/sci/ - Science & Math

>>10024802
You copied my answer I s2g I WILL SUE

I ended up with sqrt(1/3) as my answer, btw

>>10024802
nigger it's obviously (sqrt(2) - 1)/2

I solved it in 10 seconds just looking at it

>>10024812
You're leaving out the space between the smaller circle and the corner of the square

sqrt{r(b)^2+r(b)^2} = r(b) + 2r(s) + x

>>10024802
I'm getting (2-sqrt(2))/(2+sqrt(2)) but that's all mental math so may have messed something up.

The answer is pi/2

[math]r = \sqrt{\frac{{A}+{x}}{\pi}[/math]

Where A Is the area difference between the circle and square and x is the area difference between A and the area of the small circle.

>>10024820
Not claiming you're wrong, but would you mind pointing out where I went wrong? I'll type out my work in a sec

Test
[eqn]\frac{a}{b}[/eqn]
Is LaTeX broken?

the diameter of the small circle should be sqrt(2)-1.

>>10024802
[eqn]\frac{sqrt{2}-1}{2}[/eqn]

>>10024830
Turn off ad-blocker

>>10024838
No, doesn't amount for white space between R+2r and corner

>>10024823
fixed
[math]r = \sqrt{\frac{{A}+{x}}{\pi}}[/math]

>>10024838
nvm I'm wrong.

I got 3 - 2sqrt(2).

>>10024802
(sqrt(2) - 1) / (sqrt(2) + 1).

>>10024845
OK, what are A and x in terms of R?

>>10024850
>>10024823

>>10024852
Doesn't answer the question.

>>10024853
brainlet detected

>>10024842
Second this.

>>10024856
Still doesn't answer the question

>>10024802
Draw a line from the center to the corner.
1+r+sqrt(2)*r=sqrt(2)
r=(sqrt(2)-1)/(1+sqrt(2))

>>10024849
This

>>10024824
I see now that I accidentally squared things individually instead of applying it to the entire side of the equation. Reworking things now.

>>10024802
(2-sqrt(2))/(2+sqrt(2))

>>10024868
After reworking things and no longer squaring individual values, it was about this point where I got lost

>>10024889
What is x and where does the second line come from?

(Sqrt(2^2 + 2^2) - 2 ) ÷ 4

holy shit, never again will i make the mistake of thinking everyone on sci isn't a fucking brainlet.
it is so obviously sqrt(2) - 1

>>10024838
>>10024842
>>10024893
All same, all wrong
>>10024873

>>10024896
>t.

>>10024896
forgot the /2 its (sqrt(2)-1)/2

>>10024896
Yeah, you shouldn't think that as long as you're here...

>>10024891
x is the distance between the edge of the smaller circle and the corner of the square
The second line comes from using the Pythagorean theorum, making a triangle from the center of the larger circle to the corner of the square, and as it bisects the smaller circle, I just used 2r instead of d.

>>10024849
yah this, which simplifies to 3-2*sqrt(2)

square root of 2
minus 1
divided by 2

>>10024802
does anyone know where to get more of these problems

>>10024802
Pic related. The relation of the blue radius has same relation to the length blue line as red radius to orange line (where the lines touch the grey lines of course).

All you have to do is to calculate funcions of both lines, and make r1/l1 = r2/l2 connection.

Line formula is f(x) = ax+b
Let me start: a of both lines is -1 because it's symmetric sircle.
The crossing point of line 1 and circle 1 is at (-sqrt(2) /2 , -sqrt(2) /2), of the line 2 and circle 2 is (sqrt(2) /2 , sqrt(2) /2).

Now you get funcions of both lines which are calculated like this:
f(-sqrt(2)/2)=-sqrt(2)/2*-1+b=-sqrt(2)/2*
-> b=-sqrt(2)
--> f1(x)=-x - sqrt (2)
f2 is similar to f1 and leads to f2(x)=-x+sqrt(2).
Pretty easy, right? Now we just need to calculate 2 points where those lines touch the x=1 and y=1 lines.

