Am I right?

You are correct. I guessed 1/5 but homeboy said prove it and this was the only way I could

Yes. All the solutions I've seen are more elegant than my cluster fucked

ez.
[math]1. b > a [/math]
[math]2. b = frac{1}{2}[math]
[math]3. a^2

1/1 because the entire green square is green

ez.
[math]1. b>a [/math]
[math]2. b=\frac{1}{2}[/math]
[math]3. a^2< \frac{1}{4} [/math]
[math]4. \frac{1}{5}[/math] is the only choice less than [math]\frac{1}{4}[/math]

The slope of the lines that cross the square and form the green square is 1/2. Find each line's rule, find two intersection points, calculate the distance between said two points, use said distance to calculate area. There ya go lad.

Actaully the slope of two of the lines is 1/2 and the slope of the other two is -2.

plz r8 my art

>4.1/5 is the only choice less than 1/4

90 hr in mspnt

>tfw this proof is valid

Visual proofs are awesome.

Am I stupid or is it impossible to solve this without assuming that the trapeziums + their respective triangles have the same area as the central square?

You're stupid

Simple -- just calculate the areas.

Start by assuming the square is 1 by 1.

Let A be the larger right triangle (with sides 1, 1/2, and sqrt(5)/2), which has an area of 1/4.

Let B be the smaller right triangle (with hypotenuse 1/2).

B is similar to A, because they share two common angles. Comparing their hypotenuses, we find that the sides B are 1/sqrt(5) times the sides of A. Squaring this, we find that the area of B is 1/5 the area of A.

Let C be the quadrilateral. Looking inside A, we find that it's composed of 2 copies of B and 1 copy of C. But A has the same area as 5 copies of B -- thus, the area of C must be 3 times the area of B.

Summarizing our findings:

area(A) = 1/4
area(B) = 1/20
area(C) = 3/20

The non-green areas contain 4 copies of B, and 4 copies of C, with a total area of:

4*(1/20) + 4*(3/20) = 16/20

Thus, the green area is:

1 - 16/20

which simplifies to 1/5.

Here's a diagram.

>cometely intuitive problem
Why are you brainlets using math for this

Veeky Forums is all retarded undergrads.

KEK

Why does this make me laugh so much?

somebody solve this using double integrals

Each of the large triangles is a quarter of the large square so the two above and below the green square are 1/2 together. That leaves green square and side things (3/4 of green square each). If side things together are 3/2 a green square then green square is 2/5 of middle band. 2/5 of 1/2 is 1/5

Because it's true but it assumes a hidden "axiom": that one of the given choices is correct. If none of the given options were true, this heuristic blows up in your face. This is why I avoid multiple choice tests. Smart people can make educated guesses like this, side-stepping the problem.

Why don't we just break out a ruler to get the volume of each square and put them into a fraction?

Nah you have to print it, cut the thing out with scissors, weigh it, then cut out and weigh just the green part to find the fraction. Wa la

assuming these figures are representational, given their existence on a euclidean plane, this problem can be solved entirely synthetically without the need of any nonsense algebra.

My Monte Carlo simulation says it's 0.19999.... so none of the solutions in the OP

Bruh, what do you think 0.199... equals?

If it's stupid but it works

except they aren't

Except that isn't a visual proof. Were it properly drawn, K wouldn't be inside the triangle. The angles XAK and KAY are visibly unequal for example.

Neither are BN and NC equal in length. X should lie somewhere between A and B but Y should be drawn on the extended segment below C. That whole drawing is wrong.

say the sides of the white square are 2 long,
then the sides of the green square are less than 1, so the area of the green square has to be less than 1/4. 1/5 is the only option less than 1/4

...

Thats the point, visual proofs suck because they are often misleading

Algebraic proofs are misleading too. There is a reason why mathematicians have been focusing on constructibility for millennia: you have to prove that the line you draw actually bisects the angle, for example.
You posted the visual proof equivalent of one of those do some calculations get 2 = 1 things.

>all these algebraic proofs
disgusting.

With the image given we see that the red triangles are similar, and so are the blue shapes. Since the angles that I've drawn on the left form a flat line (180 degrees) we know that we can combine one blue and one red to form a square, since three of the angles can be shown to be 90 degrees, by being opposite to 90 degree angles. Thus, we have 5 squares of all equal sides, meaning the green is 1/5.

This is what intuitively jumps out at you, and it's ridiculously easy to show.

this is how I did it in my mind, am I a brainlet?

artist here. eyeballed it it looks to be 1/5 or it's one hell of an optical illusion

The first proof that's both well-written and well-illustrated. Nice work, user.

