PUZZLE TIME

[Anonymous No.6406933]

It's your favorite day, /sci/!

Post other /sci/ related puzzles if you have them.

>inb4 "does .9999... = 1?"

[Anonymous No.6406937]

well if the box isn't touching the ground anymore i would guess there is no more pressure applied to the ground

[Anonymous No.6406939]

>>6406937
I'm not sure if you read the question correctly.

[Anonymous No.6406941]

The weight of the box on the ground remains the same until the tenth balloon lifts it.

[Anonymous No.6406942]

does .3333... = 1/3?

[Anonymous No.6406947]

>>6406939
Oh, I see

Well I would say, yes. Because it's like a car jack. When you lift a car the wheels decompress, so pressure must be coming off of the car, and so same with the balloons and the box, the box still has elasticity and can decompress, only at a much smaller scale then the wheels.

[Anonymous No.6406946]

>>6406942

[Anonymous No.6406958]

>>6406947
It's different in that a car has multiple parts, and the jack as it rises cannot fall, unlike the balloons which (until balloon ten) are lowered by the box's weight and gravity.

[Anonymous No.6406961]

Each balloon subtracts a small (about a tenth) of the weight of the box, until the combined weight reaches 0 or less, causing it to take off.

So the pressure deceases a bit for each ballon.

Kinda obvious, is it not?

[Anonymous No.6406962]

>>6406933
of course it is. this is serious babby tier physics and i feel bad for any kids in this thread who get this question wrong for any reason.

[Anonymous No.6406970]

>>6406958
I don't know, but I'm getting a strange feeling in that direction. It seems like the weight of the box decreases, but not necessarily the pressure applied to the ground.

I'm not just dead sure of this yet, though.

[Anonymous No.6406973]

>>6406970
dude. pressure is force divided by area. if the weight is dropping, the pressure is dropping.

[Anonymous No.6406978]

>does .999... = 1?
>does .333... = 1/3?

I'm putting this to bed right now. These questions, without more context, are senseless.

You're asking if a conceptual number is the same as a real amount. It's like comparing two totally different languages, or states of being; it doesn't actually make any sense to try to work it out, or to put it in an equation like that.

[Anonymous No.6406977]

ok, i'll bite. the pressure the box applies to the ground will decrease from balloon one onwards. as the balloons inside (given they're helium baloons or whatever) will counteract gravity until their combined forces overcome gravity and lift it skywards.

newtonian physics yay.

[Anonymous No.6406987]

a box with 9 balloons in it would be easier to lift up than a box with none, so I would say yes.

[Anonymous No.6407019]

You can solve this in 5 seconds with a free body diagram.

Does anyone remember the thread where a guy posted a picture of a car that ran off a ramp and crashed into a barn? And then we had to figure out how fast it was going not knowing the geometry of the problem?

THAT was a good puzzle thread.

[Anonymous No.6407046]

[Anonymous No.6407162]

>>6406973
I don't know what hes doing on /sci/ ignore him.


Yes the boxes weight decreases as you add balloons as they decrease the average density of the box, and as it's volume is constant.
Therefore decreases its average mass, with each balloon displacing air of equal volume which weighs more.
As weight (the force exerted on the ground by the box) is proportional to mass, this means the force against the floor is also decreasing.

[Anonymous No.6407182]

>>6407046
part 1. a series of instructions
part 2. a series of instructions

questions like this are why almost any philosopher who is not really a mathematician is worthless

[Anonymous No.6408467]

0 0 0 =6
1 1 1 =6
2 2 2 =6
3 3 3 =6
4 4 4 =6
5 5 5 =6
6 6 6 =6
7 7 7 =6
8 8 8 =6
9 9 9 =6

make these expressions true without using any numbers

[Anonymous No.6408473]

>>6408467
(0!+0!+0!)! =6
(1+1+1)! =6
2+2+2 =6
3!+3-3 =6
sqrt+4+sqrt4+sqrt4 =6
5+5/5 =6
6+6-6 =6
7-7/7 =6
(sqrt(8+8/8))! =6
sqrt(sqrt(9*9))+sqrt9 =6

[Anonymous No.6408475]

>>6406942

Yes, because .9999... = 1

so .999... / 3 = .333...

Now .999... / 3 = 1 / 3

so .999... / 3 = .333.... = 1 / 3

[Anonymous No.6408477]

>>6406961
>Each balloon subtracts a small (about a tenth) of the weight of the box, until the combined weight reaches 0 or less, causing it to take off.
>So the pressure deceases a bit for each ballon.
>Kinda obvious, is it not?

Exactly

[Anonymous No.6408479]

>>6408467
0+0+0<=6
1+1+1<=6
2+2+2=6
3+3+3>=6
4+4+4>=6
5+5+5>=6
6+6+6>=6
7+7+7>=6
8+8+8>=6
9+9+9>=6

[Anonymous No.6408480]

>>6406978
>You're asking if a conceptual number is the same as a real amount.

wtf is a conceptual number

[Anonymous No.6408484]

>>6408473
>(0!+0!+0!)! =6

clever

[Anonymous No.6408613]

I like this simple geometry puzzle. (ABC is isosceles; find angle x).

