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Warp wrote:
Is the set of points inside a unit square larger than the set of points on a unit line?
(Before someone instinctively answers "of course", remember that eg. the set of rational numbers is actually exactly as large as the set of integers. Adding a dimension doesn't necessarily increase the size of an infinite set.)
By your phrasing, I think you're asking whether the cardinality of [0,1] x [0,1] is larger than [0,1]. The answer is no. Dimensionality (or more strictly, the cartesian product of two sets) doesn't increase the cardinality of a set. There are some pathological curves in differential geometry that manage to fill the whole plane, and some fractals that induce, as the name suggests, "fractional" dimensions, while fractional cardinality numbers don't exist.
One bijection between R and R2 is to get the decimal representation of x and y in a point (x,y) and map them to something like 0,x1 y1 x2 y2 x3 y3 ...
If you make a consistent choice between choosing 1 or 0,9999999.. you can get only one real number for every two x and y.
In real analysis though, the "size" of a set, depending on the context, can mean its measure, and in this case, you can say a square is larger than a line, in the sense that a line (or any enumerable set of lines) has null measure in R^2. One example of this is that, if the discontinuity points of a function in a domain in R^2 is a line (or more generally, has null measure), the function will still be Riemann integrable in that domain, because the places where discontinuity happens are "too small" or "negligible".
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I solved it!
First of all, let's prove that
Let's manipulate the sum a little to find:
Using this recurrence, we can prove our desired equality by mathematical induction.
Now, using Liebniz's expansion:
^It's actually \sum{j=1}{i-1} instead of \sum{j=1}{n-1}
So, your expression is simply the sum of the reciprocals of the first n squares. Euler was the first to prove that this series converges to pi^2/6 .
EDIT: Too late!
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There's nothing I hate more than series. I tried to find some recurrence relations in that sum, but didn't find something useful. It probably has something to do with the Euler-Mascheroni constant.
Randil, are you sure that the limit of this series can be found by elementary means? (i.e, is this a textbook problem or something similar?)
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Randil wrote:
1. Set g(x)=f'(x). Since f(x) was differentiable, g(x) is continuos
Not true, g doesn't need to be continuous. (Students keep writing this on the test...)
EDIT:
I forgot to put a counter example. Consider
f(x) = {x2sin (1/x) , x!=0 ; 0, for x=0}
See that f'(x) = 2 x sin(1/x) - cos (1/x) , x!=0
For x=0, f'(0) = lim x->0 (x2sin (1/x) / x ) = lim x->0 (x sin(1/x)). sin (1/x) is bounded and x tends to zero, so the limit is zero.
However lim x->0 f'(x) doesn't exist, because lim x->0 (2 x sin(1/x))=0 and lim x->0 (cos(1/x)) doesn't exist and f' is not continuous on 0.
The error in 2 was pointed by petrie911.
Also, nice proofs for 3b and 3c ! I did something a little different though, because that's one way to demonstrate that the Taylor series of sin x and cos x converges to those functions for all reals, thus I didn't use them for the proof.
b) Let a(x) = f(x) - sin x and b(x) = g(x) - cos x . Verify, similarly to (a), that a(x)^2 + b(x)^2 = 0 , and then, a(x)=b(x)=0. Implying that f(x)=sin x and g(x) = cos x
c) s(x) the first series and t(x) the second. Just check that s(0)=0, t(0)=1, s'(x)=t(x) and t'(x)=-s(x). By (b), they must be equal to sine and cosine.
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I found this a little boring, despite it being short enough and well executed.
However, since it seems that the most accepted opinion in the site is that movies like Mike Tyson's punchout are amazing, this is a good addition, so Yes.
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A little of real analysis
1) Consider a real function f [a,b] -> R , differentiable in all its domain, with f'(a)<f'(b). Prove that, for every k such that f'(a)<k<f'(b), there exists an x in ]a,b[ satisfying f'(x)=k .
