Wikipedia:Reference desk/Archives/Mathematics/2012 February 10

= February 10 =

Minimum area of a quadrilateral
The maximum area has been resolve above at Reference desk/Mathematics which made me curious as to the resolution of the minimum area of a quadrilateral.

Assume a convex quadrilateral with its longest side = a with successive and arbitrary sides = b, c, d and with a < (b + c + d). What would be a method to find the minimum area quadrilateral. hydnjo (talk) 01:41, 10 February 2012 (UTC)
 * I don't think there is a minimum. Imagine a really flat trapezoid. Widener (talk) 02:05, 10 February 2012 (UTC)
 * If you count degenerate quadrilaterals (and you imply that you can with the condition a ≤ (b + c + d) rather than a < (b + c + d)), then the minimum area is obviously zero. Widener (talk) 02:10, 10 February 2012 (UTC)
 * Hmm. Of course, notice that a > (b + c + d) is never possible anyway, so the condition a ≤ (b + c + d) is redundant actually. Widener (talk) 02:12, 10 February 2012 (UTC)


 * Yeah, I just added a < (b+c+d) so as to limit discussion about the degenerates (a ≥ (a + b + c) quads and it almost worked ;-) hydnjo (talk) 02:42, 10 February 2012 (UTC)
 * Oh, you changed the condition to from a ≤ (b + c + d) to a < (b + c + d). Be aware that it is still possible to create a degenerate quadrilateral which satisfies a < (b + c + d) Widener (talk) 09:15, 10 February 2012 (UTC)


 * There's the problem of do you allow self intersecting quadrilaterals, and if so do you count one of the areas as positive and the other negative which would be quite usual in maths so the minimum absolute are would always be zero. Otherwise it looks like the minimum would be give by one of the triangles like a-b,c,d where two adjacent sides were made to coincide. Dmcq (talk) 09:41, 10 February 2012 (UTC)
 * If you do use the interpretation which allows negative areas, then a self intersecting quadrilateral will not be convex. Widener (talk) 10:53, 10 February 2012 (UTC)
 * Am I misreading your question? Are the lengths a,b,c,d allowed to vary (subject to the conditions you mentioned)? Widener (talk) 10:28, 10 February 2012 (UTC)
 * I had thought at first that a was fixed but that b,c,d were allowed to vary (since you used the word "arbitrary" to describe them). Widener (talk) 11:56, 10 February 2012 (UTC)
 * I think we need more information. Precisely what is allowed to vary? Widener (talk) 11:00, 10 February 2012 (UTC)

If you keep all lengths fixed then you get a Four-bar linkage, a little off topic but a fascinating study.--Salix (talk): 11:20, 10 February 2012 (UTC)
 * That's assuming that the arrangement of the four sides is also fixed i.e. the only thing which varies is that one angle, in which case you can find the area of the quadrilateral as a function of the angle and then use optimization techniques such as or including basic calculus to find the angle which generates the minimum area, and then use that angle to find the area of the smallest quadrilateral. Widener (talk) 11:47, 10 February 2012 (UTC)

To review, the sides of the convex quadrilateral are successively a, b, c, and d of arbitrary length where the longest side is side a. Also, a < (b + c + d). In a previous question I learned that the maximum area is when the angles are such as to form a cyclic quadrilateral and then the area and the interior angles can be calculated and expressed in terms of the sides. After that discussion (which satisfied my need to know) I became curious as to the minimum area quadrilateral with the assumption that all of the interior angles are variable so long as the quadrilateral remains convex (no crossovers). Sorry for any confusion about the givens. hydnjo (talk) 22:40, 10 February 2012 (UTC)


 * Your use of the phrase "arbitrary length" may be causing some confusion. Are you asking, "Given positive values a, b, c, & d such that a is the largest value and a < (b + c + d), what is the minimum area of a convex quadrilateral whose sides are, in order, a, b, c, & d?"? -- 182.232.144.48 (talk) 03:28, 11 February 2012 (UTC)


