Mathematics desk
< January 26	<< Dec \| January \| Feb >>	January 28 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

January 27

Calculating the area of a triangle while on a coordinate system

Say there are three points on a cartesian coordinate system that form a triangle. What is the formula that would express the area? The normal 1/2*base*height doesn't seem like it would apply here and Triangle#Using_coordinates is too advanced. My programming professor linked to this site as a hint but I fail to see how that helps. Angles are not known and how is it possible to add two points together? 198.188.150.134 (talk) 07:48, 27 January 2010 (UTC)[reply]

Heron's formula gives the area in terms of the length of sides and the semi-perimeter (half the total). You can find the length of each side by using Pythagoras' theorem, especially this section on each pair of co-ordinates. If your points are in 3-D space, you need to extend the Pythagoras formula (just square the x-difference, y-difference and z-difference, add them together, and take the square root for the length of that side). Dbfirs 08:02, 27 January 2010 (UTC)[reply]

You can link to a section using internal links. Also, your link doesn't work because the | character is extraneous in external links. -- Meni Rosenfeld (talk) 08:22, 27 January 2010 (UTC) [reply]

Thanks, I've corrected this. It worked in "popup" but I hadn't tried clicking! Dbfirs 00:59, 28 January 2010 (UTC)[reply]

Triangle#Using_coordinates isn't "too advanced", it just covers several different cases and uses determinant notation to write it more compactly. The formula you need (if the triangles are in a 2D plane) is

{\textrm {Area}}={\big |}(x_{A}-x_{C})(y_{B}-y_{A})-(x_{A}-x_{B})(y_{C}-y_{A}){\big |}/2

where

|t|

is the absolute value of t. -- Meni Rosenfeld (talk) 08:16, 27 January 2010 (UTC)[reply]

Thank you, you guys are awesome! I love Wikipedia!! 198.188.150.134 (talk) 08:24, 27 January 2010 (UTC)[reply]

Effective procedure

What's the effective procedure used to determine if an expression is legal in proposition/first order logic? I don't mean decidability, just mechanically checking whether a random expression is a wf. Money is tight (talk) 09:35, 27 January 2010 (UTC)[reply]

Well, the details would depend on the details of your syntax. In practice people rarely bother making those completely precise anyway. But once you do, it's a tedious but not difficult programming exercise in whatever computer language you happen to know. I don't think anyone's going to want to write out the details here, and it's not likely to be any easier for anyone else than it is for you. --Trovatore (talk) 09:39, 27 January 2010 (UTC)[reply]

Actually, come to think of it, there is more to be said if you're really interested in the details. Whole books have been written on parsing — is The Dragon Book a bluelink? Also, surely we have an article on context-free grammar and Backus-Naur form. --Trovatore (talk) 09:43, 27 January 2010 (UTC)[reply]

"The Dragon Book" isn't a bluelink, but "Dragon Book" is. 131.111.248.99 (talk) 15:51, 27 January 2010 (UTC)[reply]

Not to mention the parsing article, in particular parsing#Types of parser and subsequent sections. — Emil J. 11:16, 27 January 2010 (UTC)[reply]

Thanks for your answers. I'm a bit confused about the semantics part of propositional/first order logics. I never thought clearly about them until now. For example if we just use 'not' and 'material implication' as the connectives in truth functional propositional calculus, it's certainly intuitively clear how each wf can be assign a truth value with regard to a given interpretation. However the way we can construct a wf is not unique, so it annoys me at the back of my mind that the history of constructing a wf matters in defining it's truth value (two different constructions can give the same wf). I'm sure this is not possible, but don't know how to prove it. I think a wf can be more faithfully represented as a parse tree (uniquely), but I don't know exactly how. Money is tight (talk) 07:45, 28 January 2010 (UTC)[reply]

Well-formed formulas are uniquely readable (they are given by an unambiguous context-free grammar), that's one of the key requirements on any sensible definition thereof. How to prove that depends on details of the definition. Generally, if the syntax involves brackets, such as in

F ::= p_n | ¬F | (F → F)

the proof is a straightforward matter of counting the brackets. If the definition uses bracket-free notation, as in

F ::= p_n | ¬F | →FF

it is slightly more complicated, the key step in the proof is to show that no formula is a proper initial substring of another formula. It used to be good manners that introductory books on logic included a proof of unique readability of its formulas, but many authors nowadays choose to sweep it under the rug. — Emil J. 11:48, 28 January 2010 (UTC)[reply]

Enderton's book does have a full proof of unique readability. For those who don't know: this means that each formula has a unique corresponding parse tree. The definition of the truth value of a formula in a particular structure is by induction on the structure of the parse tree of the formula.