How so, I'm under the impression my application of Pythagoras right in the 3rd post is still the clearest proof her.e

You autistic fucks, it's a multiple choice, you don't have to prove shit. Even a micro-brainlet could figure out that looks like 1/5 and it's certainly not 2/5, 1/4 or 1/3

Fucking dumbest thing I've ever seen and it's watertight as long as we assume we aren't being duped by the multiple choice. What the fuck.

>well-written and well-illustrated
>well
>-
>written

Actually the answer is 1/3

The pixels on your screen are RBG. Only 1/3 of it is actually green.

Except it's not. First off you lay the squares side by side without showing why you're allowed to do that, or that they're equal. Second, after skipping that step (where you're essentially finished with the proof) you instead go off on a tangent by using Pythagoras.

>Since the angles that I've drawn on the left form a flat line (180 degrees) we know that we can combine one blue and one red to form a square.

EHHHHH. Wrong. You know that they will combine to form a rectangle, not necessarily a square.

Your "proof" is b8, m8.

Did you not see the part after the comma beginning with "since"?

>since three of the angles can be shown to be 90 degrees, by being opposite to 90 degree angles.

Newsflash: A rectangle also has angles 90, 90, 90 and 90.

Just because all 4 angles are 90 doesn't mean it's a perfect square, brainlet.

This is about where you invoke symmetry and concluce the proof

Fine, I glossed over it and should've made it clear. I included it in my "similar shapes" prelude. The marked sides are of equal length, as can be clearly seen.

So nobody else sees it as sin(arctan(2))*cos(arctan(2))/2*4/4?

This is brilliant. Honestly impressed. Way to think outside the box, user.

how do you know for sure b > a ?

by the triangle inequality

It looks like 1/5
One of the answers is 1/5
QED

but b and a are part of a trapezoid

a trapezoid is just a square and a triangle innit

>look at picture fore five seconds
>can easily see the answer is 1/5

do people actually have difficulty with stuff like this?

that was really good
saved

no dumb fuck. the interesting part is always the proof

Why don't you look at this for 5 seconds and tell me the answer Mr. Magic Math Faggot:

[eqn]1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} [/eqn]

Don't even bother replying in more than 5 seconds.

>solveable by eye in 5 seconds
>pseudo/sci/entists need a thread dedicated to working it out

>not just doing the infinite series converging to ln(2)

shit question, stay pleb

this so much.
you take one look at it and realize that the total area consists of the green square and 4 additional squares of the same size that are simple re-arranged a little.

Rate my working out, lads.

...

>infinite series
>infinite

You had 14 fucking minutes to look at it.

Pssh, look at you fools. You're all just a bunch of children in the sandlot to me. Step aside and let a real mathematician work, if you're not too scared.

Autism, I choose you!

bumperino

This

not really the triangle inequality.
it's because b is the hypotenuse of a right triangle, one of whose legs is a.

The region is given by
[eqn]0 \le 2y-x \le 1, \;\;\; 1 \le y+2x \le 2[/eqn]

substitute u = 2y-x, v = y+2x

integrate over 1 by 1 square in uv-space

jacobian is
[eqn]\frac{1}{\begin{vmatrix}2 \; -1 \\ 1 \;\;\; 2 \end{vmatrix}} = \frac{1}{5}[/eqn]

piggots

Took me a minute but you can solve that in your head with the origami method

Just need to realize all triangles are analogues of themselves and a lot of edges have the same length

Top pip. Everyone else is a babby brainlet too scared to calc

>needs calc to do an elementary geometry problem

I have found a true piggot

somedy pls print it and weight both the whole square and the little green one.
only then can we be sure

Are you literally retarded?

What we have to do is get a glass of water and then put the whole square in and measure the increase in water. Then we take the whole square out and put just the green square in and again measure the increase in water. Then compare the increase in water.

what the fuck is a piggot?

It is a catchphrase this motherfucker right here made up and only he keeps saying it over and over.

In the past he would define it in every post by saying piggot = pig + faggot but I guess now he thinks everyone gets his retarded phrase.

Don't take too seriously, the guy is actually severely autistic. I have confronted him in many threads (he posts here daily multiple times) about how he keeps trying to force a meme down our throats. He is a total faggot. His meme has no spice but for some reason he thinks it is worth to waste our time posting it.

Every time I read "piggot" I only 1 milisecond of my life I could have spent reading something better. Fuck him.

it's this guy

You are so retarded it feels bad to know that you are probably majoring in the same shit I am. Fuck you.