Points for l1 are
(1, -1-sqrt(2)) and (-1-sqrt(2),1)

Points for l2 are
(1, -1+sqrt(2)) and (-1+sqrt(2),1).

Now just calculate the length with pythagoras of both lines and we get
l1= 4,828...
l2= 0,828...

and now back to start, r1/l1=r2/l2
<->
r2=r1* (l2/l1) = 1* (0,828.../4,828...)

And we get the solution:
r2=0,17157....

square root of two

subtract one

divide by two

this is literally middle school math what is wrong with you people, get the fuck off my board

>>10024962
https://www.youtube.com/user/MindYourDecisions/videos

>>10024993
How could we have missed this??! Dipshit.

>>10024802
OK, so the answers been here for quite some time and it's unrefuted. Done with this shit and out. Stop posting wrong answers. Stop posting answers. Stop posting.

Literally every one here should be able to do this. Unless you dropped out of middle school.

>>10025019
You don't need all these gay triangles.

Check out this algebra:
(sqrt(2)-1)=x,x/(x+2)=y
x=0.4142...
y=0.1715...

>>10024981
This looks well thought out and is actually easy to follow. No idea if it's correct, but I'm thinking it is. Thanks, anon.

>>10025027
Gah! I should have drawn in more lines!

>>10025019
[2-(1+sqrt(2)/2)] / (1+sqrt(2)/2)
=3-2*sqrt(2) = 0.17157... ok you are right.

Ok fuck me, it took me 15 minutes to find out and calculate this shit >>10024981
while you could just solved it in under 1 minute.
Damn.

>>10025035
Anon, thanks but better do it like in the other pic related, saves you much more time.
The blue vertical line has same relation to radius of big circle as orange vertical line to radius of small circle which is literally what I did but way faster.

>>10025019
>>10025059
Damn, I really need to brush up on my compass-math type stuff. Triangles, ratios, correlating the lines and angles and all that. It's quite helpful, it seems

>>10024802
Phytagoras theorem

>>10025095
brainlet

>>10025095
Except there is a gap between the small circle and the corner and people above did it right already

>>10024802
>I'm a brainlet
2r=√2-1

>>10025388
*I meant r√2+r=√2-1

>>10024802
its just a basic level geometric series.the big circle has a radius of 2 the small circle has a radius of 2k then the other smaller circle 2k^2 and the sum is equal to 1+√2. you just do 2/(1-k)=1+√2 and find k

>>10025454
>>10024802
or 3-2√2 if you wish

>>10025454
>>10025462
i wanted to say you wanted to say that you need infinitely circles to feel and it forms a infinite geometric series but I guess /sci/ got it my answare is the same as the triangle dude

>>10024896

>>10024889
[math]
\sqrt{2}=1+ \sqrt{r^2+r^2}+r = 1+ \sqrt{2}r+r = 1+r(1+ \sqrt{2}) \\
r= \dfrac{ \sqrt{2}-1}{ \sqrt{2}+1} = \dfrac{( \sqrt{2}-1)^2}{2^2-1^2}= 2-2 \sqrt{2}+1=3-2 \sqrt{2}\approx 0.17157
[/math]

>>10024889
[math]
\sqrt{2}=1+ \sqrt{r^2+r^2}+r = 1+ \sqrt{2}r+r = 1+r(1+ \sqrt{2}) \\
r= \dfrac{ \sqrt{2}-1}{ \sqrt{2}+1} = \dfrac{( \sqrt{2}-1)^2}{2^2-1^2}= \frac{2-2 \sqrt{2}+1}{3}=\frac{3-2 \sqrt{2}}{3}=1- \frac{2}{3} \sqrt{2} \approx 0.05719
[/math]

>>10025462
that's 200% too big

>>10024802
0.25

These two line segmentations are similar. now just calculate the lnegths and set their ratios equal and solve for 2r

>>10025727
Messed up. red line was supposed to be longer.