2) Let f R -> R be a real function. f is non-decreasing in the interval ]-infinity,a[ for some a in R, and f's image in this interval is bounded from above. Show that the one sided limit LIM f(x) x->a- exists.
3) Consider two real functions f,g R -> R satisfying f(0)=0 , g(0)=1 , f'(x)=g(x) and g'(x) = -f(x) for all x in R.
(a) Show that the following equality holds for all reals f(x)^2 + g(x)^2 = 1
(b) Prove that f(x)=sin x and g(x)=cos x
(c) Conclude that
sin x = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...
and
cos x = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ...
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I think he's attempting a category similar to JXQ's run (or the Maxim no warps run). He uses door warps and the shuriken glitch, but avoids the zipping tricks in klmz's eleven minute movie.
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I couldn't find a software that could do a triple integral on something that's not a box, so I implemented one myself (probably not as precise as it could be): http://pastebin.com/k8RTmBsR
This code is able to integrate n-dimensional integrals, experienced programmers should be able to quickly modify it to calculate the gravity field at any point. The present code outputs:
Point A
Ratio 0.05 : ( 0.2795 , 0.0000 , -0.0000 )
Ratio 0.10 : ( 0.5272 , -0.0000 , -0.0000 )
Ratio 0.15 : ( 0.7439 , 0.0000 , -0.0000 )
Ratio 0.20 : ( 0.9320 , -0.0000 , 0.0000 )
Ratio 0.25 : ( 1.0927 , -0.0000 , 0.0000 )
Ratio 0.30 : ( 1.2260 , -0.0000 , -0.0000 )
Ratio 0.35 : ( 1.3317 , -0.0000 , -0.0000 )
Ratio 0.40 : ( 1.4095 , 0.0000 , 0.0000 )
Ratio 0.45 : ( 1.4585 , 0.0000 , 0.0000 )
Ratio 0.50 : ( 1.4778 , 0.0000 , -0.0000 )
Ratio 0.55 : ( 1.4661 , 0.0000 , 0.0000 )
Ratio 0.60 : ( 1.4228 , -0.0000 , -0.0000 )
Ratio 0.65 : ( 1.3465 , -0.0000 , -0.0000 )
Ratio 0.70 : ( 1.2377 , -0.0000 , -0.0000 )
Ratio 0.75 : ( 1.0949 , 0.0000 , 0.0000 )
Ratio 0.80 : ( 0.9189 , 0.0000 , 0.0000 )
Ratio 0.85 : ( 0.7124 , 0.0000 , -0.0000 )
Ratio 0.90 : ( 0.4792 , 0.0000 , -0.0000 )
Ratio 0.95 : ( 0.2233 , -0.0000 , 0.0000 )
Point B
Ratio 0.05 : ( -0.0300 , -0.0000 , -0.3079 )
Ratio 0.10 : ( -0.0981 , 0.0000 , -0.6236 )
Ratio 0.15 : ( -0.1916 , 0.0000 , -0.9412 )
Ratio 0.20 : ( -0.3034 , -0.0000 , -1.2608 )
Ratio 0.25 : ( -0.4283 , -0.0000 , -1.5823 )
Ratio 0.30 : ( -0.5622 , 0.0000 , -1.9055 )
Ratio 0.35 : ( -0.7019 , 0.0000 , -2.2300 )
Ratio 0.40 : ( -0.8442 , 0.0000 , -2.5553 )
Ratio 0.45 : ( -0.9870 , -0.0000 , -2.8808 )
Ratio 0.50 : ( -1.1280 , -0.0000 , -3.2058 )
Ratio 0.55 : ( -1.2656 , -0.0000 , -3.5296 )
Ratio 0.60 : ( -1.3981 , -0.0000 , -3.8511 )
Ratio 0.65 : ( -1.5244 , -0.0000 , -4.1698 )
Ratio 0.70 : ( -1.6433 , 0.0000 , -4.4846 )
Ratio 0.75 : ( -1.7543 , 0.0000 , -4.7950 )
Ratio 0.80 : ( -1.8568 , -0.0000 , -5.1000 )
Ratio 0.85 : ( -1.9502 , 0.0000 , -5.3991 )
Ratio 0.90 : ( -2.0346 , -0.0000 , -5.6915 )
Ratio 0.95 : ( -2.1094 , -0.0000 , -5.9767 )
C++ has an awful habit of printing -0.0000 . I told it to evaluate the zero values to check its correctness. The components of the vectors, when multiplied to G*\mu*R, where G is the gravitational constant, \mu the torus density and R the largest radius, will yield the acceleration caused by gravity at that point. We can see that the torus world is very unstable at B, with some nasty pulls towards the center. I plotted A's x components and could get a great fit with a parabola:
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Warp wrote:
I had an idea: If you are calculating the gravity of a spherical object, the radius of the sphere doesn't really matter (as long as it's smaller or equal to the distance between the center of the sphere and the test point). In other words, if you eg. wanted to calculate the gravity on the surface of Earth, you would get the same result if you assumed that Earth was compressed into a point (retaining its mass) but the test point was still at the same distance from this center (in other words, the old radius).
