/sci/ - Science & Math

[Anonymous No.5994801]

Let's have some fun:

You have a perfect sphere of radius one, imagine it in the corner of a room. It doesn't touch with the actual corner, there is a void. What is the equation for the largest sphere that will fit in this void assuming this is the corner in the first octant?

pic related even though it's two-dimensions. a cross section

[Anonymous No.5994810]

Just put some expanding foam behind it brah, it'll figure it out on it's own.

[Anonymous No.5994815]

[Anonymous No.5994816]

the distance from the origin to the centre of the large sphere is root2.

So (root2 - 1) / 2 is the radius of the small sphere.

[Anonymous No.5994820]

>>5994816
the distance is sqrt(3) not sqrt(2)
this is three dimensions

and no, you can't just half the distance from the origin to the large sphere to get the radius because then the small sphere would be touching the origin and that's impossible.
and you still don't have a center

[Anonymous No.5994842]

>>5994801
(sqrt(2)-1)^2

[Anonymous No.5994844]

>>5994842
no.

[Anonymous No.5994853]

(sqrt(3) -1)/(sqrt(3) + 1)

[Anonymous No.5994858]

>>5994842
divide by 2 for raidus

[Anonymous No.5994862]

>>5994858
>>5994853
>>5994842
>>5994816
/sci/ has gotten pretty bad

[Anonymous No.5994867]

>>5994853
this one is correct

[Anonymous No.5994870]

>>5994867
explain please

[Anonymous No.5994874]

>>5994872
1/10

[Anonymous No.5994872]

>>5994867
no it's not. it's a quantity. what does one real number tell us about a sphere?

[Anonymous No.5994876]

>>5994874
0.1 isn't the answer either you bumbling tart

[Anonymous No.5994877]

>>5994872
nice way to shit up a thread. now we'll have 9000 posts about whether irrational numbers exist or not

[Anonymous No.5994878]

>>5994876
0/10

are you OP, mad that someone answered your puzzle?

[Anonymous No.5994880]

>>5994870
if r is the small radius then consider the point they touch

sqrt(r^2 + r^2 + r^2) + r = sqrt(3) - 1

solve to get my answer at >>5994853

[Anonymous No.5994885]

>>5994880
how did you derive this?

[Anonymous No.5994892]

>>5994885
>i give derivation
>asks for derivation
So OP, this is a h/w problem? yes?

[Anonymous No.5994893]

>>5994892
that's not a derivation. how's that formula related to the image?

[Anonymous No.5994895]

>>5994892
OP here. I wasn't the one asking you to derive it, even though you didn't derive anything.
You're wrong and you didn't even try to solve past the radius. This is not homework, it's a problem I give to my students for fun though so it might be considered such.
Please keep the comments civil.

[Anonymous No.5994897]

>>5994893
>that's not a derivation. how's that formula related to the image?
>if r is the small radius then consider the point they touch

Can't you fucking read anon?

[Anonymous No.5994896]

>>5994893
It's pretty obvious to see how.

The fact that you've been disingenuous by presenting a h/w problem as a fun puzzle means I'm not going to help further.

[Anonymous No.5994898]

>>5994895

>>5994853 is wrong OP?

[Anonymous No.5994900]

((x+1.716)^2)+((y+1.716)^2)+((z+1.716)^2)=(1.716)^2 is the equation of the sphere with a radius of 1.716 units.
>>5994877
your attempt to derail the thread failed.

[Anonymous No.5994901]

>>5994897
and how did you come up with that equation involving the radius?

[Anonymous No.5994902]

3i+sqrt(2)

[Anonymous No.5994903]

>>5994900
made a mistake. it will be
((x-1.716)^2)+((y-1.716)^2)+((z-1.716)^2)=(1.716)^2

[Anonymous No.5994904]

>>5994895
To get a formula for a sphere "beyond the radius" is trivial, and stop samefagging. Stop claiming it is wrong to bait me to help you with your h/w with an explanation.

[Anonymous No.5994906]

>>5994801
OP check my solution >>5994903

[Anonymous No.5994907]

>>5994900
Nope

This is a well known problem, anon had it right here >>5994853

[Anonymous No.5994908]

>>5994904
can't explain it can you?

[Anonymous No.5994910]

>>5994900
why do you think the small sphere has radius 1.716 and is centred at (1.716, 1.716, 1.716) ?

