you are given 3 exactly same sticks. randomly break one piece from each stick (note this is not breaking one stick into three), and you use them to form a triangle. is it more likely to get an acute triangle, or an obtuse triangle?

Or no triangle at all?The probability of an acute triangle is given by the volume insideand a, b, c are between 0 and 1The probability of an obtuse triangle is 3 times the volume inand a, b, c are between 0 and 1.Maybe some of the inequalities are redundant.It's not obvious to me which one is larger, although I'd guess it is the latter, but right now I don't have time to do the calculation.

QuoteOriginally posted by: EBalOr no triangle at all?indeed that's the only tricky part of the question half of the time you don't get an triangle!(so if you asked someone to guess which one is larger then asked him to actually calculate the prob, he might be surprised that it is below 1/2)the rest of the calculation is simple, because a^2=b^2+c^2 is just the surface of a cone.(for reference, the probability of forming no triangle : an obtuse triangle : an acute one is 1/2 : pi/4-1/2 : 1-pi/4. they are exactly the areas if you divide a unit square into three parts by its diagonal: x+y=1, and a quarter of the circle: x^2+y^2=1)

another stick-breaking problemi guess everyone already knows the following problem: if you randomly break a stick into three, what is the probability of these three forming a triangle? now instead of breaking into three, let's break it into four. the question is: what is the probability of these four parts forming a trapezoid?http://en.wikipedia.org/wiki/Trapezoid

maybe i'm way off the mark, but if you take the two general inequalities for the existence of a traingle, and using a monte carlo type simulation, i.e by making arbitrary cuts in the sticks and then averaging, the density is just under 0.33... thoughts?

QuoteOriginally posted by: lcam6375maybe i'm way off the mark, but if you take the two general inequalities for the existence of a traingle, and using a monte carlo type simulation, i.e by making arbitrary cuts in the sticks and then averaging, the density is just under 0.33... thoughts?could you elaborate a little more about which problem you are simulating? are you randomly breaking one stick into 3, or as the original problem, getting one part from each of the three exact same sticks?

I think that the reference is to breaking the one stick into three.

ok, for that question the answer is 1/4 if you choose the two breaking points independently and uniformly on that stick.(this is equivalent to you break the stick into two and randomly pick one with a relative chance in proportion to their length, then break that part into two. but it's not equivalent to randomly choose one piece and then break into two)there is a nice geometric solution, but i will give the more common algebraic solution:so equivalently, we are choosing two independent random numbers x and y ~u[0,1].if x<y, to form a triangle, the length of each piece has to be smaller than half of the total length (e.g., a<b+c, and a+b+c=1, you get 0<a<1-a, which is 0<a<1/2; similarly for b and c). so 0<x<1/2, 0<1-y<1/2, 0<y-x<1/2, i.e., 0<x<1/2 and 1/2<y<1/2+x. if y<x, by symmetry, we should get the same answer.so the probability of forming a triangle is.

QuoteOriginally posted by: wileyswok, for that question the answer is 1/4 if you choose the two breaking points independently and uniformly on that stick.(this is equivalent to you break the stick into two and randomly pick one, then break that part into two. but it's not equivalent to choose the longer piece and break into two)To be equivalent to picking two uniform variates, you need to do the "randomly pick one" after you "break the stick into two" in proportion to the two segment lengths.

yes you are right. thx. edited

An eloquent result, however not intuitively obvious at first, tks for the correct answer also.

np. btw, for an elegant geometrical solution (i first read it from Martin Gardner's column), see:http://www.cut-the-knot.org/Curriculum/ ... ty.shtmlit also discussed various versions of the problem i stumbled to state earlier as Traden4Alpha pointed out.so anyone discussing the trapezoid problem?----- ----- ----- ----- -----any broken part has a length of no more than 1/2 is sufficient and necessary to form a trapezoid.there is a similar geometrical approach by considering a tetrahedron, and one gets 1-4*1/8=1/2.