"Basic" geometry, Angles

>>7924511
sweet, that's one of those "tough-easy" problems, like the Langley triangle problem (slightly different angles). Hint: lots of isoceles triangles and an equilateral triangle. (Two angles equal <==> two sides equal.) You need to construct one point to solve it.

You can get BDA and EDC (via BCA), with those you can get the small angle opposite alpha. You can get the opposite angles in the center cross via DBA and BAE, then you have both angles in the little triangle that contains alpha, so then you can get alpha from that.

>>7924533
>EDC (via BCA)
nvm I dun goofed this doesnt work

20°

Solve for all the easy angles, you'll have 4 minor angles unsolved.

Then. use all interior angles in a triangle must equal 180 to set up 4 linear equations.

Solve 4 linear equations involving 4 linear variables (you only need to solve for alpha, no need to sub back in for the others unless you wish to check your work.)

In the end, I got alpha = 5 degrees.

>>7924511
This is the worst goddamn diagram I have ever seen.

Not to scale is one thing but 50 degrees as an obtuse angle is just retarded.

>>7924560
I remember it was a solution with two digits so not this. You don't need linear algebra for this.

>>7924511
The problem is unsolvable, or rather it has infinite solutions.

Let the center point be F.
Then let angle FED (which is alpha) = a, let angle CED = b, let angle FDE = c, let angle EDC = d, then we have a system of equations.
a + b + 0 + 0 = 150
0 + 0 + c + d = 140
a + 0 + c + 0 = 130
0 + b + 0 + d = 160

The determinate on the matrix on the left is 0, therefore it is not invertable and has no unique solution to the matrix-vector equation Ax=b. How ever we can find a solution depending on one variable.
The column vectors d * [ 1,-1,-1,1] + [ 0,140,160,-10] grants a solution in terms of d. Therefore alpha = a = d - 10 = 150 - b = 130 - c .

>>7924511
Why are even loli this intelligent in Japan

>>7924523
OP here

That's the good direction to take I think.

I don't know where there is an equilateral triangle though.

>>7924692
That system of equations is only necessary but not sufficient. There is only one solution that fact becomes obvious if you try to draw that shape on paper.

>>7924533
Already tried to figure out this way (with complementary angles) and it's just not enough.

>>7924560
OP

I tried this but it seems that all the equations are redundandts (meaning by that I can just find that alpha = alpha or 180 = 180,.. whatever the equation system is).

>on /sci/
>wasting your time on synthetic geometry,
>that will never be applicable to actual math (cont.)
>problems, just pleb-tier gamedev and engineering
lmao

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>>7924564
OP

maybe this will be helpful

>>7924692
OP

It's not infinite solutions.
An issue could be that the figure is not makeable, I mean that if you tried to draw it you simply couldn't (like a triangle which has an angle of 80°, and another of 95°).

But it's actually a drawable figure, so there is an unique solution.

>>7924511

Are you sure this is even sovable, using only elementary geometry?

wow this board is even dumber than expected.
It's an isocele triangle you dumb fucks!

>>7925028
>isocele

Proof?

see >>7924808

Doesn't look isocele to me. Provided you are refering to the inner-triangle we are trying to solve.

>>7924759
>>7924852
>These posts.

Jesus fucking christ.

so, what is considered elementary geometry and what is trigonometry?

>>7924511

It is unsolvable, not due to the drawing. Such a triangle is just not possible.

>>7925128

I just finished drawing it...

pls explain? granted it doesn't look like in OPs picture, but still drawable, I think someone even posted a drawing in here if you scroll up.

>>7925137

Not the drawability, the problem is that it requires certain obtuse angles to be shown as acute angles, it messes the whole thing up.

>>7924808
>You can bisect the triangle into two right angled triangles.

Nice assumption, bro.

>>7925151

he can fuckwit, the two sides are the same. Only problem is it wont help

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It's pretty easy to get to this point, but I think the people saying it's impossible are correct

I tried literally every triangle relation, possible: each and every one of them give the redundant "a + (the other mising angle) = 130.
Plus 50 is portrayed as obtuse, this triangle is more than likely just inconsistent.