[Anonymous No.6408617]

Who was free body diagram? babby's first kinematics

[Anonymous No.6408626]

Challenge: Derive the Lorentz Transformation

Hint: consider the law of transmission of light

[Anonymous No.6408636]

>>6406933
Wait, is the whole debate about 0.999... = 1 basically a paradox or just a trap for non /sci/ folks?

I just realized that 0.333...+0.333...+0.333... =/= 0.999... It's = 1.

[Anonymous No.6408649]

>>6408636
0.999... = 1 is a real thing, you idiot.

http://en.wikipedia.org/wiki/0.999...#Algebraic_proofs

[Anonymous No.6408663]

>>6408613
I'm familiar with that one. It's pretty great. You should point out though that the reader should stick to just geometry methods, otherwise it can be bruteforced with trigonometry fairly easily (though without any elegance).

[Anonymous No.6408687]

>>6406941

No it doesn't, the balloons would be pushing up on the roof of the box and removing pressure from the ground

[Anonymous No.6408693]

Find the closed form answer of ζ(3)

[Anonymous No.6408714]

>>6408613
Fuck, I'm stuck at the sum of the two top angles of the middle triangle being equal to 110.

[Anonymous No.6408721]

>>6408649

>page started in 2013
>about 50 edits since then

something tells me that it was a bunch of /sci/ nerds trying to prove a point

this is the downfall of wikipedia, when people can try to validate false information just by creating a page for it.

[Anonymous No.6408725]

ζ(3) = Σ n^-3
Σ n^-3 = Σ (1/n)^3 = 1/(1-3)

[Anonymous No.6408753]

>>6408725
0/10

[Anonymous No.6408890]

Help me out here, because I think there isn't enough information to solve. Filling in the numbers gives us the angles shown and then we end up with four unknown angles but this is what I came up with.

angle ADE = a
angle AED = b
angle DEC = c
angle EDB = x

This gives us the system

c + b = 130
a + b = 160
x + a = 140
x + c = 110

Assuming no value can be negative in this system, x has a range of 0 to 110. Unless I'm missing something there is no exact solution to the problem.

[Anonymous No.6408922]

>>6407046

[Anonymous No.6408929]

>>6408636
>or just a trap for non /sci/ folks?

Took you long enough.

It's something so simple you can troll newfags with it / people who don't know anymore science than pop-science will consider it worthy of arguing over to prove their knowledge

Keeps them outta other threads

[Anonymous No.6408939]

>>6408890
Looks like you can start solving multiple variables by making lots of equations. I forgot what it was called.

DEC + EDB + 70 = 180,
DEC + AED + 50 = 180,
ADE + EDB + 40 = 180,
AED + ADE + 20 = 180.

That's 4 equations and 4 unknowns. Is that enough?

[Anonymous No.6408941]

>>6408890
I'm pretty much stumped, too, but seeing as it can be solved with trig, there's at least a solution.

[Anonymous No.6408952]

Hey Nurgle you Fat Fuck.

This is the changer of the ways speaking.

If you send the perfect Objectivist Girl to the Dutch "Abonai" convention.
I'll whisper the answer of P vs NP into her ear.


I'll be that annoying fuckwat with a white hat.
You know who.

~That Damned Blue Bird Of Hope/Happiness~
~Every problem is solved efficiently by making it somebody ells's problem~
~Every Objectivist has a SEB field~

[Anonymous No.6408966]

>>6408693
no

[Anonymous No.6408968]

>>6408939
nope. using a linear algebra approach the grid ends up being

1 0 -1 0 30
0 1 1 0 130
0 0 1 1 110
0 0 1 1 110

which becomes
a - c = 30
b + c = 130

in other words, it is not bound, there is no single definite solution to the problem. Work it out for yourself, and reconfirms the range since

x + c = 110

[Anonymous No.6408971]

>>6408480
It's a number.
It's not real.

Checkmate atheists.

[Anonymous No.6408976]

>>6408721
So you're positing the page shouldn't exist as it's bogus/wholly false information and thus wagering it will eventually be removed.

I'll take that bet.

[Anonymous No.6408979]

>>6408971
What's real number? Is zero real?

[Anonymous No.6408990]

>>6408968
Using trig I got 30 degrees.
No idea how to do it with basic geometry.

[Anonymous No.6408995]

>>6408890
hint: construct another point and use properties of isosceles triangles all over the place.

[Anonymous No.6409002]

>>6406978
le epic le troleing

[Anonymous No.6409006]

>>6408990
I'm not sure how you would. I don't see any length given anywhere, or are you doing brute force knowing one side is equal to the other?

>>6408995
There's only one valid place to create a KNOWN isocseles triangle and it really doesn't connect to anything useful for the purposes of finding angle X.