It probably works the same with the torus, at least from the point of view of A (and perhaps even B): If we shrink the minor radius of the torus (but retaining the torus' mass) without moving A, the result would still probably be the same. We could thus shrink the minor radius to zero, in which case our problem has been reduced to a 2-dimensional one: Now we only have to calculate the gravity of a circle (at a distance of the old minor radius) rather than torus. This should remove at least one of the integrals (if not two, I'm not completely sure), making the problem significantly simpler.
Scepheo wrote:
Also, p4wn3r, I don't see how the ratio between the big and large radii matters. Even if there is little (or no) difference between the big and small radius, the reduction to a ring changes the gravity for no point.
It does change, if it didn't, all integrals wouldn't depend on the small radius, and they do.
When you calculate gravity from the Earth, you can reduce it to a point because of several symmetric simplifications. Gauss's law (it's mostly for electromagnetism, but it applies to gravitation with some changes, since the nature of the forces is, for this case, the same) states that if we trace an enclosed surface, the gravitational flux through that surface (that is, the integral of the scalar product of the field to the differential of area of the surface) depends solely on the amount of mass inside that surface.
Because of that, if you're at distance d, trace a sphere with radius d centered at the earth's center. The gravitational flux depends only on Earth's mass, but if we assume that it's a perfect sphere, we can imply that the modulus of gravitational field is the same in all directions (and even, parallel to the differential of area of our d-radius sphere). Thus, the whole integral can be expressed as (area of sphere)*(grav field). Calculating the field this way, it'll be the same as a punctual mass.
Now, look at the torus, it doesn't have said symmetry, even from A. Trace a sphere centered at the torus's center. We can certainly see that all points in the same latitude have symmetry in relation to the field's modulus, but to simplify our calculation we need to have symmetry in the entire surface. So, this reduction doesn't work.
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Well, for the small radius much smaller than the larger one, you could approximate it to a ring. In this case, it's clear that the gravity in A points to the surface of the torus.
Take, for example, a point inside the ring that's not the center and draw a line through it, measuring the length of the two segments d1 and d2. If you rotate this line by an angle theta, you'll get two lines inscribed in a circle with an angle theta between them. These lines enclose portions of the ring in each side, the mass of these portions is proportional to the length of the arc, which, for small enough theta, can be viewed as proportional to d1 cos(theta) for one side and d2 cos(theta) on the other. Since the gravitational force is proportional to the inverse of the square of the distance:
F1 is proportional to cos(theta)/d1 and F2 to cos(theta)/d2. Thus, if d1 <d2>F2 and the largest attraction is on the closest side, so the resultant gets to the surface. So, the idea of things getting trapped in the center of the torus makes no sense xD.
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Warp wrote:
p4wn3r wrote:
Assume without loss of generality that A lies on the x-axis and B's y-coordinate is zero.
I didn't quite understand. Wouldn't that mean that both A and B are on the x-axis? Perhaps you meant that A is on the x-axis and B on the y-axis?