[Anonymous No.5994909]

>>5994908
I understood >>5994880, and it seems good, so stop bitching faggot OP.

[Anonymous No.5994912]

>>5994908
I gave all the explanation anyone who set such a question would need, but not enough for someone with a h/w question that was beyond them.

[Anonymous No.5994915]

>>5994912
I am not OP. I'm >>5994885. I asked for a derivation. You did not give me a derivation.

[Anonymous No.5994916]

>>5994910
i made a mistake. it will be .1716 . i actually scaled the solution up by a factor of 10

[Anonymous No.5994917]

>>5994909
OK. Where did that equation come from?

[Anonymous No.5994918]

>>5994917
From an elementary understanding of the geometric situation

[Anonymous No.5994919]

>>5994918
Which is?

[Anonymous No.5994921]

>>5994916
It's still wrong,

yours is from (sqrt(2) -1)/(sqrt(2) + 1) which is the 2 dimensional version

replace your sqrt(2) with sqrt(3) and you'll be in agreement with the correct solution (sqrt(3) -1)/(sqrt(3) + 1)

[Anonymous No.5994923]

>>5994919
a sphere tangent to three orthogonal planes tangent to a smaller sphere tangent to the same three orthoganal planes

[Anonymous No.5994925]

>>5994921
thanks anon

[Anonymous No.5994927]

>>5994919
All you have to do is admit you set a h/w problem, apologise for all the same fagging and bull;shit, and i'll link you do a detailed explanation

[Anonymous No.5994929]

>>5994927
I can't because I am not OP.

>link
lel. This is what it comes down to huh?

[Anonymous No.5994932]

>>5994929
anyone with any google-fu could find several explanations to this well known high school problem

[Anonymous No.5994935]

>>5994929
>I can't because I am not OP.
what are you doing then?

[Anonymous No.5994941]

>>5994935
I'm just curious about he problem. Conventionally, a derivation is supplied in the answer. And, normally, a mathematician isn't reticent to provide it when asked.

>>5994932
Oh, ok.

[Anonymous No.5994943]

>>5994941
>Conventionally
No, most maths above HS the reader is expected to do a small amount of work.

>mathematician isn't reticent to provide
I am totally happy to help with h/w problems, unless OP pretends they are not h/w problems.

[Anonymous No.5994947]

>>5994943
We are not in a classroom setting.

I am not OP, if you're still trying to imply that.

[Anonymous No.5994957]

>>5994947
>We are not in a classroom setting.
My point entirely. Mathematics correspond with brevity as they expect some intelligence in their readers to join the dots.

>I am not OP, if you're still trying to imply that.
I don't believe you, and neither of us can prove it either way, so my belief is unassailable.

[Anonymous No.5994970]

did you know in an infinite dimensional space, the sphere inside the outer sphere would be the same size at it.

[Anonymous No.5994971]

>>5994970
>at it
*as it

[Anonymous No.5994988]

>>5994957
>My point entirely. Mathematics correspond with brevity as they expect some intelligence in their readers to join the dots.

Don't say you gave a derivation like >>5994892, though.

[Anonymous No.5994990]

>>5994988
That post is pretty adequate to me, you say what r is, and then some obvious geometrical formula follow.

[Anonymous No.5994993]

>>5994988
>i'm dumb, so i'll blame the book.

[Anonymous No.5994994]

>>5994990
Not that guy, but I'm pretty curious too.
Do you start with:

rbig = rsmall

?

[Anonymous No.5994995]

>>5994993
If it's a book that says it derived something when it didn't, then it's a crappy book.

[Anonymous No.5994996]

>>5994994
rBig is given in question as 1 you faggot

so rBig = rSmall would mean rSmall = 1

i cannot believe how many dumb faggots come to /sci/. why not go to forums where high school mediocres discuss their homework?

[Anonymous No.5995001]

>>5994996
Unlike that guy, I'm not afraid to admit that I'm stupid. Yeah, I did miss that part about rB being 1.

But can't you expand the r from a formula?
Also, tangent touch formula? I forgot how that works...

[Anonymous No.5995002]

>>5994990
>That post is pretty adequate to me, you say what r is, and then some obvious geometrical formula follow.

Yeah, I don't think so. Feel free to make me look like an idiot.

[Anonymous No.5995003]

>>5994995
Nope.