[Anonymous No.6409033]

>>6409006
>There's only one valid place to create a KNOWN isocseles triangle and it really doesn't connect to anything useful for the purposes of finding angle X.
From knowing that BE and BC are the same length you can determine the lengths of basically all of the lines in the lower portion of the diagram. If you just grind out expressions, you'll eventually have enough to shit out the angle you want. It's just laborious.

[Anonymous No.6409038]

>>6409033
*relative lengths, relative to BC; just set BC=1 or something

[Anonymous No.6409044]

>>6409033
>BE=BC

Oh sit how did I not see that?

[Anonymous No.6409048]

Couldn't you just do something like:
>Set BC=1
>Since ABC is isosceles, can work out y-coordinate of A
>A's x-coordinate is just 1/2
>B's ccoordinate is x=0, y=0
>The line BD can be expressed analytically as just y=sin(60)*x
>the line AC can be expressed analytically using its gradient -sin(80) and its y-intercept (which will be 2 times A's height)
>Solve for the intersection of those 2 lines

[Anonymous No.6409049]

>>6409033
> BC = BE
Wait how is that true? I mean it LOOKS pretty close but no one ever accepts that as proof.

[Anonymous No.6409054]

>>6409049
See those two angles which are both 50 degrees...

[Anonymous No.6409055]

>>6409048
Oh, and similarly solve for the position of E; then just express the slope of ED relative to the slope of BD as an angle.

[Anonymous No.6409097]

>>6408626
It's pretty easy, you start by calculating the deformation of the electric field around a point charge if the point charge has some velocity v.

You go on to work out the (deformed) shape (and length) of the path that another opposite charge would take in that field's presence (i.e. an orbit).

From there it more or less drops out that you have scaling factors for the length of objects (objects are smaller if their electron cloud shapes are smaller), and for the time that physical processes take (oscillation times within deformed potentials). You can reformulate Physics in the "local time" of any moving body; and combining these scaling factors results in the familiar "Lorentz factor" to convert between frames.

It's especially interesting that apparatus will always *measure* themselves to be at rest, and *measure* light to have a relative velocity of c, since they and their apparatus always deform accordingly. Despite, of course, the fact that they are not at rest, and that light only moves at c in one frame.

http://en.metapedia.org/wiki/Relativity_theory_of_Lorentz

>inb4 muh Einstein
>inb4 muh subjectivity
>inb4 muh positivism

[Anonymous No.6409153]

>>6409097
I note that the webpage I linked there doesn't comment much on the first part of the derivation, and simply introduces the length-contraction as per Lorentz' *initially* ad-hoc claim. A more comprehensive treatment was given in J.S. Bell's "Speakable and Unspeakable in Quantum Mechanics", chapter 9.

https://dl.dropboxusercontent.com/u/3219541/textbooks/bell_speakable_unspeakable.pdf

[Anonymous No.6409565]

>>6409097
>I can imagine a maaaaagical measurement apparatus that can maaaaaaagically measure a 'true' 'real' universe
>So I don't need to admit that the universe we live in is indistinguishable from a relativistic universe!
>I am *so SMART*

[Anonymous No.6409566]

>>6409565
Go to bed, Oscar.

[Anonymous No.6409568]

>>6409565
>muh instrumentalism
>The universe doesn't actually exist, only my measurements do
>No actual mechanism occurs by which I receive these measurements, they simply magically enter my posession!
>The universe not being real is already a foregone conclusion, and just like with the existence of the Christian God, anyone taking a counter-stance is simply being a contrarian denialist le edgy hipster atheist ecks dee

[Anonymous No.6409577]

This should spark some debate.

[Anonymous No.6409580]

Anyone have the rod in the bowl puzzle?

[Anonymous No.6409582]

>>6409565
Well yeah, obviously there's no 'real' universe, and everything's Einstein-relativistic. I mean, how can mirrors be real if our eyes aren't real?

[Anonymous No.6409584]

>>6409577
>you pick a coin at random, toss it, and record the result
>after 10 tosses
You didn't specify whether I get a new coin for each toss.

[Anonymous No.6409588]

>>6409584
or if the tossed coin leaves the pool of future picks

[Anonymous No.6409590]

>>6409580
fuck me its way too late to think about this

[Anonymous No.6409617]

>>6409580
Pls anons

Pls

It tilts to the right, don't let this take you all over again

[Anonymous No.6409615]

>>6409577
The interpretation can either be "what is the probability of eleven consecutive tails" or "what is the probability of a tails".

[Anonymous No.6409623]

>>6406941
opposite post

[Anonymous No.6409642]

>>6408479
oh lol

[Anonymous No.6409656]

Heres some interesting problems

[Anonymous No.6409668]

>>6409656
>if the oil and water are mixed than the pressure at the bottom must increase......
wut?

