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September 8

Sixty Degree Triangles

There is a lot of literature about the Right Triangle and Primitive Pythagorean Triples. I’ve seen material on how to produce these integral triples and some of the patterns involved in listing with them.

Has any work been done in a similar vein for triangles that have a sixty degree angle?

I got to thinking about this topic by considering that there are only a few angles with rational degree that also have rational trigonometric functions; and the fact that the law of cosines is very similar the Pythagorean relationship.

So, are there any triangles with a sixty degree angle and three sides of integer length? --MathMan64 00:02, 8 September 2006 (UTC)

The Pythagorean Theorem is a special case of the law of cosines.

I'm pretty sure that it is impossible for a triangle with only one angle of 60° to have all three sides rational, let alone integral. Take the ratio of this 30-60-90 triangle:

       30°
       /|
      / |
     /  |
 $2x$  /  | $x{\sqrt {3}}$
   /    |
  /_____|
60°  $x$  90°

At least 1 of these sides is going to be irrational, beacuse of the ${\sqrt {3}}$.

But of course, if a triangle can have more than one angle of 60°, then an equilateral triangle with 3 60° angles can have 3 integral sides.

--Ķĩřβȳ♥ŤįɱéØ 00:34, 8 September 2006 (UTC)

Is there a positive integer solution to

$4a^{2}=3b^{2}+c^{2}$

with a and b unequal? Melchoir 03:28, 8 September 2006 (UTC)

The short answer is no.

The long answer is :

$x^{2}+y^{2}=4x^{2}$

$y=x{\sqrt {3}}$

With the conditions that both x and y are integers.

If x is an integer then y must be an irrational number.

So the answer is no.

Ohanian 04:29, 8 September 2006 (UTC)

I couldn't get an equation that looked like that. Where does it come from? Melchoir 04:32, 8 September 2006 (UTC)

         30°
         /|
        / |
       /  |
 $2x$  /   | $y=x{\sqrt {3}}$
     /    |
    /_____|
  60°  $x$  90°

Ohanian 04:54, 8 September 2006 (UTC)

Ohanian, you misunderstood the question. One angle has to be 60° but the others needn't be 30° and 90° (we have already seen that the 30°-60°-90° triangle can't do it). As for the original question, the equation I got was $c^{2}=a^{2}+b^{2}-ab$, with plentiful solutions, such as 3, 8, 7 (the 60° is between 3 and 8). I could have made some mistake though. -- Meni Rosenfeld (talk) 05:13, 8 September 2006 (UTC)

My mistake. If one of the angle is 60 degrees and the other two can be anything then I think you can use the Law of Cos .

$c^{2}=a^{2}+b^{2}-2ab\cos(60),\,$

$c^{2}=a^{2}+b^{2}-ab,\,$

with the condition that a , b and c are positive integers.

One simple solution is c = a = b = 5

I think there could be infinite number of solutions

Ohanian 08:02, 8 September 2006 (UTC)

*Sigh*. Do you even read the other replies? Yes, we know we can use the Cosine law. Yes, we know the equation is $c^{2}=a^{2}+b^{2}-ab$. Yes, we know that equilateral triangles solve the problem, and were specifically looking for other solutions. And though I haven't mentioned it, the rule $a=2t+1,b=3t^{2}+4t+1,c=3t^{2}+3t+1$ generates infinitely many non-trivial solutions (and there are others). -- Meni Rosenfeld (talk) 09:26, 8 September 2006 (UTC)

Thanks, Meni. Now can you give me some idea of how you got the three parametric equations for a, b and c? --MathMan64 17:55, 8 September 2006 (UTC)

I started by generating (using a computer) triplets of integers satisfying the equation; I picked a particular sequence of such triplets:

3, 8, 7

5, 21, 19

7, 40, 37

9, 65, 61

2t + 1 clearly gives the first side of a triplet in this sequence; It isn't difficult to find formulae for the other sides. The same ideas can be used when exploring pythagorean triplets, only the equation is a little different. You probably know that the general pythagorean triplet is given by $a=m^{2}-n^{2},b=2mn,c=m^{2}+n^{2}$; I suspect a similar formula can be used to generate all the solutions to our current problem, but finding it will require additional work. -- Meni Rosenfeld (talk) 21:14, 8 September 2006 (UTC)

