I really have a feeling this can be completely solved

I really have a feeling this can be completely solved..

I mean, you do basically have the positions of every cornerpoint

Other urls found in this thread:

thinkzone.wlonk.com/MathFun/Triangle.htm
twitter.com/SFWRedditGifs

Nevermind, got it

The angles that make up the x in the middle are 130 + 130 + 50 + 70, which adds up to more than 360, yet it they are pieced together to form a circle.

130+70!=180

The figure given in the OP is an impossible construction in the Euclidian plane, because the central angles in this version of the original problem as-stated (OP) are in any event inconsistent.

It is telling that most of the angles in this picture are not given using the degree symbol, whereas others are (albeit with a different, inconsistent font). The image may therefore have been doctored for troll effect. But let us first suppose that every number refers to a degree measurement. First, since 50 =/= 70, the opposing angles not being equal, this is impossible in Euclidian geometry, and we are clearly supposed to consider a prosaic Euclidian plane figure.

Even if the numbers without "degree" symbols refer to any other unit/system of angle measurement, we still require that the above two angles are equal. And in the problem as-stated, they aren't, which leads (in the first case) to nonsense like the spurious "190 (degree?)" straight angles in the "solution". Again, even if the non-degree angles are not themselves degrees, this does not rescue the "solution": we still required that the two opposite angles are equal in the solution (they aren't), and so the analysis stops.

I seem to remember that there is a genuinely consistent version of OP's problem, and I may post about such a version later.

OP here, no troll
Got that picture from somewhere else, the ones with degree symbols are original, the others made by me.
Ah fuck, that is why it is not working out at all.

But yeah, I have also thought about this constuction being impossible, yet I do not really understand why.

Update, now it should work out

Still wondering about the impossible construction though...

Let us now instead consider an internally consistent variant on the OP, rejecting the OP's original problem as stated entirely, for the reason given by multiple anons. We include a "pretty-good" scale drawing which gives a real feel for the actual prosaic geometry involved. Notice that the large triangle is isosceles, and that the middle intersection point is asymmetrical/off the perpendicular bisector. Accepting the base angles (in this treatment) and the resulting 50/130/50/130 internal angles leads to a beginning internal consistency of the figure.

O-oh

So it is basically unsolvable, right?

I got 15 degrees.
EDC+BDC+(AEB+alpha)=180 -> EDC=150-alpha
EDC= 20, AEB=30
50+alpha+BDE=180
BDE=120-alpha
ADB+BDE+EDC=180
ADB=40, sub EDC&BDE
40+120+150-2*alpha=180
2*alpha=30
alpha=15

check it for me

(Me)
FYI, I don't bother doing the eqns in my head.
I just did the easy parts then figure it the rest out on paper.

not a possible construction in the euclidean plane.
130 + 130 + 50 + 70 = 260 + 120 = 380.

check the answers

Still not possible. If you label each unknown angle, you end up with 130 = x + y = 180 which is nonsense.

Let me first of all stress that every number refers to degree measurements, and nothing else. I just don't feel like writing out/representing the degree symbol ad nauseum.

The figure given in the OP is a troll-absurdity on what is an otherwise valid and internally consistent elementary geometry problem, the "right" variant of which I have seen before and am working through.

Working it out, we observe that there are two node-points about which angles immediately evidence themselves, for multiple consistent reasons which are obvious and which the reader is invited to check. The result is the intermediate stage of pic related; we then label the remaining internal angles as [eqn] \alpha - \delta [/eqn], as in pic related, in order to treat of them. This leads to four equations in four unknowns:

[eqn]

\alpha + \beta = 150 \\

\gamma + \delta = 140 \\

\alpha + \gamma = 130 \\

\beta + \delta = 160 \\

[/eqn]

IIRC this system is a necessary, but insufficient condition for identification of [eqn] \alpha [/eqn] , which certainly does not "vary", even though confused thinking on this point may lead one to suppose that it does. Even considering the quadrilateral which contains all of [eqn] \alpha - \delta [/eqn] just leads to a redundant statement of the above system.

The way I remember it is that there's some little other trick/observation to actually finish, and the finish is legit. But let me crunch again.

What the fuck happened here

Again, sorry I fucked up there. The ones without the degree symbols were filled out by me. Calculated with 380 instead of 360

literally just change the 70 to a 50. It should be 50 anyways due to opposite angles.

(Me)
Forget this, I messed up.
nvm

Alright cool. Just a bit confused.
Don't know if OP, but I may have read that as coming off a bit aggressive.

Nah, not

Also, nice get

*taking the bait*

The OP is pretty clearly enjoying an ebin troll since he's basically outright stated multiple times now that he's made a goof, yet still provides vague language inviting us to consider the problem as originally stated, although it has been dispensed with. He is even alternately entertaining the possibility that he is wrong, in order to string the thread along, yet he is still also couching the thread in his own initial erroneous ruse-terms and play cute, all of which is intended to string the thread along to his tune. One has to ask why the OP would have changed any information from "the original" problem to begin with, as he's said he has, and of course the answer is obvious: to enjoy a ruse.

dammit op

Since OP is a fucking retard and provided the wrong degree measurements, here's the problem ("Langley's Adventitious Angles") as it was originally presented in Mathematical Gazette in 1922.

There is a solution using only elementary geometry, though the process is fairly non-obvious and involves adding multiple lines not in the diagram.

The addition of this item means that the thread now contains three distinct problems: OP's initial absurdity, the bit I'm in the middle of, and this historical version, which is distinct from the version I've been working, and which has been posted multiple times before on Veeky Forums.

The diagram and wiki write-up are slightly confusing at first, but after a moment of reading it becomes clear what's what. I don't like the "overlapping" diagramming/shading of the lower angles, nor should anyone else.

thinkzone.wlonk.com/MathFun/Triangle.htm

I hate this kind of bullshit. Even the honest version is an idiotic problem.

First of all, it's not a problem that can be solved with "elementary geometry". You either have to know or discover more advanced identities than are ordinarily taught to people.

The trouble is, people learn identities, then learn their derivation, and then they start to think, "Well, I got a little help, but I totally could have figured that out for myself." when in reality, they're things that only *some* of the best geniuses noticed.

So they do these asshole things like make this problem, and suggest they can be solved with "elementary geometry", to people who learned elementary geometry, but didn't learn what they needed to solve the problem.

Some other user had the answer some time ago. Shown in pic.

Sorry, wrong pic.

Those angles in the middle are completely wrong, the problem is bunk.

GOD it took me hours to solve this. at some point I had the solution in front of me, i thought about it but somehow I didn't realise.
thanks, user

Yeah this whole this is jacked. I mean yeah I get it "not drawn to scale" but that does NOT mean you just make 90 degree angles look like 160 or vice versa. This is the kind of crap that makes people hate math in school "Gotcha" questions and "trick" questions.

this lol

is kuro good at math?

she's black, probably not.

kuro = 黒 = black
nop
is correct

i'm black and i can convert degrees into radians. top that.

i'm blue and i da ba dee da ba di