It's indeed hard to see when the coordinate system is not drawn. The origin is the center of the torus and A is at (d-R,0,0), while B at (d,0,R). The z-axis passes through the hole in the torus, being perpendicular to the plane of its largest circle.
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I finished, B is a lot harder, because it has few symmetry, also there was a typo in the formula in my previous post. Assume without loss of generality that A lies on the x-axis and B's y-coordinate is zero.
I cannot guarantee that the integrals have finite form primitives, perhaps someone could fill some values and evaluate them numerically to see the direction the vectors point?
EDIT: Does anyone want to know the math behind this? I think it's very annoying...
EDIT 2: I keep confusing R with r!!!!
In the last two integrals, r^2 (sin fi - 1) read r^2 sin fi - R^2. And in the third, r^3 (sin fi - 1) = r^2 ( r sin fi - R)
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Probably, there's no way to escape the triple integral...
If I missed no calculations, for point A, the only function which has a computable primitive is when we use spheric coordinates. Let \mu be the density of the torus (assumed to be constant), d the distance of the center of the revolved circle to the center of the torus and R the radius of the torus, the value gravitational field at A in the radial direction (the only one which is not zero) is the beautiful integral:
No way I'm taking it further, it's very time-consuming, I'll try point B later...
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Claiming for SD encoding
EDIT: It's encoded, but my internet connection is too unstable and uploads keep breaking. I'm waiting for it to stabilize, meanwhile, anyone feel free to encode this movie.
EDIT 2: My connection is still bad, but since archive.org uploads are less likely to break, I struggled a little to put it there: http://www.archive.org/details/tasGbaMegaManZero3In5644.55ByHellagels
Also, Yes vote.
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Pointless Boy wrote:
Another valiant ad hominem attack. "You're just reading wikipedia, I'm not, therefore you're stupid and I'm right!" But to answer your question, no, I don't care what wikipedia says about the nature of science because it really has nothing to do with the discussion to begin with, even though you think it does. Insofar as you think science has anything to do with anything, you strangely seem to demand that readers of a children's book have a vast, encyclopedic knowledge (knowledge exceeding that which can be found in crappy wikipedia, even!) of whatever it is you think is important just to understand the book! How insane!
(Takes a deep breath)
You have a short memory, let me remind you what happened. Someone posts a short story, you say why you don't like it, someone contests you, you say again why the story is bad and that you don't like Harry Potter. In the middle of your text, you say "this principle is not scientific, it has no meaning at all". Then I respond to you "it does have meaning and is scientific". You reply that it's not, I say "you think it's not because you have a misconception of science", after that you say "science doesn't mean anything, I'm talking about Harry Potter" and continue with this.
First of all, when you brought that argument to support your idea, it depended mostly on your claim that it was not scientific. I only posted at that specific subject to tell you that it was, but you quickly turned it to "this principle doesn't mean Harry Potter is good, it doesn't matter if it's scientific or not", when I have repeatedly said that I haven't read the books and neither was interested in discussing whether it's good literature or not.
I just tried to correct one thing you said, I gave up because you insist on bringing down a statement I never said in the first place. What the hell makes you think I consider encyclopedic knowledge necessary to understand Harry Potter? I never expressed my opinion about it!!! I am in no position to do so because I haven't read the book!!!
All this obsession seems that you're only arguing on proud and don't care whether the things you write are plausible arguments that lead to constructive criticism, nor want to learn anything, and only wants to get everyone here tired to death until they stop arguing with you. If that's all that matters, you have succeeded with me, I've already wasted too much time reading your posts.
P.S.: Feel free to call it ad hominem or any fancy latin term you want, but no one in these forums has any reason to believe a random video game site user's monologue to be relevant, especially when he doesn't care about the validity of his claims and is well known for starting polemics for absolutely no reason.