>>10024802
>brainlets can't solve an 8th grade geometry problem
>state of /sci/ in 2018.

>>10024802
(1+r)/(1-r) = sqrt(2)

>>10024938
Yes

>>10024958
you're thinking that the outer edge of the small circle is hitting the corner. it doesn't

>>10024802

>>10024802
Just use trig for fuck's sake. Geometry is dead for a reason.

1 = (1+r) cos(45°) + r
[1-cos(45°)]/[1+cos(45°)] = r = tan^2(22.5°) = 0.17157....

>>10024802
The patrician method

>>10024802
1-sqrt(2)-(x-sqrt(2x))=x
solve for x

sqrt(2) = 1.4142135624
... - R = c

c = 0.4142135624
... ÷ ([2+c] = 2.4142135624 ) = r

r = 0.1715728753

>>10027842
fucking hated engineering graphics but lol'd at this

>>10024802

>>10025730
whoa. Nice

>>10027874
>[math]\frac{\sqrt{2}-1}{\sqrt{2}+1} = r[/math]

>>10027852
dammit
sqrt(2)-1-(sqrt(2x)-x)=x
sqrt(2)-1-sqrt(2)*sqrt(x)+x=x
sqrt(2)-1=sqrt(2)*sqrt(x)
[eqn](\frac{\sqrt{2}-1}{\sqrt{2}})^{2}=x[/eqn]
I hope that's right.

>>10024802

>>10028883
oop i made an error

\frac{\left(\sqrt{2}-1\right)}{\sqrt{2}+1}

>>10024802
I FOUND IT TEACHER

>>10024802
If the pic is your problem then you and almost everyone here is a brainlet.
Radius of the big circle is the same as half of the big square lenght, so we can make a quarter square with it.
It's diagonal will be sqrt(2).
Wich means radius of the small circle is (squrt(2)-1)/2 aka difference between diagonal and radius divided by 2.

>>10024802
r = (sqrt(2) - 1) / (sqrt(2) + 1)

>>10029334
LMAOOOO BRAINLETTTT

(root(2)-1)/(3-root(2))

>>10029334
WROOOOOOOOONG

>>10025672
>=(2√−1)222−12=2−22√+13=3−22√3=1−232√≈0.05719
are you retarded? terrible aritmetic skills, go bac to school

>>10029334
fucking look at you

>>10024848
right

the fact that this thread has so many replies ensures the fact how try hard mathematicians and "intelligent" you all try to be.
While solving a high school problem and giving yourself an intelligence elevation.
"yea! i can do this, damn i am one hell of genius"

>>10030082
some answers are just people trying to find a creative and elegant solution. which I'd say isn't too stupid

>>10029368
this is correct and follows directly from line 12 here
>>10024889
you got lost because you didnt see that sqrt(x^2+x^2)=sqrt(2x^2)=x*sqrt(2). lines after 13 are wrong btw

>>10030090
Thank you - I appreciate it!

>>10030082
It's only slightly fewer replies than a "[ a/b(c+d) = ? ]" thread would get.

(sqrt(2) - 1)/2

The lenght of square is 2 so the diagonal is 2sqrt(2) (pythagore theorem) and the diagonal is also one diameter of the big circle and two diameters (4 radius) of the small circle.

so you have (2sqrt(2)-2)/4

Took a couple of minutes in geogebra

It's quite short when you do it in a coordinate system.

>>10031255
You don't need coordinate system for anything here. You can simply remove the axes and it would still be the same.

Oh my god, the legend is true, /sci/ is overrun by swaths of drooling brainlets. Imagine 99 replies to such a trivial problem, that's it, im leaving this stupid place

>>10031366
>no fun allowed

>>10027842
>>10031104
based

>>10028722
sad that I had to scroll down this far to find the best method.

>>10024802
[math] \dfrac{\sqrt{2} - 1}{2} [/math]

>>10032338
Oh shit, forgot about the small corner. Nevermind.