 * Yes, a much more concise statement than my version - thanks. hydnjo (talk) 13:33, 11 February 2012 (UTC)


 * Then Salix's mention of Four-bar linkage is entirely on topic, as is Widener's optimization suggestion. While you have only one degree of freedom here, you will likely have to consider several cases, as illustrated in the "Types of four-bar linkages" diagram. -- ToE 05:38, 12 February 2012 (UTC)
 * This KMODDL page does not address minimum area, but it works through the different types of four-bar linkages. -- ToE 05:57, 12 February 2012 (UTC)
 * If we are strict about it be it being convex then Dmcq's solution does not work. However, I've a hunch that the solution may be a triangle with lengths a+b, c, d where two adjacent edges form 180° with each other. Looking at the linkages two edges being co-linear seem to be the critical point where the configuration changes from convex to non-convex. Grashof condition may also come into play as it seems to be import in the possible configurations. Note that boundary value for Grashof's S+L = P+Q, can be solved by a+b=c+d which can flattern giving zero area.--Salix (talk): 23:39, 12 February 2012 (UTC)

modulo arithmetic
Suppose n is an integer in the range 0-1000 inclusive. Let p, q and r be qual to n reduced modulo 7, 11 and 13 respectively (e.g. if n=20, then p=6, q=9, r=7). I believe that as 7, 11 and 13 are all prime, and 7*11*13=1001, then the combination of (p,q,r) will be unique. Is there a proof of this, other than by trial and error? Also, is there a formula for establishing n given p, q and r?  An  optimist on the  run!  15:07, 10 February 2012 (UTC)


 * Chinese remainder theorem. Gandalf61 (talk) 15:12, 10 February 2012 (UTC)


 * Sorry, but that article's way over my head. Is there a simpler answer to this?  An  optimist on the  run!  18:03, 10 February 2012 (UTC)


 * Have you read the section Chinese_remainder_theorem? If you can work through that example, then you can solve your own question, and also determine whether it's unique. If you cannot work through that example, we can go from there. SemanticMantis (talk) 21:24, 10 February 2012 (UTC)


 * ... and specifically for the 7, 11, 13 case, you can find n from p, q and r by calculating
 * $$715p + 364q - 77r \ \text{mod} \ 1001$$
 * So, for example,
 * $$715 \times 6 + 364 \times 9 - 77 \times 7 = 7027 = 20 \ \text{mod} \ 1001$$
 * Gandalf61 (talk) 09:15, 11 February 2012 (UTC)


 * I've tried working through the example, but floundered at the post where it said "(1/3) (mod 4) = 3 (mod 4)". Can I ask Gandalf how he derived the numbers 715, 364 and -77 in the above example? Thanks.  An  optimist on the  run!  13:32, 11 February 2012 (UTC)


 * For the first coefficient, 715, you want a number that is a multiple of 11 and 13, and leaves a remainder of 1 when divided by 7. To be a multiple of 11 and 13 it has to be a multiple of 11 x 13 = 143. So you can check multiples of 143 in turn:
 * $$143 = 3 \ \text{mod} \ 7$$
 * $$2 \times 143 = 286 = 6 \ \text{mod} \ 7$$
 * $$3 \times 143 = 429 = 2 \ \text{mod} \ 7$$
 * $$4 \times 143 = 572 = 5 \ \text{mod} \ 7$$
 * $$5 \times 143 = 715 = 1 \ \text{mod} \ 7$$
 * and that's how you get 715 (you can short-cut the last part a little if you know how to find multiplicative inverses in modular arithmetic). In the same way, the second coefficient has to be a multiple of 7 and 13 and leave a remainder of 1 when divided by 11, so you check the multiples of 91 until you find
 * $$4 \times 91 = 364 = 1 \ \text{mod} \ 11$$
 * and -77 is a multiple of 7 and 11 and leaves a remainder of 1 when divided by 13. Gandalf61 (talk) 13:56, 11 February 2012 (UTC)
 * Checking multiples mod 7 of 11&times;13 in turn, even easier:
 * 11&times;13 = 4&times;6 = 24 = 3 mod 7
 * 2&times;3 = 6 mod 7
 * 3&times;3 = 9 = 2 mod 7
 * 4&times;3 = 12 = 5 mod 7
 * 5&times;3 = 15 = 1 mod 7
 * Bo Jacoby (talk) 16:16, 11 February 2012 (UTC).
 * Thanks for the explanation - it seems to make sense!  An  optimist on the  run!  23:45, 11 February 2012 (UTC)