And, seeing your reaction, I think you actually read it in Wikipedia :P
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@Pointless Boy:
Let me guess, you read Wikipedia's article on scientific theories, right? I could tell you that defining what is scientific or not is not so straight-forward as you may think, and is one of the greatest questions in epistemology, and that the (unfortunately bad) Wikipedia article is almost exclusively the empiric view of science, but as I correctly assumed, you're not interested in philosophy, a surprising fact given the topic's title.
I'm not making incoherent statements at all, it's just that you are obsessed with a subject, at the point of discussing with me parts of a book I already told you I didn't read, and when anyone tries to discuss how pointless your argumentation is, you avoid the subject by saying you were not addressing that matter and adding more stuff to make it as confusing and tiring as possible.
TASVideos ill needs a writer such as you, perhaps one day I'll see you start something more useful than an empty quote war. Until that day comes, I hope this thread gets locked for at least a week.
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Pointless Boy wrote:
p4wn3r wrote:
It's not useless blather, it's an ad hoc principle added to a scientific theory to make it consistent. After reading some parts of the articles, it seems that general relativity alone allows in some special cases that objects interact in such a way that would lead to contradictions inside the theory. Thus, they impose such a principle to abolish these paradoxes.
No, it is a made up concept that means nothing and makes no predictions. Relativity is not a "scientific theory" in the domains in which the Nabokov/Novikov/Nolikov consistency principle would apply, because in those domains it means nothing and makes no predictions.
Also, if one would take strictly what you wrote, no serious scientists would consider science at all to be real, since the problem of induction states that causality cannot be observed nor deduced within a finite set of experiments. Additionally, I find it funny how you consider such questions as artifacts of mathematics when they are more closely related to epistemology than math itself.
Except I did not consider the question you seem to think I considered. I said that fake fakery is not "real" in the sense that it says nothing about the universe. (It makes no predictions.) We all understand that relativity says something about the universe. (It makes predictions.) In that sense it is real enough, which was eminently clear from my previous statements. In the domain in which the Nabokov/Novikov/Nolikov consistency principle would apply, relativity is not "real" because it says nothing about the universe. (It makes no predictions.)
It seems you have a misconception of science then, as both of your replies make no sense. I'm not sure if I should have used the expression "no sense" in a sentence, because I run the risk of you bringing up other evidently clear points of your view of semantics.
I'd love to take the matter further, but it looks like you're more interested in writing long winded posts about why you don't like a book rather than understanding a physical principle.
Seriously, I really don't care whether you consider Harry Potter a good series or not, I haven't read any of the books and have only seen the first movie, it's just that through two pages your overstated point is that anything which is irrational according to reality (or, more generally, one possible interpretation of the reality proposed by the author) is automatically devoid of meaning.
Notice that the notion of what is true or a rational behavior within a situation depends on one's own view of reality, your whole text is just an attempt to rephrase your opinions, making them look like based on a solid, incontestable vision of science, fiction or whatever, that cannot possibly be purely objective, just to make them seem universally accepted. You'd be really lucky to convince someone arguing that way, this is just sophistry.
Also, you'd only change your mind if someone showed that something has meaning inside your own assumptions, and this is impossible since no one can fully understand your conceptions. This discussion can't possibly lead anywhere.
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Here is a WIP: http://dehacked.2y.net/microstorage.php/info/689668224/POKEBLUE.vbm
We had got a little farther than this, up to Seafoam, but will now have to redo one part. Although it's only around six minutes, the annoying part is that the L40 Tentacool DV: 7 encounter to T-fly Kabuto, the rarest in the entire run, was in these six minutes. We were really close to conclusion, almost no hard parts left, but because of some route trouble, this time-consuming encounter will have to be manipulated again and people are gonna have to wait a little more until the submission.
The actual problem in the route was very strange. It seems that, while facing Zapdos allows us to get Snorlax, if we attempt this with Articuno (a faster alternative), the Articuno encounter would destroy the bug, and the Snorlax encounter wouldn't happen.
I'm sorry for the late wip, I had forgot how old the movie in this thread was...