The relationship to Pythagorean triples is a good guess. We want zeros of the polynomial a²+b²−ab−c². Notice that each term has total degree 2, so this is a homogeneous polynomial, a conic in the projective plane whose points are (a:b:c), and we can approach this through algebraic geometry. If we set c to 1 and consider real solutions, we have an ellipse (a²+b²−ab = 1) tilted 45°, with major axis √2 and minor axis √2⁄3. The point (0,−1) is on this ellipse, and so a line through that point is expected to intersect in one other point. Parameterize the line by its slope, so that its equation is (b+1)/a = s. This will intersect the ellipse at (2s−1,s²−1)/(1−s+s²). Going back to the homogeneous form, we obtain the parametric equations a = 2s−1, b = s²−1, c = 1−s+s². For a and b to be positive, we need the portion of the ellipse in the first quadrant, meaning 1 < s < ∞. Restricting s to integers does not give all integer solutions, and slopes approaching infinity are a minor nuisance. Instead, change the pivot point to (−1,0), obtaining parametric equations

$a=1-t^{2}\,\!$

$b=2t-t^{2}\,\!$

$c=1-t+t^{2}\,\!$

Now the first quadrant restriction is 0 < t < 1. To obtain all integer solutions (up to a common integer factor), we need only take all rational values of t in that range. That is, let t equal n/m, and multiply through by m² to clear the denominators. We obtain

$a=m^{2}-n^{2}\,\!$

$b=2mn-n^{2}\,\!$

$c=m^{2}-mn+n^{2}\,\!$

Now for any positive integer n, and any integer m greater than n we obtain an integer solution. Furthermore, if we also insist that n and m have no common factor except 1 (they are coprime), then we should obtain each possible integer shape exactly once. If we wish to exclude mirror image duplicates (swapping a and b), then instead of requiring m > n we should require m ≥ 2n (derived from a ≥ b).

In fact, we can generate nearly equilateral triangles by letting m = 2n+1, thus simplifying the equations to a = 1+4n+3n², b = 2n+3n², c = 1+3n+3n². Small examples are (21,16,19), (40,33,37), (65,56,61), and (96,85,91).

Finally, it's interesting to compare our equations in n and m with those given above for Pythagorean triples; they are almost identical. --KSmrq^T 02:04, 9 September 2006 (UTC)

Thanks to both of you. That's exactly what I was looking for. Now I'll try to do the same thing on 120 degree triangles. I also bet that using KSmrq’s idea for nearly equilateral triangles, and by changing the m = 2n+1 equation by little bits I can generate different sequences for which I can find triples of t equations like the ones that Meni gave.

--MathMan64, 9 September 2006 (UTC)

converting

how to convert 500 kilometers to inches thanks, Bradly

Try googling 500 kilometers in inches. :) —AySz88\^-^ 04:33, 8 September 2006 (UTC)

1 inch = 25.4 mm = 0.0254 m = 0.0000254 km

thus

1 km = 1 / 0.0000254 inches

From here onwards it's pretty easy. So I leave it to you as an exercise.

Ohanian 05:01, 8 September 2006 (UTC)

A related question: When driving on US roads, I've encountered signs giving distances in feet and other signs using miles. Do people get used to this, do they just know that 1/4 mile is less than 2000 feet?--gwaihir 07:43, 8 September 2006 (UTC)

I haven't gotten used to it, and I live here! Generally the signs that use feet have the form "lava pit in 200 feet", so the exact distance isn't as important as the general knowledge that there's a hazard ahead, coming up soonish. Melchoir 16:52, 8 September 2006 (UTC)

In general, feet are used when you have less than 30 seconds to react to the sign, while miles are used when you've got more than 30 seconds. --Serie 19:45, 8 September 2006 (UTC)

I just got a GPS unit, and it gives distances > 800 yards in miles, smaller distances in yards. Since using it, (and having it announce "after 400 yards, turn left"), I have gotten a lot more familiar with how far a given distance of yards is. My dad, who was somewhat of an athlete, always used to speak in terms of football fields (100 yards) but that was never a meaningful value for me. Now I have a much more intuitive feel for such distances. --WhiteDragon 21:23, 8 September 2006 (UTC)

Finding Spatial Derivatives

Okay, a few of you may remember me, a while back I asked a question about 3D gaussian blur for a 3D fluid solver I'm writing. I must again thank you all, the gaussian diffusion algorithm has been successfully implemented... unfortunately I have yet to program a UI in which to test it, so we shall see.

At any rate, I am currently at the point in my program that will advect a level set function through the domain to track the surface of a liquid. The standard semi-Lagrangian methods I use to advect the gas cause unacceptable mass dissipation when dealing with liquids. (this problem has been well-documented by Ron Fedkiw). The most common solution I've seen is the Particle Level Set Method, which places massless marker particles at the interface to correct errors that appear in the level set grid. However, I don't feel like doing that (for now).