Here is the solution:

https://www.duckware.com/tech/worldshardesteasygeometryproblem.html

>>7924511
seems like AB || ED so its the Z ankle...

>>7924807
>on sci
>not being able to figure out something that seems to be so easy
>"it's useless anyways"
if that helps you sleep at night

>taking the b8

>>7924744
dubs don't lie

>>7924902
OP

I think the not on scale drawing and that phase are just here to makes you doubt

>>7925084
OP

My bad, a good example would be 125° and 80°

>>7925128 >>7925137
>see >>7924808

20°

>>7924511
bump

>>7925161
...ADE...

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>>7924808
Green lines are parallel.

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from an old thread

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Linear algebra causes one to fall into a trap by admitting of "infinitely many" solutions per gaussian elimination. But the geometric figure, being fixed by its angles, does not. The figure has no parts which may "slide", thereby admitting of infinitely many solutions.

Not an answer, I've just been thinking about this. Embarassingly I didn't even notice that the thing was isosceles until later, wanting instead to doodle with Gaussian elimination.

The diagram post using a perpendicular bisector just above, appears at-a-glance to be correct, but I'm going to put this down for the moment and come back to it later when I'm bored.

If you recognize the proper Isoceles as opposed to the trap Isoceles than its pretty easy to come up with 70. You can use systems of equations and trig ratios, but that would just over complicate everything

I thought the triangle towards the center was isosceles and I came up with 65° which is incorrect

>>7927452

(as you've realized) no, because the segments which drop inside the larger triangle dowards to either side, to generate the triangle you are referring to, these two segments are not symmetric in their angles with respect to the larger triangle, which /is/ isosceles.

As an above diagram indicates, this "50-130" point of intersection is offset from the perpendicular bisector of the larger triangle.

>>7924511
Look at that stupid loli I bet she secretly enjoys being called stupid

>>7927807
lmao

>>7925161
I get to this point as well.
What am I doing wrong?

>>7927832
>>7925168

Hmmm. Never mind.

>>7927807
all lolis are cute and smart
shut up

70 degrees. using vertical angle theorom (vertical angles are congruent.) and the fact that all the angles in a triangle add up to 180, and that BAC and ADC are similar triangles.

>>7927859
Not true, by the way, you pedophile
Lolis are still female and female idiocy transcends all dimensions

>>7927882
Not true, by the way, you misogynist virgin.
Young people when given IQ/math tests usually score higher on average, granted they're adults.

>>7927913
K
But if lolis r cute n smart why do they always end up with dicks in their mouths

>>7927919
because it's manga and not real life. i prescribe you go out and breath some fresh air right now

>>7927921
No thanks. You need it more than me

>>7927935
fuck you asshole

>>7927948
Uer nan was touching her yams while you were touching yourself to the latest loli doujin

>>7924511
alpha = DÊA = 80
BDE = 50
EDC = 90
DÊC = 70

>>7927919
>But if lolis r cute n smart why do they always end up with dicks in their mouths
Because they're cute

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>>7924511
Insoluble

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a is 30

>>7930208
wrong

>>7924511

80 degrees, sci is fucking retarded

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>>7924511
Good job stealing /a/ content faggot

>>7927882
>Not true, by the way
t. grannycon who likes 80 year old oppai piggu

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>inb4 70
Fuck you OP

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>>7930912
You can't know these are 80's, you have no proof of any paralellism or similar triangles atm, so you can't get these other two numbers either

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>>7924511

took my a little but I had real fun doing it actually

>>7930939
You are correct; that solution is not correct. You can, in fact, get the solution algebraically.

At this point I assume that you agree that all of the other angles are correct.

Call the unknown angles [math]\alpha, \beta, \gamma, \delta[/math] going in a clockwise fashion.

Observe that [math]\alpha + \delta = 150[/math] because of the 30° angle to the left of [math]\alpha[/math] and [math]\delta[/math].