[Anonymous No.6409677]

>>6409668

Oil is less dense than water. When you mix the oil-water mixture, in the column above the scale there will be higher ratio of water than there was before, hence a higher mass and greater pressure?

[Anonymous No.6409682]

>>6409677
Oh, yes, I see.

[Anonymous No.6409694]

>>6409580
imagine two empty containers, one with a ping pong ball inside, the other with a steel ball hanging over it. Which one weighs more? Yes, the ping pong ball container. Now add the same amount of water to both containers. The ping pong container will still weigh more. About the thought that the force of the ping pong string pulls up on its glass container, yes it does, but that does not effect the total weight of the system for the ping pong ball is made of air and offers no extra buoyancy to the container

I think I'm right?

[Anonymous No.6409701]

>>6409694
Not quite. You're forgetting about the buoyant force on the steel ball.

[Anonymous No.6409705]

>>6409694
Not a bad approach, but you need to include the buoyancy of both orbs, even though only one is buoyant enough to float.

[Anonymous No.6409758]

>>6409577
With replacement, it's simply 11/20.
Without replacement, it's 1/2?

[Anonymous No.6409824]

>>6409577
You pick a coin. You start off with 1/10 probability (1:9) that it's unfair, and flip it ten times, getting 10 tails. This has likelihood 1 if unfair, 1/1024 if fair, for a likelihood ratio of 1024:1. Multiplying gives 1024:9, or a ~99% probability that the coin is biased. Which in turn means the chance of a tails on the next flip is about 99.5%

Alternatively, the question meant something else or I can't maths.

[Anonymous No.6409835]

>>6409824
>not optimising your PNGs

[Anonymous No.6409837]

>>6409824
This is correct. Almost nobody that one.

[Anonymous No.6409939]

>>6408479

lmao

[Anonymous No.6409974]

>>6409824
>>6409837

that`s cool

[Anonymous No.6410998]

I can't seem to find the original problem, there was an exterior circle we were supposed to find the radius of. Anyone have the original for this triangle problem?

[Anonymous No.6411048]

>>6408613
http://www.gogeometry.com/LangleyProblem.html

[Anonymous No.6411049]

>>6408479
Thats new, nice one.

[Anonymous No.6411063]

>>6408636
0.999... = 1 for people who don't know how to divide. This is to be expected given the nonsense presented as "real numbers."

> one goes into one zero times
This is what passes for math since Cauchy.

[Anonymous No.6411583]

>>6410998
I remember this, it was a screenshot from some japanese game show. I remember attempting a constructionist approach but it turned out that it requires a non-constructible number somewhere in there.

[Anonymous No.6411645]

Find the centre of mass of this shape using only a straight edge. No measuring compasses, hanging stuff, etc etc.

[Anonymous No.6411650]

>>6411063
>one goes into one zero times remainder one
fixed

[Anonymous No.6412266]

>>6408473
You should be able to do it without using sqrt().

[Anonymous No.6412268]

A guy has a yard stick, and he is running towards a barn that is, slightly less than a yard in length, at .8c. Does the yard stick fit in the barn?

[Anonymous No.6412279]

>>6412268
Depends on whose frame of reference.
From the stick's point of view, it does not fit in the barn.
From the barn's point of view, it does.
Any more relativity "paradoxes", /sci/?

[Anonymous No.6412597]

>>6412279

Wait, that seems wrong.

The stick shrinks (from the barns reference frame) and thus ...

Wait no, I just got it. I missed the bit where the barn was slightly shorter. I thought they were equal in length.

Rewording slightly:

A stick (1 yard)
A barn (1 yard)

.8c relative velocity

From the sticks reference frame, the barn is shorter (so it doesn't fit)

From the barns reference frame the stick is shorter (so it does fit)

...

Huh.

Interestingly, my immediate first thought (that you'd be wrong) was wrong even with the misinterpreted question. I wonder what I was thinking.

[Anonymous No.6412729]

lets start with a = b + c where a, b and c are positive and a-b is positive

lets multiply it with a - b

a^2 - ab = ab + ac -b^2 - bc

then we will subtract ac

a^2 - ab - ac = ab - b^2 -bc

which is same as

a (a - b - c) = b (a - b - c)

now lets divide it with a - b - c

and i becomes

a=b

now its your job to find what i did wrong or did i just explode math by dividing with zero

[Anonymous No.6412736]

>>6412597
>From the sticks reference frame, the barn is shorter (so it doesn't fit)
>From the barns reference frame the stick is shorter (so it does fit)

Yes, and the important part to note is that both of these things can be true without causing any contradictions.