>>10032343
To account for the corner you just duplicate the step though, with the new hypothenuse being [math] \dfrac{\sqrt{2} - 1}{2} [/math] and r being the height of the small right triangle.

>>10025095
imagine having to go through life being this guy

calculate surface of big circle
calculate surface of square
subtract surface of circle from surface of square
divide by 4
and so on and so on

easy

>>10032349
>surface

>>10032352
what's the right word?
my English is bad, I'm Chinese

>>10032354
Sucky sucky fucky fucky.

>>10032354
Calculate the area of the big circle?

>>10024900
Think you're correct cause I got the exact same answer and everyone else seems to have a different asnwer.

Are you all serious? That's literally an elementary school problem
2*sqrt(2)-x=2
x=2*sqrt(2)-2
r=x/4=sqrt(2)/2 - 1/2

>>10024802
(sqrt(2)-1)/2

>>10024802
https://www.imaginary-institute.com/blog/2014/07/23/a-little-geometry-problem/

(sqrt(2)-1)/(sqrt(2)+1)

How does this fucking thread get a hundred posts

>>10032408
4chan embraces crappy threads.

help

I did it!!!!!

1-sin(45°)-(cos(45°)-cos(45°)^2)/(1+cos(45°))

A = 1
B = sqrt(2)+1
D = sqrt(2)-1

C = (sqrt(2)-1)/(sqrt(2)+1) = 0.1715728753...

>>10028722
This honestly, took me less than a minute to solve.

>>10033402
also pretty elegant

>>10033551
I think my first attempt was more elegent >>10033196

>>10033607
wrong

Nigga just use a ruler...

>>10024802
Wow, you're either a brainlet, not old enough to even be online, or just trying to bait. How can you even work on this for "too long"? The radius of the big circle is one, so the distance from the center of the square to the corner is 2^(1/2) because Pythagorean theorem. Subtract 1 from that, and you're left with the diameter of the small circle. Divide the diameter of the small circle by 2 and you are left with it's radius, r=((2^(1/2))-1)/2. I'm a brainlet and even I could figure that out.

>>10033607
you dumb fuck

>>10033623
Preach! These black mufuggin crackas be dumb as hell! Lawd hav mercy....

Three minus two times the square root of two.

>>10024820
>>10024824
You guys seriously can't eye out that it isn't pi/2? Yikes

>>10035176
What do you mean the small circle isn't 57% wider than the larger circle???

Let S be the length of a side of the square, R the radius of the big circle, and r the radius of the small circle.

S = 2R

By Pythagoras' theorem:
(2R + 2r)^2 = 2S^2 = 2(2R)^2 = 8R^2

=> 2R + 2r = sqrt(8)R = 2sqrt(2)R

=> R + r = sqrt(2)R

=> r=(sqrt(2)-1)R

When R=1, r = sqrt(2)-1

>>10035567
wrong

>>10035567
woops should be 4r instead of 2r, giving r=(sqrt(2)-1)/2

>>10035576
wrong

[sqrt(2)/2-1]/2 = 0.207

>>10035567
>>10035576
>>10035588

>>10024802
Half of sqrt(2) - 1

>>10035887
Nvm thats the brainlet answer, its 3 - 2sqrt(2)

(Sqrt(2)-1)/(2*sqrt(2))

>>10024802
The small circle is inside a square that is 1/36 of the square of the original circle by inspection.

>>10036106
Well, between 1/64 and 1/36.

>>10024802
Fucking piece of shit circle. The answer is 3-2(sqrt(2)). The small circle is inside a square similar to the original circle i.e. it's the same thing with the circle in the corner just smaller. I then used proportions to solve for the diameter of the small circle and divided by two.
Took me 30 minutes. Feeling like a huge brainlet right now.

>>10024802
Are you all literally fucking retarded? It's a fucking right angled triangle. Sweet mother of god...