Similarly,
[math]\alpha + \delta = 150 \\
\beta + \gamma = 140 \\
\alpha + \beta = 130 \\
\gamma + \delta = 160.[/math]

We now have 4 equations and 4 unknowns. We can arrange them in a matrix to solve for the 4 angles.

[math]\left[\begin{array}{cccc|c}
1 & 0 & 0 & 1 & 150 \\
0 & 1 & 1 & 0 & 140 \\
1 & 1 & 0 & 0 & 130 \\
0 & 0 & 1 & 1 & 160 \end{array}\right] \implies \left[\begin{array}{cccc|c}
1 & 0 & 0 & 0 & 67.5 \\
0 & 1 & 0 & 0 & 62.5 \\
0 & 0 & 1 & 0 & 77.5 \\
0 & 0 & 0 & 1 & 82.5 \end{array}\right].[/math]

>>7930888
To be fair this problem has been posted on /sci/ countless times with countless pictures. This is the first time I've seen OP's picture but is also the first time I've seen this problem drawn so not to scale that half the posts are trying to figure out if it's even drawable.

I don't think I've ever seen someone manage to solve this problem before the solution is linked though.

With regards to your picture, I think it's strange to expect the reader to know what a C^* algebra is but not know what a category is.

While on the topic of fun geometry problems, does anyone have the "kindergarten math" problem?

>>7924523
>>7924511
The only reason this is difficult is the stupid "do it with your hands tied behind your back" restriction.

It says "elementary geometry", but it is actually impossible with what most people will consider to be elementary geometry. You have to get into obscure identities which most people don't learn and have no use for because they learn the far more straightforward and powerful tools of trigonometry. Now, in principle you could discover these identities for yourself, starting from the basic geometry most people learn (or at least are exposed to) in high school, but now you're talking about the kind of shit geniuses spent years of their lives struggling to get straight.

So really, what it should say is, "solve it using esoteric geometry now only of interest to mathematicians, which you probably have never been exposed to, but which I'm calling 'elementary geometry' to fuck with you and unfairly make you feel stupid for not being able to do it".

>>7931068
They aren't esoteric identities. Just complicated techniques.

Not only is it insoluble, it is physically impossible. Some have been saying they've drawn it or created it programs like Geogebra. However these are just simulations. If you attempt to construct the figure physically, as I just have, you will find that it can not be done. I have been trying for the past hour to build the figure out of popsicle sticks and elmer's glue so that I could measure the angle with my 3D protractor, and have been utterly unable to do so (it keeps falling apart, the popsicle sticks break, I get one stuck to my hand, etc). I was frustrated for a while, thinking that it was my own fault, but after much reflection I realized it is because of a fallacy inherent in the problem itself.

tl;dr it's a trick question. Nice try OP, but you're not getting one by me.

>>7931051
>I think it's strange to expect the reader to know what a C^* algebra is but not know what a category is.
What? The problem expects the reader to understand what both are. It only defined what an opposite category is.

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>>7930888
>mfw some nerd is mad that the content is "stolen" from "their" board

>>7931235
Lol, still. The opposite category is literally the second thing you learn after getting the definition of a category.

I think it would be weird to expect a reader to know what a category is and not know what an opposite category is.

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It's 60 degrees you fucking idiots

check my math

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>>7931329

kek, he's right guys

>>7924511
Bump

Am I the only one who noticed she holds the pointer with her right hand and pulls it to the right with her left hand facing up? Who the fuck does that?

>people dont realize that there is no unique solution
:3?

>>7931329

why 130-a and a+10 and not 135-a and a+5 ?

i got 65 degrees sampai

>>7927185
This is correct, but I've never seen this property of isosceles triangles before.

>>7932759
>This is correct, but I've never seen this property of isosceles triangles before.

Perhaps because you've never crossed the Pons Asinorum.

>>7932759
Me neither

>>7925161
How is the angle closest to C only 20? Shouldn't it be 110?

Unsolvable, as one of the angles have to be both 30 degrees and 50, following other contradictions

>>7924511
>/sci/ Is incapable of solving the problem.
I mean... Okay, it's a really tough one, and none of your "Calc III" and pure math will help you here, but:
>/sci/ doesn't understand a given solution.
Urgh...