[Anonymous No.6412744]

>>6412279
Imagine a disk of radius R rotating with constant angular velocity <span class="math">\omega [/spoiler]
The reference frame is fixed to the stationary center of the dick. Then the magnitude of the relative velocity of any point in the circumference of the disk is <span class="math">\omega R [/spoiler]. So the circumference will undergo Lorentz contraction by a factor of <span class="math"> \sqrt{1-(\omega R)^2/c^2} [/spoiler]. However, since the radius is perpendicular to the direction of motion, it will not undergo any contraction. So
<div class="math"> \frac{ \mathrm{circumference}}{ \mathrm{diameter}} = \frac{2 \pi R \sqrt{1-( \omega R)^2/c^2}}{2R} = \pi \sqrt{1-(\omega R)^2/c^2} </div>.
This is paradoxical, since in accordance with Euclidean geometry, it should be exactly <span class="math"> \pi [/spoiler]

[Anonymous No.6412754]

>>6412729
By definition a - b - c = 0, it's in the axiom of the problem bitch tits

[Anonymous No.6412755]

>>6412744
>center of the dick
I know you're a homosexual

[Anonymous No.6412756]

>>6412744
the no pi term is just a scaling factor anon-kun

[Anonymous No.6413417]

>>6411645
Is nobody going to attempt this?

[Anonymous No.6413437]

>>6411645
I sort of just guessed

[Anonymous No.6413473]

>>6408467
Why not add 10 to that list?
[sqrt(10-(10/10)]!= 6

[Anonymous No.6413551]

>>6412756
oops. engineering mistake
>>6412755
I copied it as best I could from wikipedia

[Anonymous No.6413600]

>>6411583
Oh man, really? I tried so hard to figure this out. I think we had at least two threads on it. Did you save your work? I'd love to see the resolution of this.

[Anonymous No.6413623]

>>6411645
First i estimated center if the entire object extended as far as the bottom right corner. Then i assumed center would move down and left if only the bottom right extended that far, so draw straight line down to left

Then estimate center of the shape if bottom right was cut off. Then assume center would move down and right if adding on bottom right
draw line down and right


find intersection of lines

[Anonymous No.6413627]

>>6413623
forgot image

[Anonymous No.6413715]

>>6413623
In the right general area, but no guessing allowed!

[Anonymous No.6413719]

>>6413437
Not even close

[Anonymous No.6413725]

>>6413715
come on, if i printed it out, took out a pencil and a straghtedge, i could make it perfect

Is there a better method than mine?

[Anonymous No.6413758]

>>6413725
>First I estimated
>Then estimate

Yes, there's an exact method.

[Anonymous No.6413761]

>>6411645

[Anonymous No.6413762]

>>6413758Estimated in this case meant that if i wanted to draw in the missing parts and then draw in the diagonals using a straightedge to find the centers i could, but I'm lazy

[Anonymous No.6413768]

>>6413761
This could well be correct, I'm not totally sure I don't really understand what you've done over on the right.

I have noticed this though:
How did you get the horizontal lines bisecting the angles at points 1 and 2?

No compasses allowed!


This is defiantly the closest yet, mind.

[Anonymous No.6413770]

btw i swear if the anser is balance it on the straightedge or some shit like that i'm gonna be annoyed

[Anonymous No.6413774]

>>6413762
Ah ok, so you found the centre of the full rectangle, including the the missing bit, and the centre of the left hand rectangle, without the bottom of the L, then...? What gave you the lines you then drew diagonally, the ones that cross at your centre?

[Anonymous No.6413777]

>>6413770
No no no, I promise it's a simple 2D geometric construction.

[Anonymous No.6413783]

>>6413774
Each of the lines is a line from each center to the opposing corner

so line from larger shape's center would go to bottom left corner

and line from the shape minus the small section's center would go to bottom right corner

[Anonymous No.6413786]

>>6413768
I didn't use a compass, I transported the angles from the figure below. But I did estimate the weights to make those triangles, considering the big rectangle is 5 times bigger than the smaller one.

http://fr.wikipedia.org/wiki/M%C3%A9thode_du_dynamique_et_du_funiculaire

I have no idea how you call that in english

[Anonymous No.6413793]

>>6413783
Ah I see now. In that case, no that doesn't give the correct centre of mass.

[Anonymous No.6413813]

>>6413786
I see. I haven't seen that graphical "Méthode du dynamique et du funiculaire" before, it looks quite old fashioned...

Anyway, while I'm sure this gives a correct centre of mass, by measuring the bases of those triangles at the bottom, (one having base 5 units, the other base 1 unit), you haven't just used a straight edge. No length measuring allowed!

[Anonymous No.6413844]

>>6413813
then I have no idea how to do it, I know it is on the line joining the centers of the two squares. But I don't know how to find out where on that line

[Anonymous No.6413850]

>>6413844
You could always make two DIFFERENT squares...

[Anonymous No.6413876]

>>6413850
Good job making me feel stupid

[Anonymous No.6413890]

>>6413876
I wasn't quite sure how to give a hint without totally giving it away.

It's funny, the first time someone showed it to me, I did exactly the same thing as you. Got the first line, then couldn't for the life of me work out what to do next.

When they pointed it out I got the same feeling of "How did I not see that?"

[Anonymous No.6413916]

>>6413876
>>6411645
now that this has been solved the next puzzle is...
TO DO IT AGAIN. IN A DIFFERENT WAY.