An elegant solution for a more advanced book is to define a formula as a (type of) finite tree, rather than a (type of) finite string. This definition bypasses the entire parsing issue, makes it clear how to do a proof by induction on formulas, and makes it easy to generalize to infinitary formulas. — Carl (CBM · talk) 12:46, 28 January 2010 (UTC)[reply]

... or to circuits. — Emil J. 13:06, 28 January 2010 (UTC)[reply]

Thanks to you both! My book just defines wfs using the standard recursion (which is to me totally ambiguous), although it does provide a more rigorous formulation using sequences (but the sequence constructing a given wf need not be unique hence my question). Can you tell me the name of Enderton's book? Money is tight (talk) 23:44, 28 January 2010 (UTC)[reply]

It's A Mathematical Introduction to Logic, ISBN 0-122-38452-0, homepage. It's not my absolute favorite textbook, but it is better than many. Nice points including being clearly written and being modern enough not to not use outdated terminology. — Carl (CBM · talk) 23:59, 28 January 2010 (UTC)[reply]

Three little geometry puzzles

Some geometry puzzles I came across:

There is a triangle ABC with a triangle DEF inside it. Apart from the fact that D is on AB, E is on AC and F is on BC, you know nothing else. Now the question is, what is $\angle BDC+\angle AED+\angle CEF$ ?
There is a triangle DAB where C is a point on line DB and E is to the left of B, beyond A. A line is drawn from A to C. Given that BC = CA = DA, you're supposed to prove that angle DAE is three times angle CBA.
Triangle PQR is isosceles with PQ = PR. S is along PQ and K is along QR. A line is drawn from S past K to T, which is on the extension of line PR. Given that RT = KR, you're supposed to prove that angle PST is three times angle RTK. 4 T C 13:41, 27 January 2010 (UTC)[reply]

OK, figured out the second one (finally!). Two more left. 4 T C 13:52, 27 January 2010 (UTC)[reply]

Hmmm, in the first one they said "you know nothing else". Which means the answer isn't dependent on the angles in the triangles, or else they would have provided the angles. So you can assume the angles are whatever you like! If I assume the triangles are equilateral, the answer shoots into your face: 180 degrees. 4 T C 14:04, 27 January 2010 (UTC)[reply]

But is that allowed? 4 T C 14:05, 27 January 2010 (UTC)[reply]

Well anyway, only the last one is still a problem. 4 T C 14:07, 27 January 2010 (UTC)[reply]

I'm pretty certain the 1st problem you've given cannot be solved. I just draw the diagram, worked out the angle for the same case as you (and I don't get 180°, but rather get 210° - ∠BDC is 90°), but realised that if you change the shape to e.g. make it isosceles with AC shorter or longer the answer is different. I stopped trying to solve them at that point.

And I've put back in your original questions and replies, as without them mine makes no sense; and more generally if you post questions and solutions here they are meant to be shared with other readers and editors, so I would say it's impolite even to remove even your own questions once answered. --JohnBlackburne^words_deeds 14:21, 27 January 2010 (UTC)[reply]

Re. last problem, There's no guarantee that T will be on the same side S on the line KS as you say it will be. To me that invalidates the problem right there.--RDBury (talk) 16:48, 27 January 2010 (UTC)[reply]

If you assume that T is where you're expecting it to be, so R is between P and T, then let b = angle KTR = angle TKR. Then angle PRQ = 2b = angle PQR. Angle QKS = b so angle PSK = 3b, Q.E.D.--RDBury (talk) 17:11, 27 January 2010 (UTC)[reply]

I've just worked out that if T is not where you expect it to be, so P is between T and R, then angle angle PST is three times angle RTK - 180 degrees. So the problem should really be to show that angle PST is three times angle RTK up to a multiple of pi.--RDBury (talk) 17:25, 27 January 2010 (UTC)[reply]

1. The first one has answer: anything between zero and full circle (360°). If vertex B is close to C, E is somewhere in the middle of AC line (not close to its endpoints) and D is close to A then all three terms in your sum are close to 0°. On the other hand, if A is close to BC, then ∠BDC is close to a straight angle (180°), and if D and F are both close to B, then ∠AED+∠CEF is close to 180°, so the whole sum is close to 360°. --CiaPan (talk) 08:14, 28 January 2010 (UTC)[reply]