All math books above high school expect the reader to do the trivial derivations themselves. And many expect the reader to do harder ones too.

[Anonymous No.5995005]

>>5995003
>trivial
How is it trivial?

[Anonymous No.5995006]

>>5994995
Do you not know what sqrt(r^2 + r^2 + r^2) relates to in the problem? That is pretty much spelling it out.

If anon was being mean he'd have just written that as r sqrt(3).

He was being very generous.

Face it you're just retarded.

[Anonymous No.5995007]

>>5995005
Sorry, should have said trivial to anyone with a three digit IQ.

[Anonymous No.5995008]

>>5995006
>Do you not know what sqrt(r^2 + r^2 + r^2) relates to in the problem?

OK. What is it?

[Anonymous No.5995009]

>>5995008
Sorry, unless you admit you're OP, admit it's a h/w problem and apologise for lying, I'm not gonna help.

[Anonymous No.5995011]

>>5995006
<span class="math">((3)r^2)^0.5[/spoiler]

I am legit fucking curious, and I'm angry that I can't understand where this came from.

[Anonymous No.5995012]

>>5995011
Shit, messed up my code.

<span class="math">(3r^2)^(0.5)[/spoiler]

[Anonymous No.5995013]

>>5995009
>Sorry, unless you admit you're OP, admit it's a h/w problem and apologise for lying, I'm not gonna help

The time I say I'm OP would be the first time I'd be lying.

[Anonymous No.5995014]

>>5995012
...Fuck it.
Also, there IS a way to tell if I'm OP.
Used 4chanx before?

>>5995012
>>5995011
>>5994801

[Anonymous No.5995015]

>>5995014
There's the proof.

[Anonymous No.5995017]

>>5995011
The coordinate of the centre of the inner circle is (r, r, r)

Sqrt(r^2 + r^2 + r^2) is the distance from the centre of this circle to the origin.

[Anonymous No.5995019]

>>5995017
Where the hell did this come from?
I'm doing my first year in Chem Engineering, and I have no clue about this.
I'll look this up later. Thanks.

[Anonymous No.5995025]

>>5995019
Did you not know that to get the distance between two points in three space you use sqrt[(b1-a1)^2 + (b2-a2)^2 + (b3-a3)^2]

[Anonymous No.5995029]

>>5995019
If you've never done coordinate geometry, just use Pythagoras (in two dimensions the distance from the origin to the center of the circle is the same as the hypotenuse of a right triangle with sides r and r.)

[Anonymous No.5995032]

>>5995029
>hypotenuse of a right triangle with sides r and r.
This problem is in 3 space

[Anonymous No.5995037]

>>5995015
Oh yeah, such a nice proof. If only it wasn't easy for even a complete retard to edit the HTML (without even having to understand HTML) by just fucking right-clicking the link and change it before taking a screenshot...

It's like, people go on saying "A picture isn't a proof, it might be photoshoped", well, a screenshot of text in a forum is like a hundred times easier and faster to fake than a pop star being naked.

[Anonymous No.5995044]

>>5995025
assuming the retard doesn't know that, how the hell does he think he can deal with the explanation of a problem like this. it's like teaching algebra to someone who hasn't even got arithmetic

>2a 3a^2 = 6a^3
>but why is 2 times 3 equal to 6 ?

[Anonymous No.5995047]

>>5995032
>in two dimensions

Expand the problem into three dimensions. You still use pythagoras.

There's no way in hell you didn't do this in your engineering maths class, I did that when I was fucking 14.

[Anonymous No.5995050]

>>5995047
Yes, but it is no longer a triangle. Some retards don't know that the euclidean metric in three space is like pythagoras

r^2 = x^2+y^2+z^2

[Anonymous No.5995082]

the nth circle which follows this pattern, has a radius of x*((3-2sqrt(2))^n), where x is the largest circle's radius. So the first one will be 3-2sqrt2

[Anonymous No.5995084]

>>5995082
*This is in 2D though.

[Anonymous No.5995111]

radius = (1-root(2)) / 2

Pythag isn;t hard

[Anonymous No.5995133]

>>5995111
It's 3d space though.

[Anonymous No.5995152]

>>5995133
Your 2d pick cannot be correct if it's a cross section.

You'd need a cross section of the corner rather than the sides.

Anyway, the answer is (1-root(3))/2 for a cube case.