I promise the second solution is just as simple a construction

[Anonymous No.6413961]

>>6408613

x=40?

[Anonymous No.6414008]

>>6413961
Work

[Anonymous No.6414020]

>>6408971
How can numbers be real if i isn't real?

[Anonymous No.6414023]

>>6414008
>that hand-writing

Computer Scientist?

[Anonymous No.6414275]

>>6408613
ok now what

[Anonymous No.6414279]

>>6414275
After working the problem to this point I doubted that there was a unique solution to the problem. Then I saw the solution, but wasn't convinced. Particularly I doubted that there was a point F on DC that was equidistant from E and B and EFB had an angle of 60 degrees, but after I reviewed it, I came to the conclusion that the point no only existed, but had to exist.

[Anonymous No.6414283]

>>6414275
forgot about ex angle theorem

[Anonymous No.6414286]

>>6408479
My sides

[Anonymous No.6414288]

Find the total area of the grey circles.

I saw this posted about a week after /sci/ was created and never found the solution.

[Anonymous No.6414355]

>>6414288
Good one. I called the radius of the biggest grey circle r. I got r = R/4 now, but this doesn't look right. May be a calculation error though.

[Anonymous No.6414369]

>>6414288
Call the radius of the largest grey circle small r.
Purple line has length r
Red line has length R/2 + r
Yellow line has length R - r

Call the center of the largest (half)circle (0, 0). We now that the center of the largest grey circle has y = r now. Call the x coordinate u and solve for it using the triangles and Pythagoras:

looking at the triangle formed by the yellow and purple line and the bottom on the halfcircle, we get:
u = sqrt((R-r)^2 - r^2)

Doing the same for the red and the purple line:
R/2 + u = sqrt((R/2 - r)^2 - r^2)
so
u = sqrt((R/2 - r)^2 - r^2) - R/2

substituting both in u = u yields:
sqrt((R-r)^2 - r^2) = sqrt((R/2 - r)^2 - r^2) - R/2

(here is when I went wrong in my last post, forgot the R/2 part)

Now, take a square maybe? I haven't worked this out yet.

[Anonymous No.6414376]

>>6414369
Correction, I used a minus instead of a plus in the Pythagorean stuff of the red and purple line.

Final equation is supposed to be:
sqrt((R-r)^2 - r^2) = sqrt((R/2 + r)^2 - r^2) - R/2

[Anonymous No.6414386]

>>6414376
If we work out the squares and brackets, we get:
sqrt(R^2-2Rr) = sqrt(R^2/4+Rr) - R/2

take the square of both sides (reversible operation because both sides are positive):

R^2 - 2Rr = R^2/2 + Rr - R * sqrt(R^2/4 + Rr)
R^2/2 - 3Rr = R * sqrt(R^2/4 + Rr)
divide by R
R/2 - 3r = sqrt(R^2/4 + Rr)
square
R^2/4 - 3Rr + 9r^2 = R^2/4 + Rr
9r^2 - 4Rr = 0
r(9r - 4R) = 0
9r - 4R = 0
9r = 4R
r = 4R/9

[Anonymous No.6414387]

>>6414386
I probably should have done a more general version of this, so that we can apply the same technique for the smaller squares (because only the quantity R/2 changes into some bigger part of R.

[Anonymous No.6414397]

Wouldn't you need to take a sum somewhere? The circles seem to get infinitely small.

[Anonymous No.6414401]

>>6414397
Yes, this is not the final result. I just found the radius for the largest grey circle. No concrete idea how to continue. Except doing the same but now with aR instead of R/2, but this is a lot of work...

[Anonymous No.6414782]

>>6414401
>>6414397
>>6414387
>>6414386
>>6414376
>>6414369
>>6414288


Guys, guys, guys! All the tangents to the lines connecting the centres, each being straight lines of length r(n)+r(n+1), cut the axis at 1 point! Can we do something with this?

[Anonymous No.6414806]

>>6406933
I've only done first year physics, but I'll give it a shot. Pressure = Force/Area, right? So the downwards force due to gravity acting upon the box. But when balloons are added, they exert an upwards force on the box, so the new downwards force of the box on the ground is F = mg - qb, where b is the rising force of a single balloon and q is the number of balloons. Since the area of the box that is touching the ground does not change during this time, the pressure changes linearly until the 10th balloon is added, making the box rise.

Someone pls respond

[Anonymous No.6414809]

>>6406941
Weight is a force. Pressure is Force/Area.

[Anonymous No.6414833]

>>6414782
I have this so far:
Consider the shape formed by the 2 red triangles. because they're tangent, the right angle is justified.

Clearly, the length L is constant for all n. Where n is the number of the circle counting anti-clockwise from the largest one.

So using this fact and a simple formula for the are of each triangle, we can eliminate L to form a recurrence relation on r, the radius of each circle.