3. Let γ = ∠RTK = ∠RKT = ∠SKP, β = ∠TRK, ε = ∠QRK and φ = ∠RPQ.
From the sum of angles in the triangle RTK: β = π–2γ and its supplementary angle is ε = π–β = 2γ.
From the sum of angles in PQR: φ = π–2ε = π–4γ.
Finally from the sum in SKP: ∠PST = π–φ–γ = π–π+4γ–γ = 3γ, q.e.d. --CiaPan (talk) 15:52, 28 January 2010 (UTC)[reply]

sequence space

Consider the sequence space of all complex sequences with the metric $d(x,y)=\sum _{j=1}^{\infty }{\frac {1}{2^{j}}}{\frac {|x_{j}-y_{j}|}{1+|x_{j}-y_{j}|}}$ . Let M be an infinite set. Suppose that M is compact(By compactness only sequential compactness is to be understood regardless of it implying the open cover definition in metric spaces). I wish to show that there are numbers $\gamma _{1},\gamma _{2}\cdots$ such that for all $x=(\xi _{k}(x))\in M$ we have $|\xi _{k}(x)|\leq \gamma _{k}$ . How should I proceed. Thanks-Shahab (talk) 15:09, 27 January 2010 (UTC)[reply]

Fix k. If there is no such

\gamma _{k}

, you can find x₀, x₁, x₂, ... in M such that

|\xi _{k}(x_{n})|\geq n

for every natural number n. Then you can apply sequential compactness. — Emil J. 15:39, 27 January 2010 (UTC)[reply]

Ah yes. Thanks. Now how do I go about proving the converse-Shahab (talk) 16:01, 27 January 2010 (UTC)[reply]

What do you mean by converse? If M is a set such that there exists numbers

\gamma _{1},\gamma _{2}\cdots

such that etc., then in general M need not be closed, let alone compact. — Emil J. 17:55, 27 January 2010 (UTC)[reply]

Can you give an example of such an M (and M has to be infinite). This is an unsolved question in my book, so either I am missing something here, or there is a misprint.-Shahab (talk) 18:28, 27 January 2010 (UTC)[reply]

For example, let M = {x_n | n = 1, 2, 3, ...}, where ξ₁(x_n) = 1/n, and ξ_k(x_n) = 0 for k ≥ 2. You can take γ₁ = 1, γ_k = 0 for k ≥ 2. However, the sequence {x_n} converges to the zero function, which is outside M, thus M is not closed. For another example, the set of all sequences x such that |ξ_k(x)| < 1 for every k is not closed either, by a similar argument. — Emil J. 18:53, 27 January 2010 (UTC)[reply]

Smooth structures (manifolds)

Resolved

I am doing a homework problem where I am supposed to show a collection of 2 charts defines a smooth structure on S^n, the n-sphere. The 2 charts are just the sphere minus the north pole and the sphere minus the south pole. Any way, I think I am done but I just need help in understanding. I showed that the two charts are smoothly compatible. But, a smooth structure needs a maximal atlas. This is probably not maximal, right? I do have a theorem that tells me that every smooth atlas is contained in a unique maximal smooth atlas. So, in reality, what is true is that these 2 charts form a smooth atlas though not a smooth structure by themselves. However, they are contained in some unique smooth structure. Is this correct?

Yes, this is not maximal, and yes, it is contained in a unique maximal one.

Also, when this theorem says a "unique maximal smooth atlas", does this mean there is one unique maximal atlas for each manifold that has a smooth atlas? Or does it mean there is a unique one specific to the smooth atlas in question? That is, could a manifold have two different maximal atlases? Thanks. StatisticsMan (talk) 15:26, 27 January 2010 (UTC)[reply]

The second: a smooth atlas A is included into a unique maximal altas B, which is just the set of all charts φ compatible with A (that is, such that A ∪{φ} is still a smooth atlas. But there can be two non-compatible atlases A and A' on the same set M: and they will give rise to different smooth structures). So the existence of the largest super-atlas B containing A is quite obvious. The role of B is just a simple way to attach canonically an atlas to the smooth manifold M. Otherwise you should define a differentiable structure as a "class of compatible atlases", which would make the definition more abstract and heavier. However, all the information about the differentiable structure is already encoded in the atlas A, and in fact you can do everything with A (including the definitions of the tangent bundle, of smooth maps, &c). BUt note that in fact the first you wrote is also true: a smooth manifold by definition has already some defined smooth structure, so there is a unique maximal atlas etc; a manifold can't have two maximal atlas by definition, but a set can - and thay may even be topologically compatible.

