Wikipedia:Reference desk/Archives/Mathematics/2006 December 3

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Probability question (triangle pts)
I have a reasonable grasp of probability, but soon reach a limit. I came across this on an old exam paper (maybe not remembering the exact words, but the gist is there): "A rigid plane lamina of arbitrary size and shape has 3 points chosen at random. What is the probability that if legs are attached there, the lamina will form a stable table?"

It's clear that for stability the triangle formed by the 3 points must include the centroid, but it's "arbitrary" that troubles me:

(a) if it means that all shapes will have the same answer, just pick a simple one, e.g. straight line, which I think gives an answer of 3/4;

(b) if it means that the answer must be "averaged out" over all possible sizes and shapes, each having a different answer, I don't know how to begin.

Can anyone point me towards a solution?86.132.238.91 12:19, 3 December 2006 (UTC)


 * Everything being arbitrary, my answer shall be. There is 0 probability to have two points in line with the centroid ; so it lets 1/2 to have one point in the half-plane where the two others are, and 1/2 for stability. What do you think ? -- DLL .. T 16:43, 3 December 2006 (UTC)

My first thought is that a probability of 1/2, for random point selection, can come only for a point in half the area. But it's not area, it's moment which defines the centroid, so your approach seems more to do with the median. But are they distinct, for a uniform density? Nor am I sure what the median is in two dimensions.

So if the first point is irrelevant, as it can be anywhere, how do you define the half-planes? Would a straight line from (very near) the first point through the centroid divide the lamina into two equal areas? If an arbitrary split into two equal parts is made before the first point is chosen, though it's not possible to have 3 points in the same half and balance, it is possible to have at least one in each half and not balance.86.132.238.91 23:16, 3 December 2006 (UTC)
 * Is this abitrary plane a rectangle (if so the question is easy) or is it any shape???87.102.32.183 08:48, 4 December 2006 (UTC)
 * scrub that - it's not that easy for a rectangle - but it is for a circle p=1/4 if the angle between points a and b is x then the probability for stability is x/2pi therefor the averaged probability is Intergral(x from 0 to pi) of{x/2pi) divided by pi?. Can't extend this to other shapes though..87.102.32.183 09:10, 4 December 2006 (UTC)
 * Also if the plane is any shape - I'd assume the center of mass is calculated assuming constant density and thickness?
 * So taking a line through the centroid (center of mass) is it true that 1/2 the surface area is on one side - half on the other - answer not necessarily - this complicates the question - are you sure this is not a regular shape - the question seems very difficult.87.102.32.183 08:53, 4 December 2006 (UTC)


 * I agree that the answer for a circle, whether solid or a ring, is 1/4. Simulating placing 3 points on a square gave a value of 27+%, with enough trials to make this significantly more than 1/4, thus the same value will not be obtained regardless of shape. Could it be that the problem was for an infinite rigid lamina, which would effectively be circular as all directions from the centroid are equally likely, giving 1/4 for the probability of balance?86.132.238.194 17:25, 5 December 2006 (UTC)


 * I do get the same 1/4 value regardless of shape: I did some sort of ellipse-like thing, a circle, and (via random sampling: 249803/106) a square. I suppose, then, that the practical answer is just to go with a circle or so and do it.  But I think I also have a partial reason why it's independent of shape: obviously, what matters are the directions to the three points from the centroid; their distance from it is irrelevant.  So instead of picking a random point, we can simplify the problem to picking a random direction, with the caveat that directions with more points along them must be picked more often.  For a shape which is the interior of some polar function $$r(\theta)$$ centered at the centroid, the are of a slice (and thus the relative chance of selecting it) is $$\frac12r^2(\theta)d\theta$$.  Meanwhile the moment corresponding to that slice (its centroid times its mass) is $$\frac\rho3r^3(\theta)\hat\theta d\theta$$.  The requirement that the slice centroids sum to the centroid of the whole lamina ($$\vec0$$) means that $$r^3(\theta)\perp\sin(\theta+\phi)\;\forall\phi$$.  We can therefore expand $$r^3(\theta)$$ in a Fourier series containing no first-order terms; somehow this symmetry causes the integrals of $$r^2(\theta)d\theta$$ for the probability to not depend on r.  (The overall integral, if anyone wants it, is $$p=\int_0^{2\pi} P(\alpha) \int_0^\pi \left(\int_{\alpha-\beta}^\alpha P(\alpha-\beta)P(t+\pi)\,dt+\int_\alpha^{\alpha+\beta} P(\alpha+\beta) P(t+\pi)\,dt\right)\,d\beta\,d\alpha$$, where $$P(\theta):=\frac{r^2(\theta)}{\int_0^{2\pi} r^2(\theta)\,d\theta}$$ is the probability density function for picking points in a slice.  The two $$\beta$$ integrals allow for the interchange of the first two points chosen, and the $$t+\pi$$ corresponds to the fact that the third leg must be on the other side of the centroid.  Maybe this helps someone flesh out the solution.  --Tardis 19:10, 5 December 2006 (UTC)
 * ok I admit that I (so far) have lost you past the second equation.. But what about a 'spoon' shape - that is a circular section radius r (area a=pixr2) connected by a very long very very narrow piece to a much smaller circle radius r/n (area=pixr2/n2 =a/n2) if the long connector is length rn2 or greater then the centre of mass will be outside the larger circle (on the long thin bit). If n is big eg 100 then the probability that a random point will be on the larger circle will be (n2-1)/n2 (ie very great more than 0.99) therefor for shapes of this type (notably non convex) the probability of balancing is very low... Perhaps the above method only works for convex shapes??87.102.32.250 19:36, 5 December 2006 (UTC)