Greatest Common Divisor of Two Polynomials
What is the difference between finding the greatest common divisor of two polynomials and finding their GCD to some (prime) modulus p? Could I be provided with any references for where to read around the issue? Thanks. 131.111.216.115 (talk) 17:01, 10 February 2012 (UTC)


 * Perhaps this is a question that lies in territory unfamiliar to the members of the Reference Desk. Could someone at least voice their thoughts on the issue, even if they don't have an answer as such? I would appreciate any help with which you could provide me. Thanks. 131.111.216.115 (talk) 12:01, 12 February 2012 (UTC)
 * The greatest common divisor between two polynomials is a polynomial. The greatest commom divisor between two polynomials modulo a prime is a polynomial modulo a prime. Different beasts. Do I understand you question correctly? Bo Jacoby (talk) 16:19, 12 February 2012 (UTC).
 * Yes, that's what I thought. I just worried the question might have some unexpected subtlety to it. So if we wish to compute the gcd of two polynomials f and g modulo p, we reduce them to polynomials f' and g' where the coefficients of f' and g' are all integers greater than or equal to zero and less than or equal to p-1, compute the gcd of f' and g' as you would normally and then stick mod p on the end of that result and we have the answer? 131.111.216.115 (talk) 12:40, 13 February 2012 (UTC)
 * Anyone? 131.111.216.115 (talk) 14:23, 15 February 2012 (UTC)

Solving the 3-body problem if two of the objects are stationary
The simulation here is of a gravitational system with two fixed sun and one to four movable planets. The values that can be changed are the mass of the first sun, the x position that the first planet starts at, and the number of planets. The starting x value of additional planets is an offset of the starting x value of the first planet. Additional planets do not interact with each other and instead are provided for comparing the differences in paths. I want to know if a 3-body problem with two fixed objects and one movable object can be solved exactly. --Melab±1 &#9742; 20:49, 10 February 2012 (UTC)


 * You can't have two fixed objects. What is going to hold them still? They'll fall towards each other. What you can do, is have two objects in circular orbits around their centre of gravity (so they'll look fixed in a rotating reference frame) and then have a third object small enough for the effect of its gravity on the first two objects to be negligible. Then you can solve it reasonably easily. --Tango (talk) 22:37, 10 February 2012 (UTC)


 * There's no reason mathematically you can't have two fixed objects, the force on a satellite would just be the sum of the forces from the two objects individually. It may be that there could never be two real fixed objects in space, but that's something for physicists to determine. It appears from the simulation that the solution is chaotic, generally an indication that there is no closed form solution. That could be due to inaccuracies in the simulation though.--RDBury (talk) 23:18, 10 February 2012 (UTC)


 * It doesn't make much sense to solve a problem about gravity without taking into account the gravitational attraction between your two largest objects. --Tango (talk) 14:33, 11 February 2012 (UTC)


 * It is chaotic, but can the objects x and y positions be expressed as $$x(t)$$ and $$y(t)$$? --Melab±1 &#9742; 21:40, 16 February 2012 (UTC)


 * I think it is solvable in bipolar cylindrical coordinates. The problem is closely related to that of the electron in a diatomic molecule, and unless I am mistaken this is reasonably well understood. Note that by fixing two of the bodies, it is actually a two-body problem, i.e. there are only two frames relevant to the problem, albeit with an awkward force relating their motion. — Preceding unsigned comment added by 129.67.37.224 (talk) 23:58, 10 February 2012 (UTC)