[Anonymous No.5995157]

<span class="math">\left(x-\sqrt{\sqrt{2} - \sqrt{3 - 2\sqrt{2}}}\right)^2 + \left(y-\sqrt{\sqrt{2} - \sqrt{3 - 2\sqrt{2}}}\right)^2 + \left(z-\sqrt{\sqrt{2} - \sqrt{3 - 2\sqrt{2}}}\right)^2 = 2 - \sqrt{2} - \sqrt{3 - 2\sqrt{2}}[/spoiler]
Will post the solution in a next post.

[Anonymous No.5995165]

>>5995157
I made a mistake, that works only for a 2d circle

<span class="math">\left(x-\sqrt{\sqrt{2} - \sqrt{3 - 2\sqrt{2}}}\right)^2 + \left(y-\sqrt{\sqrt{2} - \sqrt{3 - 2\sqrt{2}}}\right)^2 = 2 - \sqrt{2} - \sqrt{3 - 2\sqrt{2}}[/spoiler]

[Anonymous No.5995168]

>>5995165
What the fuck is wrong with me? I meant

<span class="math">\left(x-\sqrt{2 - \sqrt{2} - \sqrt{3 - 2\sqrt{2}}}\right)^2 + \left(y-\sqrt{2 - \sqrt{2} - \sqrt{3 - 2\sqrt{2}}}\right)^2 = 2 - \sqrt{2} - \sqrt{3 - 2\sqrt{2}}[/spoiler]

[Anonymous No.5995171]

>>5994801
If we put in the largest sphere that will fit in the corner, that will still leave behind a gap in the corner, which can again be filled with its largest circle, and so on; but since the whole problem doesnt depend on scaling, the ratio between each spheres radius and the next stays constant, and so does the ratio between each sphere and the distance between its midpoint and the origin. Lets call the radius of the small circle in the OP r. Going from the origin r to the right and r upwards brings us to the midpoint of the small circle, so their distance is sqrt(2)*r.
1/(1+r+sqrt(2)*r)=r/(sqrt(2)*r) (ratio of radius to distance is constant)
=>sqrt(2)=1+r+sqrt(2)*r
=>r=(sqrt(2)-1)/(sqrt(2)+1)
The small circle is that with midpoint (r|r) and radius r. Done.

[Anonymous No.5995176]

>>5995171
Oh wait, sphere? Replace all sqrt(2) with sqrt(3) then.

[Anonymous No.5995211]

<span class="math">\left(x + \frac{2}{3}\left( -3 + \sqrt{3} + \sqrt{6 - 3\sqrt{3}} \right)\right)^2 + \left(y + \frac{2}{3}\left( -3 + \sqrt{3} + \sqrt{6 - 3\sqrt{3}} \right)\right)^2 + \left(z + \frac{2}{3}\left( -3 + \sqrt{3} + \sqrt{6 - 3\sqrt{3}} \right)\right)^2 = \left(-\frac{2}{3}\left( -3 + \sqrt{3} + \sqrt{6 - 3\sqrt{3}} \right)\right )^2[/spoiler]

Solution in the next post

[Anonymous No.5995219]

The equation of a sphere with its center on <span class="math">C\left(x_0, y_0, z_0 \right)[/spoiler] and ray <span class="math">R[/spoiler] is:

<span class="math">\left(x-x_0\right)^2 + \left(y-y_0\right)^2 + \left(z-z_0\right)^2 = R^2[/spoiler]

The sphere with ray 1 and tangent to the planes <span class="math">x=0[/spoiler], <span class="math">y=0[/spoiler] and <span class="math">z=0[/spoiler], is the sphere that has it's center on <span class="math">C\left(1, 1, 1 \right)[/spoiler], its equation is:

<span class="math">\left(x-1\right)^2 + \left(y-1\right)^2 + \left(z-1\right)^2 = 1[/spoiler]

Its closest point to the origin, is the point that intersects the line <span class="math">x = y = z[/spoiler].

The coordinates of that point are given resolving the equation:

<span class="math">\left(x-1\right)^2 + \left(x-1\right)^2 + \left(x-1\right)^2 = x[/spoiler]

and using the smallest of the two values obtained, <span class="math">1+\frac{1}{\sqrt{3}}[/spoiler]. We call that point <span class="math">A\left(1+\frac{1}{\sqrt{3}}, 1+\frac{1}{\sqrt{3}}, 1+\frac{1}{\sqrt{3}} \right)[/spoiler]

Cont.