 * I can contribute this, at least: you can get a different answer if the lamina is allowed to be disconnected. If it consists of three equal-area islands that are sufficiently small, then it balances if and only if the legs are chosen from all three islands, which happens with probability 2/9. Melchoir 01:20, 6 December 2006 (UTC)


 * Hmm. And we can obviously connect your islands with near-zero-width strips to a common tiny pylon in the center and make even a simply connected lamina with that lowered probability, although this is not fundamentally different from 87.*'s idea, just simpler/more obvious.  Technically convexity is not required if my idea ends up being rigorous; one of the shapes I tested with was not convex (but all were polar functions).  --Tardis 03:04, 6 December 2006 (UTC)

Riemann Sum
My calculus teacher has us use upper and lower Riemman sums as opposed to right and left. I've looked all over on google for a TI-83 upper and lower Riemann sum program, but all I can find are programs for left and right. Does anybody know how I'd write a program for this? Thanks a lot. Sashafklein 17:09, 3 December 2006 (UTC)


 * If you look at the article on riemman sums, the article defines it as the sum from 1 to n of f(yi) multiplied by (xi - xi-1). If f(yi) is greater or equal (it should be the latter) to the greatest value of y within the interval (xi,xi-1), then it is an upper riemann sum.  If it is equal to or lower than the lowest value of y over that interval, then it is an lower riemann sum.  Basically, for upper, where you have an increasing slope you have right riemman sums, and for intervals of decreasing slope you use left riemman sums.  In other words, for the equation y=|x|, for all values x<0, in order to do an upper riemman sum, you would use left riemman sums, and for all values x>0 you would use right riemann sums.  As for partitions at which the max/min is neither xi or xi-1, I'm not sure what you would do. --AstoVidatu 22:27, 3 December 2006 (UTC)

questions about Moufang set isomorphisms, and how I can construct one
Hello,

I don't know if there is anyone here who is familiar with the concept of Moufang sets. I myself have had a brief introduction but it was too vague and now I am confused.

1) What exactly does it mean when two Moufang sets are isomorphic? Or what is an isomorphism?

2) Suppose I have a set X, and for two elements a and b in them, I can find groups of permutations on X, $$U_{a}$$ and $$U_{b}$$, both working sharply transitive on $$X\backslash{a}$$ and $$X\backslash{b}$$ respectively. Is it possible to create a Moufang set from just this?

Thank you, Evilbu 17:31, 3 December 2006 (UTC)


 * I have no idea what a Moufang set is, so here's a generic recipe: Start by defining the concept of a morphism of Moufang sets. Presumably it will be a function on the underlying sets that preserves the additional structural properties enumerated in the definition of a Moufang set. You'll also need the general concept of function composition. Then there are several ways you can proceed to define an isomorphism: work out what it means for a Moufang set homomorphism to be an epimorphism and a monomorphism; if it is both, call it an isomorphism. Or do it via inverses: if $$f:M\to N$$ is a homomorphism and so is $$g:N\to M$$ and if $$g\circ f$$ is the identity morphism on M and $$f\circ g$$ is the identity morphism on N, then f and g are both isomorphisms and inverses of each other. --MarkSweep (call me collect) 09:50, 7 December 2006 (UTC)