Diagonal = sqrt(2)
Radius of inner circle = 1
=> r = sqrt(2) - 1

>>10036241

DIVIDED BY 2

FUCK

>>10036241
Ahh, /sci/. Where everyone comes together to not read anything in the entire thread, manages to act like a condescending twat and then make the same mistake a bunch of the other guys have made before.

>>10036280
I'm still your darling, am I?

>>10036283
If by darling you mean let's engage in a bout of hatefucking then yes

>>10024802
(1+r)^2=2(1-r)^2
1+2r+r^2=2-4r+2r^2
0=1-6r+r^2
r=3+-2sqrt(2)
First solution is nonsense so r=3-2sqrt(2)

3-2sqrt(2)

side of square is 2
>diagonal of square is 2sqrt(2)

4r+2=2sqrt2

r = [sqrt(2) - 1]/2

>>10036307
You see, the center and corner circles' diameters don't make full square diagonal

>>10036314
i dont think its this cause if they werent colinear then the little circle wouldnt have 2 pts of contact with the square

im wrong tho cause i didnt account for the smallest little bits in the corners

the answer is 2sqrt(2) < Answer < [sqrt(2) - 1]/2] lmao

>>10036314
im pretty sure they're colinear

im wrong tho cause i didnt account for the little corner bits

>>10036307
>>10036322

>>10033402
this is best soln i am big dummy lol

>>10033642
lol

>>10024802
roughly r=0,124040226

I can't read shit when you type out math in raw test
I also can't be bothered to try and explain in words what I did, just that I called the naive answer a, and the awkward shitty gap x.

tl;dr:
[math]r = 3 - 2 \sqrt{2}[/math]

>>10024802
this shit is simple pythagoras
(1+r)^2 = (1-r)^2 + (1-r)^2
surely even you brainlets can simplify this and then use the quadratic formula

Length of square = 1*2 = 2
Length of diagonal of the square c^2=a^2+b^2
c=8^0.5
c=2.83

Length of square corner to closest part of the circle = diameter of circle - length of square diagonal
= 2-2.83
=0.83
Half it to equal one corner gap
=0.414
Half it to equal the small circle radius
=0.207
=0.2

>>10037459
Fuck
Plus the white bit thats left over
I'm such a fucking brainlet

>>10024802
(big r + small r ) * cos(pi/4) + small r = big r

>>10037466
>I'm such a fucking brainlet

We already knew that when you used decimal representations of numbers and typed formulas all out plaintext

>>10024802

>>10024802
I realized that the circle was inside a triangle. The distance from the big circle to the corner is that triangle's height. Then just used basic triangle properties and then used the right triangle formula for the in-radius.
This crap took me several hours to figure out.

>>10038131
As a tutor, I have a "rule of thumb". Every sketch that my thumb can completely cover is non-existent.

The distance between the small center and the large is r+1, and it's making an angle of pi/4 with the green radius. So 1=cos(pi/4)(r+1)+r. Redistribute and plug into cosine and you get 1=(1+root(2)/2)r+root(2)/2. Then do algebra to isolate r, and simplify the radical to get 3-2root(2). Very straightforward problem, everyone in this thread is retarded.

>>10038449
It's r = (sqrt(2) - 1) / (sqrt(2) + 1) as shown above multiple times. You don't really need any trigonometry for that, you can use similarity or diagonal equation.

>>10036190
Took me 1 hour.

>>10024802
very simple:
the length from the center to the top right corner L is equal to sqrt(2).
Also L = R + r + rC with R being the large circles radius, r being the small circles radius and rC being the length from the circles center to the top right corner.
rC = r*sqrt(2).
Thus we can solve this:
L = R + r + rC
sqrt(2) = 1 + r + r*sqrt(2)
Thus we get
(sqrt(2)-1)/(sqrt(2) + 1)) = r

you know the coordinate where the two circles touch, so you can say that the little circle must be centered around the point within the upper-right area that's equidistant from the contact point and the two sides of the square. once you know that point you can just solve for the radius.

we also know that point lies along the diagonal, and that the distance from a point on the diagonal to the top side is just their difference in y-coordinates, and the distance between that point and the right-side is the difference in x-coordinates. so i think the whole thing just works out to a constrained linear system.