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Pointless Boy wrote:
Also, the Nolikov self-consistency principle is not "entirely fictional". It was formulated in regards to closed timelike curves, a potentially pathological feature in the very real theory of General Relativity. Perhaps you should actually read the article I linked to.
It is entirely fictional. It is meaningless blather based on meaningless blather that cannot be tested and makes no predictions. Moreover, relativity isn't "real" in the sense you seem to think it is. It is merely a mathematical description of our universe that is mostly correct in certain circumstances. No serious scientist considers it to be anything but that. And no serious scientists considers "closed timelike curves" to be "real" in even the limited sense that relativity is "real" in the domains in which it applies, since they are unobserved, unobservable, and nearly universally seen as artifacts of mathematics. (Again I point out that all of that is immaterial, since you can't blindly assume insanely erudite mathematical trickery about an already more or less inaccessible mathematical description of spacetime in an argument about what is and isn't consistent in a children's book featuring time travel. Any reasonable person would ask "why isn't time travel being used for something important," not scour wikipedia for obscure theoretical errata aiming to desperately explain an obvious error on Rowling's part.)
It's not useless blather, it's an ad hoc principle added to a scientific theory to make it consistent. After reading some parts of the articles, it seems that general relativity alone allows in some special cases that objects interact in such a way that would lead to contradictions inside the theory. Thus, they impose such a principle to abolish these paradoxes.
Also, if one would take strictly what you wrote, no serious scientists would consider science at all to be real, since the problem of induction states that causality cannot be observed nor deduced within a finite set of experiments. Additionally, I find it funny how you consider such questions as artifacts of mathematics when they are more closely related to epistemology than math itself.
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I wrote:
Thing is, there are infinitely many quadrilaterals with the given sides, the cyclic one, that simplifies algebra, doesn't satisfy the formula above :(.
Damn, I should really pay more attention to things I do. The cyclic quadrilateral is inscribable in a square. I'll solve this problem for the cyclic given quadrilateral (that is, besides being in a square, it'll also be in a circumference). I'm almost certain that there are infinite possible quadrilaterals for this problem.
Let's go:
juef wrote:
A quadrilateral with sides of length 223, 297, 228 and 296 is inscribed inside a square. None of its angles is right. Find the perimeter of the square.
Lemma 1: If a quadrilateral is inside a square, then 1 / (sqrt(2)* cos(a - 45º)) <= d1 / d2 <sqrt> (d1 - d2 sen a) cos t = d2 cos a sen t => tg t = (d1 - d2 sen a)/d2 cos a
This, of course if a != 90º, in this case, every t would be a solution, as long as d1 = d2.
Now, as I said in the first paragraph, d1 cannot have a larger projection in y. We assure this by limiting t to the interval [ -45º,45º]. Since the same applies for d2 and x, the angle between them has also to be on that interval: -45º <= 90º - a - t <=45º, or 45º - a <= t <= 135º - a.
If we take a an acute angle (0º<=a<90>= -45º and 45º <= 135º-a.
Thus, we need only to constrain a to [45º-a,45º].
Using tg (45º-a) <= tg t <= tg 45º and simplifying, we get the expression in the lemma.
Lemma 2: If the diagonals AC and BD in a quadrilateral satisfy (*) for a!=90º, let P be their intersection point, t = arctg((d1 - d2 sen a)/d2 cos a), u is the largest length among {AP,CP} and v the largest among {BP,DP}. If u*cos t + v cos (a + t) <= d1 cos t and u*sen t + v sen (a + t) <= d1 cos t, then the quadrilateral can be inscribed in a square.
Notice that the 1st lemma could only be used to prove that a quadrilateral doesn't fit into a square. The case where (*) is fulfilled and the diagonals don't fit into the square is when the intersection point is awfully placed. Draw the segments AC and BD. Now, project AP and PB into the rotated x-axis. Note that it's possible that the segments obey (*), but if the projection of PB is in the same direction, the sum of their lengths can be larger than the side of the square, making it impossible to fit them.