Resemblance of two small triangles and one big.Ez

>>7925161
HOLY FUCK I'M SUPPOSED TO BE DOING MY COMPLETELY IMPOSSIBLE CFD ASSIGNMENT WITH ANSYS THAT IS DUE IN LIKE 3 DAYS, AND HERE I AM, SPLURGING OVER A FUCKING QUESTION ON 4CHAN

THE ANGLES ON B CAN BE APPLIED TO E, (AND THE ANGLES ON A CAN BE APPLIED TO D AS WELL)

SO, 30 + ALPHA + 80 = 180

ALPHA IS 70.

NOW CAN SOMEONE PLEASE TELL ME HOW CAN I SIMULATE A DEVELOPING FLOW PROFILE WITHIN A PIPE IN ANSYS FLUENT

>>7933910
Just b ur self :^)

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>>7933910
You can stimulate a developing flow profile in my a-pipe any time sugar

>>7932765
Interesting. Thanks.

https://en.wikipedia.org/wiki/Pons_asinorum

>>7924511
infinite solutions.
try alpha 10, 20. always works.

>>7925187
assuming parallels without it being given

leave

>>7934844
>>7933910
>>7933773
>>7933369
>>7932693
>>7932641
>>7931374
>>7931329
Wrong.

>>7933332
how did you even get that?
(cbd + dba) = (cba) = 80
(cae + eab) = (cab) = 80
80 + 80 = 160
180 - 160 = 20

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>>7927185

>>7924692
>it has infinite solutions.

Nah, b.

alpha <= 129.999....

Alpha can be 20 or 40. There is more than one solution. My solutions are not exhaustive. The linear system they create is degenerate. I've plugged in at least two different solutions and it all works.

>>7936221
there is only one. come on man, just look at the figure. The main triangle is 80-80-20, and the angles of the bisectors at A and B are given. This determines the ratios BE/BC and AD/AC, and these determine alpha.

You are missing one equation.

>>7924511
Four of the given angles, including alpha, have infinite possible values.

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>>7931028
You solved you own matrix incorrectly.

Source for the image?

>>7936391
the prisma illya mango

>>7924511
a is 120 right?

>>7935803
OP

great but how do you show that the triangle with two angle X is isoscele

>>7931028
whatever that matrix is supposed to mean alpha is 20°

0 < a < 130

>>7936294
>You are missing one equation

Sorry the troll is over. I plugged in 20 and it worked and I plugged in 40 and it worked. There isn't only one solution.

>>7936636
The orange circle

>>7937599
There is a single solution.

>>7937599
please draw this to scale and measure 40 degrees and i'll suck your cock

>>7924511

ED and AD are parallel therefore AED = BEA

alpha = 70

>>7936823

I think this is the only correct answer

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>>7938495
>ED and AD are parallel
Pffff

Ayyyyy Lmao

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Let's generalize...

I shitposted your question in 2ch,hk/sci(russian main board) and we found answer.(i dunno english)
Draw segment DF, parallel to the ground . Then we spend the segment AB, BD intersect at the point G. From symmetry considerations it is clear that the ADF triangle is a triangle BDF ( it is possible to izi prove and strictly , but I'm too lazy to write , if anyone asks - sign for ) , that is an angle equal to the angle BDF AFD. So , BFG isosceles triangle and BG = FG. Triangles CGD and CGF are on three sides.ie the angle ACG equals the angle BCD is equal to 10. Now look at the triangles ACG and ACE. AC Side they have a common , and the adjacent angles are 10 and 20 , ie, triangles are equal and AG = CE.ABC is an isosceles triangle , that is, AF = FC, from which it follows that GF = FE. Please note that an equilateral triangle DFG , that is GF = DF, ie, EF = DF.This means that an isosceles triangle DEF , ie the angle EDF equals the angle DEF is equal to a + 30 . EFD is equal to the angle 80 , from which we find that 2 (a + 30) +80 = 180 , a = 20 .

>>7927185
>>7940991
Looks the same.