Kurisu thread?

Kurisu thread?
Kurisu thread

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youtube.com/watch?v=a-GqSWsISVs
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KURISUTINA
youtube.com/watch?v=a-GqSWsISVs

S(colored) = (1 - pi/4)*(r^2)/2

r - radius of the biggest circle

Wrong.

√7/8 + arctan((16 - 5√7)/9) - ¼ arctan(4/3 + √7/3) - π/16 ≈ 0.146381,
assuming the bigger circle has radius 1.

Can it be solved without calculus?

if r is radius of big circle
colored area is slightly < (r^2) - [(pi*r^2)/4]
10/10 approximation
yes i am an engineer

Yes.

I don't even know what this is asking. Am I retarded?

This approximation is good enough

t. another mechanical engineering student

with r being the radius of the bigger circle, the colored area is
(pi^2 / 2)*r^2

I have the strong feeling this is actually wrong tho

Solve the mystery of who pissed on her circle drawing

Find the area of the shaded portion, assuming the radius of a large circle is 1.

I meant to write
(pi^2 / 16)*r^2

still seems wrong but getting closer

Are you just guessing? The fact that π^2 is in your answer already tells you the answer is wrong.

Still waiting for non-calculus solution.

This with another term. Too much of a brainlet to find the expression for the last two tiny pieces.

(r^2)-[pi*(r^2)/4]-[(r^2)/4-pi*(r^2)/16]

not just guessing, reasoning was that the ratio of the small square to the big circular sector would have been equal to the ratio of the colored area to the small circle
that was pretty much a wild guess

if we consider the radius to be 1
my solution results in an area of 0.2146

gg user! your solution reaches 0.1610

if we consider this to be accurate we're
.1610 - .1464 => 10% error
Come on guys. I think we can approximate it to 3 sig fig
Also, props for that huge ass eq user

A = area of 1 corner of the large square that the page circle doesn't overlap with = (r*r) - (2pir^2 /4) = small square - 1/4 of large circle

B = area of 1 corner of the little square that the little circle doesn't overlap with = (r*r) - (2pi(1/2r)^2 /4) = small square - 1/4 of the small circle

C = smallest area
D = colored area
area of small square - quarter of large circle - top left B = D + 2C

D + 2C looks like a banana and that's good enough for me
I'm going to bed
You probably have to do a bit more subtracting to get rid of the 2C but I'm too tired

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if im able to do this I want kurisu to come in my dreams and pet me

Can't you just set up the equations for the two circles, calculate the intersections, then calculate the difference between the integrals between the intersections?

Why don't you just post this thread in /sci/ instead?

I once wasted two hours straight trying to solve this. The only thing I could prove was that I didn't know math.

I was too lazy to do the maths so I just wrote a simple Monte Carlo simulation in python and the area, with r=1 for the big circle, I'm getting is around 0.0365...

they are bunch of nerds

The exact same thread is on /sci/ right this very moment.

Derp. Times 4 of course, so that makes it 0.1463...

What is this and how/where do i learn it?

I'm pretty sure I saw this exact problem in some sort of pedagogy book. Iirc it was brought up as a problem middle schoolers could solve but professors could take an hour to solve, because the middle schoolers were primed by the right chapter.

Thank you user for your contributions to anime algebraic geometry or whatever this is

Steps are here.

Wow, is Sup Forums the smartest board around????

Easy. The answer is: area of the outer square * 0,0335918031327875

practical solution, I like it

I'm more interested in Azusa that Kurisu

R radius of large circle.
p radius of small circle.
Define both circles in polar coordinates:
(subscripts b is for big circle and s for small circle)
x_b=R*cos(ϴ)
y_b=R*sin(ϴ)
x_s=-0.5+0.25*cos(ϴ)
y_s=0.5+0.25*sin(ϴ)
Intersecting points are x_b=x_s and y_b=y_s
R*cos(ϴ)=-0.5+0.25*cos(ϴ)
ϴ=arccos(1/(1-4*R)) V ϴ=-arccos(1/(1-4*R))
So now you've got the borders of your integral but the little red corner (pic related) screws it all up so you should probably take the integral of the small circle between the two boundries minus the integral of the large circle + the red part (which is (0.25R²-0.0625*π)/4) but I highly doubt this can be done analytical so what's the fun?

Forgot to add the pic

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>tfw there will never be a Kurisu logic puzzle for a lowly immunologist like me

Kurisutina would never stoop to fraternizing with immunologists.

t. stats brainlet

(area of yellow space) =
(area of a quadrant of the square = 1) -
(area of the large circle inside the quadrant = pi/4) -
(area of the quadrant not occupied by the small circle divided by 4 = (4-pi)/16) -
2 * (area of the little funnel shapes formed by the intersecting circles)

(area of funnel shape) =
(area of a quadrant of a quadrant of the square = 1/4) -
(area of a section of the small circle from the top of the square to the intersection point of the circles) -
(area of a section of the large circle from the top of the square to the intersection point of the circles) +
(area of the section of the large circle that's not in the quadrant of the quadrant) -
(area of the triangle that remains in the quadrant of the quadrant after removing the circular sections and the funnel shape)

Looking at some of the other solutions, I feel like I've gone the long way around, but I did end up at 0.14638126919742421589

She is so cute when confused.

I draw it on autocad and get the area, duh
but srsly
x=square - big circle area/4
y=smaller square - smaller circle area/4
x-y approx.

I really don't see how to calculate the last 2 remaining areas to rest without calculus

green/(red+greed) = percentage of the square area that isn't occupied by the circle
=(r^2 - pi*r^2/4)/(r^2) = 1-(pi/4) (independent of r)

yellow/(blue+yellow) = same proportion
Y/(pi*r^2/4) = 1-(pi/4)
Y = (1-(pi/4))*(pi*r^2/4) = 0.1685 (for r = 1)

I got a compile error with that.

Would Kurisu be able to solve Fermat's last theorem?

The "yellow/(blue+yellow) = same proportion" isn't truly the same proportion, it's just a rough approximation to solve it 'visually'

Haha just use calc you dummies

She's even cuter when she's flustered.

im sure theres a solution there somewhere
i cant be bothered though
also stick to middle school level math

I want to bully Kurisu!

More digits courtesy of wolfram alpha

>0.1463812595303478247594795603569273444245192919532469576672245175873001006790736933111740163472883434074359118379150474426623935932353216046257302200604896107626027298545401035089971065649502090362905358013262740228050046751411...

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I wish someone translated this,

Two miracles of the universe.

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I spent an hour trying to solve this last time. Does it have a solution?

It does indeed have a solution. Multiple in fact.

Can the lines cross? Because if yes, that's easy.

It doesn't prevent you from doing so.

indeed

Are you finishing where you started? I don't see how this is easy.

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It's just like the 3x3 solution but with one more line.

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Pathologist here. I feel you.

I guess this works. Nice.

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Good job user.

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Ok

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didn't really need to use brain

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Kurisu has amazing legs.

I can't take my eyes off them.

Who couldn't?

Fair point.

Just finished the VN. Need more of best girl.

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hnnnnggghhh

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Fuck, why did I waste the entire morning trying to solve OP using only basic geometry.

Okarin looks better than Kurisu in this pic, and Kurisu looks pretty good
No homo tho

Kurisu threads make me want to crack open a math textbook for some int gains