This yields the equation <span class="math"> r(n+1)=r(n) (\frac{Ar(n+1)}{Ar(n)})[/spoiler].

I believe this ratio of areas between is constant in which case this is a simple geometric series, as we know r(1) due to this kind anon >>6414386.
Thus the solution becomes simple.

So the question is, can anyone find Ar(2)?

[Anonymous No.6414837]

>>6414833
Forgot pic.

[Anonymous No.6414839]

>>6414837
Would've been better if I'd actually labelled whet I wanted people to find...

[Anonymous No.6414849]

>>6408613
180 - (Angle a plus angle c minus 50 )= angle d
Due to straight line is 180

180 - (20 + c - 50)

C is 80 because it's a simple isosceles.

Therefore ? Is 130

Lol it's as If u guis aren't high school students that enter I as tests.

[Anonymous No.6415056]

>>6414839
any chance you could measure the radius of C2? I'm getting 4/17

[Anonymous No.6415085]

>>6414782
I don't see how you could prove this.

[Anonymous No.6415098]

Assuming that the box has equalized pressure with the ground, the weight on the ground will decrease as more balloons are added.

The same way you can float a balloon with a little weight or have it kiss the floor or descend slowly with more weight.

[Anonymous No.6415194]

>>6414806
yeah this is true, the puzzle is just stupidly easy, I think I learned this in high school physics

[Anonymous No.6415216]

Whats so difficult about the triangle problem...x(EDB)=30
Just calculate AEC=130
Then 130-50 for DEC
And 180-70+80=30 q.e.d

[Anonymous No.6415231]

Solve E=mc^2

[Anonymous No.6415319]

>>6415056
I can tell you the coordinates of J and f, but it won't be exact as I placed C1,C2,C3... by guesswork. they defiantly intersect the smaller and larger semicircles where they're supposed to, but may not meet each other as perfect tangents.

Yup, 4/17 looks right, after checking between j and the point f. Well done. How did you get it?

[Anonymous No.6415353]

>>6415319
Used http://en.wikipedia.org/wiki/Descartes%27_theorem which given the radii of three tangent circles produces the radius of the fourth circle tangent to all three. You can do this iteratively to get the radii of all the circles, which are 4/9, 4/17, 4/33, 4/57, etc. The pattern is 4/(4n(n-1)+9) starting from n=1. To get the final answer, sum up all those radii squared times pi and you get <div class="math">\frac{1}{32} \pi ^2 R^2 \left(\sqrt{2} \sinh \left(2 \sqrt{2} \pi \right)-4 \pi \right) \text{sech}^2\left(\sqrt{2} \pi \right)</div>which is ~ 0.869973 R^2.

[Anonymous No.6415362]

>>6415353
the LaTeX was supposed to be
<div class="math">\frac{1}{32} \pi ^2 R^2 \left(\sqrt{2} \sinh \left(2 \sqrt{2} \pi \right)-4 \pi \right) \operatorname{sech}^2\left(\sqrt{2} \pi \right)</div>

[Anonymous No.6415364]

>>6415362
>>6415353
goddammit well it's Out[7] in the screenshot I posted

[Anonymous No.6415518]

>>6415364
Awesome job! But this should be done without Computer! How can we calculate the recurrence relation and the sum??

[Anonymous No.6415564]

>>6413916
solve analytically
express the solution as a linear combination of the edgepoints
draw the corresponding lines

[Anonymous No.6415582]

The squares of an n-x-n checkerboard are randomly assigned the values 1 through n^2. Show that there are two neighboring squares (sharing an edge) whose values differ by at least n.

I don't have a proof yet.

[Anonymous No.6415668]

>>6415518
assuming R = 1, Descartes' theorem gives <span class="math">k_{n+1} = 1 + k_n + \sqrt{k_n - 2}[/spoiler], where <span class="math">k_n[/spoiler] is the curvature of the nth circle. I'm not sure how this leads to <span class="math">k_n = \frac{4n(n-1)+9}{4}[/spoiler] though. And I have absolutely no idea how to evaluate <div class="math">\sum_{n=1}^\infty \left(\frac{4}{4n(n-1)+9}\right)^2</div>

[Anonymous No.6415716]

>>6411645
By measuring compass do you mean a compass that lets you measure an angle or just any normal compass?

I ask because normally people talk about straightedge and compass constructions.

http://en.wikipedia.org/wiki/Compass-and-straightedge_construction

[Anonymous No.6415768]

you should be able...

[Anonymous No.6416008]

>>6415716
Sorry, I missed a comma. It should be: no measuring, no compasses (of any kind.)

Only using a straight edge. One already found a solution, but I do know there are at least 2.

[Anonymous No.6416009]

>>6416008
that should read *One anon >>6413876

[Anonymous No.6416026]

>>6415668
Well it simplifies down to <span class="math">2 \sum_{n=1}^{\infty} \frac{1}{2n^4-4n^3+9n^2+1}[/spoiler]
I don't know where to go from there. It doesn't factor nicely.