>>10038851

if i center it around 0,0, then

it's the point (c,c) where 1-c = ||(c,c) - (sqrt(1/2),sqrt(1/2))||

1-c = sqrt(2)(c-sqrt(1/2))
1-c = sqrt(2)c -sqrt(2)sqrt(1/2)
1 + sqrt(2)sqrt(1/2) = (sqrt(2) +1)c
(1 + sqrt(2)sqrt(1/2))/(sqrt(2) + 1) = c

and then subtracting sqrt(1/2) for the point on the circle gives a radius of about 0.121.

i think...

>>10038904
The answer should be around 0.172

3-2sqrt(2)

>>10038904

where am i screwing up? it's close but i made a mistake somewhere. subtracting my value of c from 1 gives the right answer.

>>10038918

yeah, i know but i'm not sure where i'm messing it up. ||(1,c) - (c,c)|| = 1-c, which should be equal to ||(c,c) - (.71,.71)||, since the center of the little circle is going to be equidistant from the sides of the square and from the point (.71,.71) where the circles have to touch.

>>10038904
>>10038925
>>10038930

ah, i found it. i have to take ||(.121, .121)|| which is indeed .17

>>10024802

>>10026959
Based anon

>>10033402
u won

1-cos(45)-(cos(45)-cos(45)^2)/(cos(45)+1) = 3-4cos(45) ~ 0.1716

Draw a line from the centre of the small circle to the horizontal radius of the large circle, and draw a line connecting the two centres.

The right triangle satisfies
(1+r)^2 = (1-r)^2 ^ (1-r)^2

Solve to get answer

Draw a line from the centre of the small circle perpendicular to the horizontal radius of the large circle, and draw a line connecting the two centres. This forms an isosceles right triangle.

The right triangle satisfies
(1+r)^2 = (1-r)^2 + (1-r)^2

Solve to get answer

2R=S(square side)=2
2R+4r=2√2
2+4r=2√2
r=(√2-1)/2
Really easy

>>10029368
why not just write as 3-2*sqrt(2)

>>10040815
I just didn't cared enough to remove irrationals from denominator.
That's how I got it:

sqrt(2) = 1 + r + sqrt(2) * r
sqrt(2) * r + r = sqrt(2) - 1
r * (sqrt(2) + 1) = sqrt(2) - 1
r = (sqrt(2) - 1) / (sqrt(2) + 1)

>>10040065
lmao brainlet you forgot about the white space in the corner

it looks like the circle is wearing a hat huhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhu

>>10024802
(sqrt( 2*(( 2 * 1 )^2 )) - 1 ) / 2
Solved in 1 minute

>>10041596
>(sqrt( 2*(( 2 * 1 )^2 )) - (2*1) ) / 4
sry fixed

( / ( - ( sqrt (2) ) 1) 2)

>>10033402
pretty but unnecessarily complicated

>>10033402
Fucking good
>>10036575
Too complicated

Am i on to something ?

Fugg i just realized o missed something

>>10024900
This is correct.

The diagonal of the encompassing square is sqrt(8) = 2sqrt(2). This distance minus the big circle's diameter (2) gives you 4 r's.
r = (2sqrt(2) - 2) / 4 = (sqrt(2) - 1) / 2

its just a geometric series, why /sci/ makees it seem so hard?

>>10044029
Brainlet

2 WINNERS, they are

>>10028722
>>10033402

>>10044580
Post your method, dumb faggot

>>10044894
It's already in the thread, brainlet.

>>10032408
And, it's still alive. I'm surprised.

>>10045386
seriously why does everyone have such an obsessive need to post their own solution when clearly it's been posted a hundred times over
is this a compensation thing