I thought it's a median for some reason.

Only if 10 = 20.

...

Yes, the answer is 20, but how'd ya get there?

This. It's just a handful of easy addition and subtraction if you keep in mind that triangles and straight lines have to be 180° total.

Just wanted to show how some of these assholes pretending the solution is obvious are wrong. The solution can't be found with elementary geometry.

See
He shows it's not so easy. You end up with the nonsense "no solution" solution if you aren't careful.

a + c + d + b = 360 - 70
Is the fifth equation guys.

Not linearly independent from a+b=130 and c+d=160

It seems really odd that you wouldn't be able to use classic geometry to solve this. It seems so doable. I'm not convinced it's not yet.

eye balling it

Actually I'm just a fucking idiot.

The problem he ran into with MATLAB is that the drawing gives you a fair amount of data to work with and he didn't use all of it. Since the rules were looser in his implementation, MATLAB just figured that there was more than one way to solve it rather than there not being any possible solution, hence the infinite returns in the matrix.

Look buddy, the answer is 20 degrees. That can be trivially verified with trigonometry. Any supposed solution that gives an answer other than 20 degrees is quadruple wrong.

Yes, he didn't constrain the problem enough to give a unique solution, thus the matrix is singular.

I'm not going to lie, I did the exact same thing before I saw someone else post the result. Unlike him, I recognize the problem has a solution.

What did you use to determine the problem has a solution that the other user didn't?

I didn't prove it was doable without trig, but the figure is fully and properly constrained such that only one solution exists, and indeed it does per 's drawing.

to that user's credit, he admitted the possibility of missing information and said "go boot up a software package to figure it out for you", which is exactly what happened, nobody has posted a manual solution yet

really want to know what that system of linear equations is now

This is a really good problem. Taking it to the tards over at /sci/ for continuation.

Except /sci/ is even more retarded than Sup Forums.

If you just want to watch a bunch of Dunning-Kruger retards flail about then I don't blame you.

>Except /sci/ is even more retarded than Sup Forums.
Mostly agree with you, but there are a few folks that are actually intelligent over there.

The real hope is that it doesn't get pruned before a solution is found. Sup Forums moves much faster.

I'm feeling a bit stupid right now, but I've got these equations so far.

jotta=alfa+10
gamma=beta+20
alfa+beta=130
beta+jotta=140
alfa+gamma=150
jotta+gamma=160

where:
alfa is alpha
beta is angle EDB
gamma is angle DEC
jotta is angle CDE

I need something more.

Don't put in carry on unless tour willing to let them put it through the scanner alone. I traveled with my tower checked in a bag and it was fine

Here's the solution guys. Link courtesy of /sci/.

duckware.com/tech/worldshardesteasygeometryproblem.html

If you like challenging (and possible) problems, don't peek.

This was the first "you should be able to solve this" that was genuinely awesome. Thanks OP.