[Anonymous No.5995252]

>>5995152
nope.
>>5995171
nope
>>5995211
nope

correct answer is a circle centre (r, r, r) with radius r where r = (sqrt(3) - 1)/(sqrt(3) + 1)

[Anonymous No.5995254]

>>5995176
Correct. Glad sci has at least three people that can do this babby problem

[Anonymous No.5995255]

>>5995219
To find the ray of the sphere we want to find, let's translate the circumference and the line so that the center of the circumference lies on the plane <span class="math">x=0[/spoiler], and let's reason in 2D.

We have a line <span class="math">y=x[/spoiler] that intersects a circumference that has one of his extremes in the point <span class="math">{A}'\left(1 + \frac{1}{\sqrt{3}}, 1 + \frac{1}{\sqrt{3}}\right)[/spoiler].

We have to find the ray of the circumference that is tangent to the axes and to the other circumference in the point <span class="math">{A}'[/spoiler].

The distance between two points it's given by the equation:

<span class="math">d=\sqrt{\left ( x_A - x_B \right )^2 + \left ( y_A - y_B \right )^2}[/spoiler]

The center of the circumference should be on the line <span class="math">y=x[/spoiler], the ray of the circumference should be equal to the distance to the axis, therefore <span class="math">d=x[/spoiler], and we know that <span class="math">x_A = y_A = 1 + \frac{1}{\sqrt{3}}[/spoiler], so we can write:

<span class="math">\sqrt{\left ( 1 + \frac{1}{\sqrt{3}} - x \right )^2 + \left ( 1 + \frac{1}{\sqrt{3}} - x \right )^2} = x[/spoiler]

or more simply:

<span class="math">\sqrt{2\left ( 1 + \frac{1}{\sqrt{3}} - x \right )^2} = x[/spoiler]

The length of the ray <span class="math">{R}'[/spoiler], and the coordinates of the center <span class="math">{C}'=\left ( {x_0}',{y_0}' \right )[/spoiler], are given by the smallest solution of the two:

<span class="math">{R}'={x_0}'={y_0}'=-\frac{2}{3}\left( -3 + \sqrt{3} + \sqrt{6-3\sqrt{3}} \right )[/spoiler]

We can translate this again to the original position and write the equation of the sphere:

<span class="math">\left ( x + \frac{2}{3}\left( -3 + \sqrt{3} + \sqrt{6-3\sqrt{3}} \right ) \right )^2 + \left ( y + \frac{2}{3}\left( -3 + \sqrt{3} + \sqrt{6-3\sqrt{3}} \right ) \right )^2 + \left ( z + \frac{2}{3}\left( -3 + \sqrt{3} + \sqrt{6-3\sqrt{3}} \right ) \right )^2 = \left ( -\frac{2}{3}\left( -3 + \sqrt{3} + \sqrt{6-3\sqrt{3}} \right ) \right )^2[/spoiler]

[Anonymous No.5995257]

>>5995255
Second last equation:

<span class="math">{R}'={x_0}'={y_0}'=-\frac{2}{3}\left( -3 + \sqrt{3} + \sqrt{6-3\sqrt{3}} \right )[/spoiler]

[Anonymous No.5995259]

>>5995257
WHAT THE FUCK

<span class="math">{R}'={x_0}'={y_0}'=-\frac{2}{3} \left ( -3 + \sqrt{3} + \sqrt{6-3\sqrt{3}} \right )[/spoiler]

[Anonymous No.5995263]

>>5995219
the x on the rhs is also wrong, it should be x sqrt(3)

HAHA FAGGOT!!!!!

[Anonymous No.5995267]

>>5995211
>>5995219
>>5995255
>>5995259
>>5995257
Not only is your method overly complicated, you have made an error that gives you the wrong answer, and you can't even LaTex

correct answers here
>>5994853
>>5995252

[Anonymous No.5995278]

Now do it in n dimensions. (hard mode, find the number of dimension for which the hyper-sphere extends out further from the origin than the sphere containing it)

[Anonymous No.5995282]

>>5995267
No.

[Anonymous No.5995289]

My solution doesn't work either ;_;

[Anonymous No.5995296]

I was close though, I guess the translation in 2D doesn't work.

