I really have a feeling this can be completely solved..
I mean, you do basically have the positions of every cornerpoint
I really have a feeling this can be completely solved
Austin Gray
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Isaac Stewart
Nevermind, got it
Aaron Sullivan
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.
John Nguyen
130+70!=180
Ian Edwards
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.
Eli Perez
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.
Blake Hughes
Update, now it should work out
Still wondering about the impossible construction though...
Joseph Baker
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.
Nolan Flores
O-oh
So it is basically unsolvable, right?
Isaiah James
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