Though an algorithm to determine a necessary and sufficient condition is more complex, if we take the largest segments in each line, project them into each axis and verify that their sum is smaller than the square's side in both of them, we can already say the quadrilateral can be inscribed. With these two lemmas in mind, let's go to the actual problem, I'll use Randil's amazing picture:
As I said before, let's assume this quadrilateral is cyclic, prove that it can be inscribed in a square and find the square's side. Starting from the bottommost vertex and going in counterclockwise order, name them A,B,C and D. Let P be the intersection point of diagonals, a the acute angle between them and PB = x.
Since the quadrilateral is cyclic, we can prove using angle inscribed in a circumference that angle DBC = angle DAC. Repeating this process for other angles, we get the similarity relations between triangles:
PAB is similar to PDC and PBC is similar to PAD.
In this way, if PB = x, we find out that PC = 223x/228 , PD = (223*296x)/(228*297), PA = 296x/297. Since x>0, the diagonal BD is larger, by a tiny amount, but it's larger.
For the angle a, let's use Brahmagupta's formula for the area A = AC*BD*sin a/2 = sqrt((s-AB)(s-BC)(s-CD)(s-DA)). By Ptolemy's theorem, AC*BD = AB*CD + DA*BC, so:
sin a = 2sqrt((s-AB)(s-BC)(s-CD)(s-DA))/(AB*CD + DA*BC) ~= 0,964932
Useful later, cos a ~= 0,26249.
Now, let's use Lemma 2 to prove it can be inscribed in a square, first, (*) needs to be true. Take d1 as the largest diagonal, it's equivalent to prove that:
(d1/d2) <sin> (1,97477/1,97470) <= 0,964932 + 0,26249 , which is true.
Now, tan t = ((d1/d2) - sin a)/cos a ~= 0,136. We get sin t ~= 0,13 and cos t ~=0,99
We now take u = x and v = 296x/297, rearranging some terms of the inequality, we need to prove that:
cos t + (296/297)*cos(a+t) <1> 0,134 <= 0,965, which is true.
sin t + (296/297)*sin(a+t) <1> 0,13 + 0,985 <1> BD ~= 372,50
The perimeter of the square is, finally, P = 4 * 372,50 * 0,99 = 1475,1.
Notice that the assumption that the quadrilateral is cyclic was essential here because:
(1) It allows using Brahmagupta's formula.
(2) It allows using Ptolemy's theorem.
(3) It gives useful similarity relations to the lengths of the segments needed to prove the lemmas.
(4) It allows us to find BD.
Without this, it's A LOT of algebrism, which is not fun to do. Also, seeing as the inequalities were easily satisfied, there are probably many more possibilities and this solution is not unique.
EDIT: I seem to have forgotten that no angle is right. Just take pairs of adjacent sides and see that Pythagoras's theorem doesn't hold for the triangle formed by them and their corresponding diagonal.
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Unless I'm missing something obvious, this seems to be a hard problem.
I could prove that for a quadrilateral to be inscribed into a square, it's necessary that:
1 / (sqrt(2)* cos(a - 45º)) <= d1 / d2 <= sqrt(2)* cos(a - 45º)
Where d1 and d2 are the diagonals of the quadrilateral and a the angle between them. This condition isn't sufficient, however, the intersection point of the diagonals needs to be considered.
Thing is, there are infinitely many quadrilaterals with the given sides, the cyclic one, that simplifies algebra, doesn't satisfy the formula above :(. The problem will only have a unique solution if only one of them satisfies the relation (if we're unlucky, a lot of them do, but again, only one may be inscribable because the condition is not sufficient). I'm a bit rusty with quadrilaterals and don't remember how to efficiently extract its diagonals given the sides.
It might be possible to get something by equaling the area to d1*d2*(sin a)/2 and to Bretschneider's formula. I'll take a deeper look into this later if I have time.
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Interesting, I just thought: I have to pick z stages out of x. To see if I have an item or not, the order doesn't matter, so I can think of only combinations. I have no item if I picked a combination of x stages that's also a combination of x-y no-item stages.