[Anonymous No.5995297]

>>5995282
I'm afraid you are a faggot, for proof that the correct answers are in fact correct try

http://jwilson.coe.uga.edu/emt725/Spheres/Spheres.writeup.html

a = 1, b = the r, gives the answer given multiple times in this thread of (sqrt3 - 1)/(sqrt3 + 1)

[Anonymous No.5995306]

In n dimensions, the diameter of our hypersphere of interest is <span class="math"> \sqrt{n} - 1 [/spoiler]

Thus for <span class="math"> \sqrt{n} - 1 > 2 [/spoiler] or <span class="math"> n>10 [/spoiler] the hypersphere extends further out on any given axis than the hypersphere who's closest point to the orgin is containing it.

Enjoy your daily mindfuck.

[Anonymous No.5995304]

>>5995297
Jesus fucking christ, I can't believe that even in the board that must be the most "intelligent" one of 4chan is full of fucktards like you and all the other ones that replied writing bullshit.

I'm out.

[Anonymous No.5995305]

>>5995282
FAGGOT

The correct answer is a circle centre (r, r, r) with radius r where r = (sqrt(3) - 1)/(sqrt(3) + 1).

[Anonymous No.5995309]

>>5995304
>I'm out

I may have an objectionable personality, but I provided a link to show you how and why you are retarded. I could have just called you a faggot.

[Anonymous No.5995313]

>>5995306
no it isn't you retard

that is the distance of the contact point between the spheres to the origin, not the diameter

[sage No.5995329]

>>5995309
I APPLIED MY SAME REASONING IN 2D FOR TWO CIRCLES AND IT WORKS, THE SOLUTION IS <span class="math">2 + \sqrt{2} - sqrt{3 + 2 sqrt{2}}[/spoiler], YOU CAN TRY IT YOURSELF WITH ANY PROGRAM THAT DRAWS A PLOT FROM AN EQUATION.
THE ONLY MISTAKE I MADE IN 3D WAS TO TRANSLATE WITHOUT CONSIDERING THE DISTANCE FROM THE y AXIS.
YOU'RE THE FUCKING FAGGOT BETWEEN ME AND YOU.
SAGE GOES IN ALL FIELDS.

[sage No.5995333]

>>5995329
<span class="math">2 + \sqrt{2} - \sqrt{3 + 2 \sqrt{2}}[/spoiler]

[Anonymous No.5995336]

>>5995329
your reasoning is wrong

TRY http://jwilson.coe.uga.edu/emt725/Spheres/Spheres.writeup.html

[Anonymous No.5995338]

>>5995329
>graphing program
hahaha

i'd rather trust actual reasoning from a university website than some faggot on 4chan drawing pictures

[Anonymous No.5995342]

>>5995329
In 2D the inner circle has radius r = (sqrt(2) - 1)/(sqrt(2) + 1) and is centred at (r, r, r)

[sage No.5995343]

>>5995336
>>5995338
>>5995342
I hope you all fucking die by anorexia.

[Anonymous No.5995346]

>>5995333
not true i'm afraid, try the link given with a = 1 and b the sphere inside the corner

[Anonymous No.5995347]

>>5995342
>(r, r, r)
*(r, r)

[Anonymous No.5995349]

>>5995343
unlikely

[Anonymous No.5995435]

The equation of the little sphere is of the form
(x-h)^2+(y-h)^2+(z-h)^2=h^2, with h>0, because it is tangent to the
coordinate planes and lies in the first octant. So we just need h. The
point on the big sphere nearest the origin is (1-1/sqrt(3), 1-1/sqrt(3),
1-1/sqrt(3)). This must be the same as the point farthest from the
origin on the little sphere. This is (h+h/sqrt(3), h+h/sqrt(3),
h+h/sqrt(3)). Solving h+h/sqrt(3)=1-1/sqrt(3) gives h=2-sqrt(3), and
so h^2=7-4*sqrt(3).

(x-2+sqrt(3))^2+(y-2+sqrt(3))^2+(z-2+sqrt(3))^2=7-4*sqrt(3)

[Anonymous No.5995453]

>>5995435
close but not quite. to get radius of small sphere r you solve

r sqrt(3) + r = sqrt(3) - 1

so r = (sqrt(3) - 1)/(sqrt(3) + 1)

and equation is (x - r)^2 etc = 3(7 - 4 sqrt(3))

[Anonymous No.5995458]

>>5995453
>3(7 - 4 sqrt(3))
dunno where that came from. I meant r^2

[Anonymous No.5995482]

>>5995336
It's weird that the example doesn't show that the smaller circle as touching the wall.
Is there a reason for that?