I have found a paper called Stable but Nondissipative Water by Oh-Young Song, Hyuncheol Shin, and Hyeong-Seok Ko of the Seoul National University which explains a method of preserving mass without placing particles at the interface. The method is called CIP Advection--it is essentially the same as normal semi-Lagrangian advection, except for that is uses a special interpolation method. The nice thing about CIP interpolation is that it uses only two grid points, whereas cubic spline interpolation uses four grid poitns. Instead of the other points, CIP uses the spatial derivatives of the two points in question. This also makes enforcing motonicity very simple, becusae all we have to do is modify the derivatives when an overshoot or undershoot can be predicted.

However, I'm not really getting how they get their derivatives. I have written a simple vorticity confinement algorithm, which uses the following method to find the spatial derivatives:

$\nabla c_{ijk}=[{\frac {c_{i+1jk}-c_{i-1jk}}{2}},{\frac {c_{ij+1k}-c_{ij-1k}}{2}},{\frac {c_{ijk-1}-c_{ijk+1}}{2}}]$

Where c is any cell,
i,j, and k are the spatial dimensions,
and the cell spacing is assumed to be 1.

My understanding is that this method is called central differencing. This is not the method Song et al use. They provide the equation:
${\frac {\partial \phi _{\xi }}{\partial t}}+u\cdot \nabla \phi _{\xi }=-\mathbf {u} _{\xi }\cdot \nabla \phi$

Where φ is the level set,
u is the velocity field at that point,
ξ is any one of the spatial dimensions x,y, or z.

At first it appeared to me that they were using an equation that requires spatial derivatives to calculate spatial derivatives... which didn't make much sense. Then it occurred to me that they might have another vector field containing the spatial derivatives... but I'm not sure how much sense that makes either, becuase of the aforementioned numerical dissipation which is the problem to begin with. Any light on this would be much appreciated! (remember that I only need something accurate enough to look convincing, as I will be making movie SFX with this, not calculating stresses on airfoils! ;) )

-Loki7488 20:58, 8 September 2006 (UTC)

Eh, guess this one's a little more tricky than I thought? Well, can anyone tell me if there is a higher order method for getting the derivatives than central differencing? Ideally, I'd like to need only the point in question and the two points on either side of it, as using additional stencils greatly increases complexity once I start extending the interpolation to 2 or more dimensions. Of course, if there is no way to do this without the extra data points, I think that I may be able to get away with using lower order methods on subsequent interpolations. Can anyone shed any light on this? -Loki7488 02:31, 13 September 2006 (UTC)

Unusual Theories of Integration on Rⁿ

Hi there,

This question is directed toward anyone who has knowledge of several theories of integration, including the theories of Lebesgue and Daniell.

The following two books each include a construction of a theory of integration in Euclidean space:

• Calculus on Manifolds by Michael Spivak (Chapter 3)
• Advanced Calculus by Lynn H. Loomis and Shlomo Sternberg (Chapter 8)

Both theories are in some sense theories of Riemann integration, but both also have something more to them. My question is, roughly, where do the integrals developed in these chapters sit in the overall scheme of integration theory?

For example, Spivak develops a straightforward theory of Riemann integration, and then observes that the integral of f over C is not necessarily defined even when C is an open set and f is continuous. He therefore goes on to extend the integral using partitions of unity in such a way as to end up with an integral defined for any bounded function over any bounded set. I have not seen this approach used anywhere else, and I am trying to figure out how this 'extended integral' compares to the more common sorts in the theory of measure and integration.

In the case of Loomis and Sternberg, the integral developed appears to be considerably more sophisticated. Although the authors claim it's 'just Riemann integration', there is clearly much more going on in this development; it seems to have characteristics both of Lebesgue and Daniell integration. One remarkable consequence of the development is that Fubini's theorem hardly needs to be proved at all; it's an immediate consequence of the uniqueness property of the integral. The authors go on to extend the integral still further in Section 13 (Absolutely Integrable Functions), and at this point I have no idea at all how this integration theory compares in power and scope to the likes of the Lebesgue approach.

If anyone can shed some light on where either or both of these developments 'fit' into the overall scheme of modern integration theory, it would be much appreciated.

Thanks!

 — merge 21:56, 8 September 2006 (UTC)

Please click my name repeatedly. Hope that this is acceptable to Wiki. If I mention my book here, it would be worse in terms of self-advertising which is against the policy. Twma 02:48, 9 September 2006 (UTC)

Sorry, but I don't see how advertising your book constitutes an answer to a question about the treatments of integration in books by other authors.  — merge 09:49, 10 September 2006 (UTC)

I do not know what is the developments 'fit' into the overall scheme of modern integration theory. I merely pointed out a direct method toward vector measures and mean-values on groups that might be RELATED to the modern integration theory (without manifolds). Twma 00:20, 12 September 2006 (UTC)