There are xCz of the former and (x-y)Cz of the latter, the probability follows right after.
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andymac wrote:
A general solution to this problem is as follows: let's say you have x stages and y items. The probability of having an item by stage z is 1 - ((x-y)Pz)/(xCz).
By xCz you mean x! / ((x-z)! z!) and by (x-y)Pz , (x-y)! / z! ?
Shouldn't the answer be 1 - ((x-y)Cz)/(xCz) ?
Let's manipulate the sum a little to find:
Using this recurrence, we can prove our desired equality by mathematical induction.
Now, using Liebniz's expansion:
^It's actually \sum{j=1}{i-1} instead of \sum{j=1}{n-1}
So, your expression is simply the sum of the reciprocals of the first n squares. Euler was the first to prove that this series converges to pi^2/6 .
EDIT: Too late!
I cannot guarantee that the integrals have finite form primitives, perhaps someone could fill some values and evaluate them numerically to see the direction the vectors point?
EDIT: Does anyone want to know the math behind this? I think it's very annoying...
EDIT 2: I keep confusing R with r!!!!
In the last two integrals, r^2 (sin fi - 1) read r^2 sin fi - R^2. And in the third, r^3 (sin fi - 1) = r^2 ( r sin fi - R)
No way I'm taking it further, it's very time-consuming, I'll try point B later...
As I said before, let's assume this quadrilateral is cyclic, prove that it can be inscribed in a square and find the square's side. Starting from the bottommost vertex and going in counterclockwise order, name them A,B,C and D. Let P be the intersection point of diagonals, a the acute angle between them and PB = x.
Since the quadrilateral is cyclic, we can prove using angle inscribed in a circumference that angle DBC = angle DAC. Repeating this process for other angles, we get the similarity relations between triangles:
PAB is similar to PDC and PBC is similar to PAD.
In this way, if PB = x, we find out that PC = 223x/228 , PD = (223*296x)/(228*297), PA = 296x/297. Since x>0, the diagonal BD is larger, by a tiny amount, but it's larger.
For the angle a, let's use Brahmagupta's formula for the area A = AC*BD*sin a/2 = sqrt((s-AB)(s-BC)(s-CD)(s-DA)). By Ptolemy's theorem, AC*BD = AB*CD + DA*BC, so:
sin a = 2sqrt((s-AB)(s-BC)(s-CD)(s-DA))/(AB*CD + DA*BC) ~= 0,964932
Useful later, cos a ~= 0,26249.
Now, let's use Lemma 2 to prove it can be inscribed in a square, first, (*) needs to be true. Take d1 as the largest diagonal, it's equivalent to prove that:
(d1/d2) <sin> (1,97477/1,97470) <= 0,964932 + 0,26249 , which is true.
Now, tan t = ((d1/d2) - sin a)/cos a ~= 0,136. We get sin t ~= 0,13 and cos t ~=0,99
We now take u = x and v = 296x/297, rearranging some terms of the inequality, we need to prove that:
cos t + (296/297)*cos(a+t) <1> 0,134 <= 0,965, which is true.
sin t + (296/297)*sin(a+t) <1> 0,13 + 0,985 <1> BD ~= 372,50
The perimeter of the square is, finally, P = 4 * 372,50 * 0,99 = 1475,1.
Notice that the assumption that the quadrilateral is cyclic was essential here because:
(1) It allows using Brahmagupta's formula.
(2) It allows using Ptolemy's theorem.
(3) It gives useful similarity relations to the lengths of the segments needed to prove the lemmas.
(4) It allows us to find BD.
Without this, it's A LOT of algebrism, which is not fun to do. Also, seeing as the inequalities were easily satisfied, there are probably many more possibilities and this solution is not unique.
EDIT: I seem to have forgotten that no angle is right. Just take pairs of adjacent sides and see that Pythagoras's theorem doesn't hold for the triangle formed by them and their corresponding diagonal.