 * If you have the two massive objects (e.g. stars) fixed, is that equivalent to having them in a circular orbit around each other and having the coordinate system rotate? Would that give the correct results?  Bubba73 You talkin' to me? 06:29, 11 February 2012 (UTC)


 * I don't know what you mean by "the correct results". You can get results that way. The fact that you have a rotating reference frame would be very significant for the results, though. --Tango (talk) 14:33, 11 February 2012 (UTC)


 * What I'm saying is, suppose you have two stars of equal mass and they are at a distance and velocity to be in a circular orbit around each other. If you have a coordinate system rotating at that rate, the two stars would be fixed in that coordinate system.  Now suppose you add an object to the system that has very low mass compared to the stars.  It is easy to calculate the trajectory of the small body in that rotating coordinate system, with the two fixed stars.  My question is: does that correspond to what would happen in the real world?  Bubba73 You talkin' to me? 15:38, 11 February 2012 (UTC)


 * I would expect so. Intuitively, I'd expect the small object could either be far away from the two large objects, orbiting their barycenter in a roughly elliptical path, or close in, orbiting each object to form a figure eight.  I would predict a range of instability between the distant orbit and close orbit.  Do simulations match these predictions ? StuRat (talk) 20:18, 12 February 2012 (UTC)


 * Free return trajectory and its references may be of interest - they are orbits of a spacecraft around the Earth and Moon and the most well-known take the form of a figure-of-eight as you describe. You missed one class of orbit, though - they could be close enough to one object that they just orbit it without much influence from the other object (eg. satellites in Earth orbit mostly ignore the moon, the Moon's orbit around the Earth mostly ignores the Sun, etc.). Oh, and there are Lagrange points that are quite interesting. --Tango (talk) 23:25, 12 February 2012 (UTC)


 * Yes, that third orbit sounds plausible. StuRat (talk) 05:58, 13 February 2012 (UTC)


 * I wrote a program for this about 18 years ago - the two stars nailed down and I would throw in a random small object. Most of the time the small object would get thrown off the screen, but I don't know if they went off to infinity or just a large orbit.  I think large orbits around the two stars are ok.  Sometimes it would go into a close orbit, but it wasn't a figure-8.  It would go around one star then pass between them, then make a U-turn and go around the other star, then pass between, make a U-turn, and go around the first star.  This was from the point of view of the rotating coordinate system. Bubba73 You talkin' to me? 00:55, 13 February 2012 (UTC)


 * I wonder if the object really would reverse, or if it only appears to do so in the rotating frame of reference. StuRat (talk) 05:58, 13 February 2012 (UTC)


 * Probably not. But my question is basically if you mapped what is happening in that rotating frame back to a normal frame, does it map back to what actually happens.  People seem to think that it does, but my intuition was not good enough to know.  Bubba73 You talkin' to me? 06:10, 13 February 2012 (UTC)


 * Changing the frame of reference shouldn't change how it actually behaves, only how we visualize it. StuRat (talk) 06:44, 13 February 2012 (UTC)
 * Rotating reference frame might help here. Robinh (talk) 20:36, 13 February 2012 (UTC)


 * Thanks. I used to understand this stuff better than I do now.  So the "fictitious forces" are just apparent artifacts of the rotating coordinate system and don't really affect the physical behavior, right?  Bubba73 You talkin' to me? 19:06, 14 February 2012 (UTC)

Mercator latitude scaling


Let's imagine I have a Mercator projection map centered on the equator and I've drawn a perfect square on it. Let's say that because of the scale of the map, the square is 1" by 1" which corresponds to 1 mile by 1 mile. Now if I get a map of a more northern latitude, and I apply the same 1" by 1" square, but while the 1" of width will still correspond to a mile, the 1" of height now corresponds with a greater value, like 1.25 miles.