[Anonymous No.6416039]

>>6415362
<span class="math">\frac{1}{32} \pi ^2 R^2 \left(\sqrt{2} \sinh \left(2 \sqrt{2} \pi \right)-4 \pi \right) \sech^2\left(\sqrt{2} \pi \right) [/spoiler]?

[Anonymous No.6416157]

>>6415668
Ill be asking my calculus teacher for this one!

[Anonymous No.6416271]

>>6416026
This looks like something that *may be solved through Laplace. Don't mind me, it just looks eerily similar to something I saw in a pre-DSP class or something.

[Anonymous No.6416450]

>>6416008
Thank you for the clarification.

[Anonymous No.6416521]

>>6415768
nineteen

[Anonymous No.6416582]

>>6407019
>Does anyone remember the thread where a guy posted a picture of a car that ran off a ramp and crashed into a barn? And then we had to figure out how fast it was going not knowing the geometry of the problem?
Do you realize that the thread in question was a claims adjuster trying to get out of paying for somebody's car by saying "hey he was speeding and breaking the law so fuck 'em"

[Anonymous No.6416586]

>>6408473
>sqrt+4+sqrt4+sqrt4 =6
that's to the 1/2 power.

[Anonymous No.6416599]

>>6408467
inline int evaluate(int x, int y, int z) { return 6; }
int main {
evaluate(0,0,0)

evaluate(1,1,1);

evaluate(2,2,2);

evaluate(3,3,3);

evaluate(4,4,4);

evaluate(5,5,5);

evaluate(6,6,6);

evaluate(7,7,7);

evaluate(8,8,8);

evaluate(9,9,9);
}

[Anonymous No.6416605]

THE WEIGHT OF THE BOX IS CONSTANT, you fucking idiots! All that changes is the force exerted by the box on the ground. Lets say the box weighs exactly 1Kg and local gravity is 10ms^-2 (to make the maths easy) when no balloons are inside the box the force exerted downwards towards the ground is 1*10 = 10 newtons. Now lets assume each balloon provides 1 newton of buoyancy in the upwards direction. With one balloon inside the box, it will exert 10 newtons - 1 newton = 9 newtons on the ground. But the weight of the box is still constant. Up until the 10th balloon where we have 10 newtons (box) - 10 newtons (balloon buoyancy) there will always be a force exerted by the box on the floor. And as a result, a pressure over the boxes area.

[Anonymous No.6416629]

>>6408467
>>6408473
[(4!) / (4+4)]! = 6

Working on 8/9

[Anonymous No.6417332]

>>6415668
I solved the recurrence one! Can't solve the sum though!

[Anonymous No.6417476]

>>6416605
>THE WEIGHT OF THE BOX IS CONSTANT, you fucking idiots! All that changes is [the weight].
That's what you're saying right now

[Anonymous No.6417785]

>>6417332
You want to share with the class?

[Anonymous No.6418060]

>>6417785
>>6417332
Yes, I can share, but it's kinda big! But I'll do it anyway! Still looking for the sum solution!

[Anonymous No.6418067]

>>6416605
>THE WEIGHT OF THE BOX IS CONSTANT

weight =/= mass

[Anonymous No.6418206]

>>6418060
Part 1

[Anonymous No.6418255]

>>6417785
Part 2

[Anonymous No.6418263]

>>6417785
Part 3

[Anonymous No.6418303]

>>6418206
>>6418255
>>6418263
I and a quick skim through and this looks legit. Good stuff!

[Anonymous No.6418315]

>>6418067
No, but it's still true as long as the box doesn't move relative to the earth (and it won't until all ten balloons are put into it).

[Anonymous No.6419549]

>>6415216
How do you know that AED=50 degrees?

[Anonymous No.6419951]

>>6419549
Because DBE is 20 and the other angle is 110 (you get 70 from the first triangle and 180-70=110) so 180-20-110=50

[Anonymous No.6419986]

>>6419549
oh sorry you meant AED
because AEC is 130
BEC=50
and DEC is 130-50=80
so you know AED is 180-50-80

[Anonymous No.6420001]

>>6419986
I'm following you up to DEC=130-50=80.

If I understood correctly, you're saying that DEC=AEC-BEC.
Since AEC=AED+DEC, it would follow that AED=BEC, which is indeed true, but I simply can't see how it can be proven geometrically.

[Anonymous No.6421195]

>>6420001
You are right, my mistake
Its not solvable x can be from 0 degrees to 110

[Anonymous No.6421294]

>>6418206
When I use the quadratic equation I get k_n=(1/R)(1+Rk_(n-1)+/-2sqrt(Rk_(n-1)-2))

You have pretty much the same thing except without a 2. Did I fuck up or did you just miswrite yours? We get the same answer of k_2=17/(4R) or 9/(4R) but our equations are different so I'm a little confused.

[Anonymous No.6422941]

>>6416026
bump. I really want to know how to evaluate that