And, the next part of the question asks me to show it is the same smooth structure as in some example. This just means I need to show they are smoothly compatible, right? In reality, this other atlas is not a smooth structure either but what it means is that they are both contained in the same unique maximal smooth atlas? StatisticsMan (talk) 15:36, 27 January 2010 (UTC)[reply]

Yes, you have to show that the new charts are compatible. Then, saying that A "is (or is not) a smooth structure" is just a matter of language -because you just defined a differentiable structure as a maximal atlas. So , yes, but you can well say " A defines a smooth structure". --pm a 21:30, 27 January 2010 (UTC)[reply]

Okay, thanks. I think it makes more sense now. StatisticsMan (talk) 22:47, 27 January 2010 (UTC)[reply]

Another triangle problem

The triangle problems above make me wonder about a geometry problem that I never solved:

Given 3 points in XY space, you define a triangle. Given three more lengths, you define the radius of three circles. Do all three circles fit inside the triangle without overlapping each other or the sides of the triangle? They can touch, but not overlap. You cannot put a circle inside of a circle.

I can brute-force a solution to this, but is there an elegant geometry solution? -- kainaw ™ 17:34, 27 January 2010 (UTC)[reply]

is there information missing from this problem? the reason I ask is that if you are talking about arbitrary, unfixed radii, then (obviously) you can make the radii arbitrarily small and fit three circles (or as many as you like), but if you are talking about arbitrary, fixed radii there's no general answer - it would depend on the relationship of the the given radii to the length of the sides and the angles of the triangle. --Ludwigs2 18:51, 27 January 2010 (UTC)[reply]

I said that you are given 3 points and 3 radii. They are not adjustable. The question is: once you have 3 points and 3 radii, will the circles fit inside the triangle? -- kainaw ™ 20:59, 27 January 2010 (UTC)[reply]

The obvious necessary condition for the circles fitting is that none of the radii can be larger than the inradius. If you assign each radius to one corner of the triangle, you can find the criterion that has to be satisfied pairwise among the three radii to make things fit. Just brute forcing it r_a/tan(α/2) + r_b/tan(β/2) + 2(r_ar_b)^1/2 ≤ c, where α, β, γ are the angles at A, B, C respectively. There's probably a nicer way to express that. Satisfying those two sets of conditions should be necessary and sufficient if I'm not mistaken. Rckrone (talk) 20:16, 27 January 2010 (UTC)[reply]

That was my initial solution, but consider a very long triangle (ie: sides 1, 10000 and slightly more than 10000 (the other side)). It is possible to have three circles placed such that their centers form a straight line inside the triangle, but they do not fit into the three corners. Because I thought of an exception to the 3-corners solution, I assumed that there would be more and more and more exceptions and I gave up on the problem. I only just remembered it because of the triangle question above. -- kainaw ™ 21:02, 27 January 2010 (UTC)[reply]

An interesting side question is for a given triangle what are the radii of the three circles that are mutually tangent and each tangent to two sides of the triangle? Rckrone (talk) 20:22, 27 January 2010 (UTC)[reply]

How to build an ellipse?

Hey, I'm trying to build an ellipse using two concentric circles and an arbitrary radius of the larger circle. I've already tried some other methods of doing this, like taking the perpendicular to the major axis at a random point on the circle, and taking a parallel to the major axis through the same radius' intersection with the smaller circle. The locus of that intersection based on the random point on the larger circle gives me an ellipse tangent to both circles at the extrema. What I'd like to find is some way to prove that that actually is an ellipse, perhaps parametrically. I've tried doing some things with c=square root(a^2-b^2) when c = the distance between the foci and a is the major axis and b is the minor axis. I've been able to locate the foci geometrically. Any help would be appreciated! --Fbv65e del — t — c // 20:10, 27 January 2010 (UTC)[reply]

Have you looked at ellipse? Centre to focus is sqrt(a^2 - b^2) if a is your big circle radius and b your small circle radius. Also the ellipse is the locus of x,y satisfying (x^2/a^2) + (y^2/b^2) = 1 . -- SGBailey (talk) 23:16, 27 January 2010 (UTC)[reply]