What's the general formula to figure out how that 1" square will scale as I go to different latitudes? I've gone over the Mercator page, but math is not my strongest suit, and applying those equations to this specific problem is causing me some difficulty.

I'm using this for Javascript, and I'm not terribly good at translating mathematical formulae into Javascript, so if you could express this in a way that doesn't involve fancy Greek letters, I'd be very grateful... thanks. --Mr.98 (talk) 20:54, 10 February 2012 (UTC)


 * It looks like the Gudermannian function article might be of some use to you. See the applications section. — Fly by Night  ( talk )  21:36, 10 February 2012 (UTC)


 * Doesn't the fact that the Mercator projection is a conformal map suggest that, for relatively small squares, the distances represented by the sides are equal? -- 182.232.144.48 (talk) 03:43, 11 February 2012 (UTC)
 * When I plot a 20mi x 20mi square in Google Maps, the deformation is really palpable, even at straightforward (e.g. non-polar) latitudes. --Mr.98 (talk) 15:14, 11 February 2012 (UTC)
 * shows no deformation for a ~20mi x ~20mi square at 60°N. Note that the tip of the marker is at equal distance from the lines at 60°N and 102°W, as it should be since 0.29 = 0.58 * cos(60°N). What's your evidence that Google Maps has deformation for relatively small squares? 98.248.42.252 (talk) 19:36, 11 February 2012 (UTC)
 * I am using a Ground Overlay in the API that has a fixed height and width. To plot Ground Overlays in Google Maps you have to specify it as a pair of lat,lng coordinates corresponding to the SW and NE corners of the image. I'm dynamically playing the Ground Overlay depending on whatever the lat,lng of the center of the map is. So the formulae for where to plot the lat,lng is to take the center coordinates and subtract or add the height or width (divided by two) to them to end up with the box. If you do this, the deformation is really palpable. It's actually not even 20mi on each end; it's more like 5mi on each end. --Mr.98 (talk) 21:31, 11 February 2012 (UTC)
 * Do you mean that the image has a "fixed height and width" in pixels and you specify the lat/lng of the corners? If so, then you are fully in control of the scaling; if you want the height and width to scale equally, you need to choose your corners taking cos(latitude) into account.  If not, then can you better explain the situation, possibly providing an image? -- ToE 01:36, 12 February 2012 (UTC)
 * You've got it. "you need to choose your corners taking cos(latitude) into account" — yes, I know, that's the entire goal of the question is to figure out the scaling factor. Using just cos(lat) doesn't seem to do it by itself — is that a percentage? Or...? I'm having trouble making sense of exactly how to use that. --Mr.98 (talk) 02:10, 13 February 2012 (UTC)
 * OK. If your display area is   high and   wide, and you wish to display a region centered on the coordinates ,  , with a range of longitude   (that is, from   to  , then you should use  .  (This assumes north-up (or down).)  Note that   here should be based on the physical dimensions, and that ratio may not be the same when calculated using pixel count if the horizontal and vertical pitch of your display differ.  (How close to 1.0 is   on most modern monitors?)  Give the formula a try and see if it gives an undistorted figure first at the equator, and then at a higher latitude such as 60 degrees.  If the first looks good and the second is still distorted, then something else is going on here. -- ToE 02:58, 13 February 2012 (UTC)
 * Also make sure that  is working in the right units.  At a latitude of 60 degrees, the value of the cosine should be 0.5.  If your   is in degrees and your   function expects radians (as is common in many languages) you will need to convert, such as with  . -- ToE 03:10, 13 February 2012 (UTC)


 * The width of the map is equal at each latitude, but the actual distance around the sphere is proportional to the cosine of the latitude. If 1" on the map is 1 mile of actual distance at the equator, then at latitude θ, 1" will be cosθ miles. As 182.232.144.48 mentioned, height and width scale equally with the Mercator projection. Rckrone (talk) 05:26, 11 February 2012 (UTC)