[Anonymous No.5995485]

>>5995482

pictures are for babbies

[Anonymous No.5995490]

>>5995485
So...the diagrammatic representation isn't correct?
The calculation seems just sound, it's just that this particular illustration grinds my balls.

[Anonymous No.5995503]

>>5995490
oh that pic, the vertical represents the z axis, which the sphere does not touch, the horizontal represents the x,y plane, which it does

that page is overly complicated. if i had pen and paper i'd write a five line explanation. ascii is a drag and i don't latex

[Anonymous No.5995508]

>>5995336
>>5995490
It's...not exactly sound.

>I can express that length in two ways. First it is the diagonal of a cube with side of length a.

How can they assume the diagonal length to be part of a cube? No matter how you scale the circle, the horizontal line of this "cube" (distance between lower b and lower a) IS not a.

Maybe I'm just having trouble seeing this in 3D, though.

[Anonymous No.5995542]

>>5995453
>>5995435

same thing, I think.

r= (sqrt(3) - 1)/(sqrt(3) + 1) = 2-sqrt(3)
r^2 = 7-4 sqrt(3)

[Anonymous No.5995558]

>>5995542
>(sqrt(3) - 1)/(sqrt(3) + 1)
oh, my bad

[Anonymous No.5995632]

drawing square surrounding the lower 4th of the larger circle and smaller circle, the diagonal would be sqrt(2).
sqrt(2) - 1 would give you the diagonal of the square surrounding the smaller circle.
using some trig you can get the height of the square.
divide that by 2 and you get the radius of the smaller circle.
since you give the radius of the larger sphere from the cross section, i am going to assume the radius i found for the smaller circle is the same as the radius of the smaller sphere.

[Anonymous No.5995919]

>>5995304
> I can't believe that even in the board that must be the most "intelligent" one of 4chan
But why people keep telling/thinking that ???

It's so obviously wrong.
This thread is a perfect example.

No matter which board you're lurking on, it's still 4chan anyway, so trolling/bullshit most of the time.

But that's also why we are here.

For /sci/ for instance, we all know that if we have serious, real questions, /sci/ is not gonna help a lot.

[Anonymous No.5996432]

>>5995278

In <span class="math">n[/spoiler] dimensions, let <span class="math">{\bf u} = (1,1,\dots,1)\,\in
R^n.[/spoiler] The sphere we want has equation <span class="math">\|{\bf x}-r{\bf
u}\|^2=r^2.[/spoiler] All we need is <span class="math">r.[/spoiler] The point on this
sphere farthest from the origin is <span class="math">r(1+1/{\sqrt n}){\bf u}.[/spoiler]
This must be the same as the point nearest to the origin on the other
sphere, which is <span class="math">(1-1/{\sqrt n}){\bf u}[/spoiler]. Solving for
<span class="math">r[/spoiler] gives <span class="math">r= (n+1-2{\sqrt n})/(n-1).<span class="math">[/spoiler][/spoiler]

[Anonymous No.5996442]

>>5996432
as the dimension gets higher, the radius of the smaller sphere approaches the same size as the larger sphere.

[sage No.5996451]

>Easy homework-eque problem
>/sci/ doesn't autosage
>/sci/ can't even solve it

Why the fuck do I even come here anymore

[Anonymous No.5996465]

GONNA TRY AND SOVLE IT NOT LOOKING AT THE THREAD OR ANY OF YOUR FAGGOT POSTS FUCK YOU
So anyway, the center of the large circle is obviously (1,1), so the bit where it touches the smaller circle is ((1-(cos45)),(1-(sin45)) = ((1-sqrt(2)/2),(1-sqrt(2)/2)).
So then I took the ratio of the distance from origin to the further edge of the large circle at a 45 degree angle, under the distance to the further edge of the small circle at a 45 degree angle, and I got 0.239.
0.239 = Rs/Rl -> Rs = 0.239 So (0.239, 0.239)

oh fuck you said sphere
Fuck it, it's probably pretty close to being the same thing.

[Anonymous No.5996483]

the diameter of the sphere will be root 2 take 1.