#### Topic: Triangle from stick slices

At once I am sorry, if a bayan. Today set a problem, very much it was pleasant. A direct stick in a random way break in two places. What probability of that from the received slices it is possible to add a triangle. Allocation of probability of position of fractures on length of a stick to consider as uniform.

#### Re: Triangle from stick slices

Hello, rg45, you wrote: R> At once I am sorry, if a bayan. Today set a problem, very much it was pleasant. R> a direct stick in a random way break in two places. What probability of that from the received slices it is possible to add a triangle. Allocation of probability of position of fractures on length of a stick to consider as uniform. Draw a square on a plane. On one coordinate the first fracture, on the second - the second. At you the area (or a little) turns out where the triangle is possible. Actually, all.

#### Re: Triangle from stick slices

Hello, alpha21264, you wrote: A> Draw a square on a plane. On one coordinate the first fracture, on the second - the second. A> at you the area (or a little) turns out where the triangle is possible. Actually, all. Well so, all the same, one area, or a little? Digit, the sister, digit!

#### Re: Triangle from stick slices

Hello, rg45, you wrote: 50 %. Or it turns out, or not

#### Re: Triangle from stick slices

Hello, rg45, you wrote: R> At once I am sorry, if a bayan. Today set a problem, very much it was pleasant. R> a direct stick in a random way break in two places. What probability of that from the received slices it is possible to add a triangle. Allocation of probability of position of fractures on length of a stick to consider as uniform. We select on a stick a segment equal to its half are long. If fractures get to it a segment the triangle turns out. Probability 1/4 independent of position of this segment on a stick. Like so.

#### Re: Triangle from stick slices

Hello, rg45, you wrote: R> Hello, alpha21264, you wrote: A>> Draw a square on a plane. On one coordinate the first fracture, on the second - the second. A>> at you the area (or a little) turns out where the triangle is possible. Actually, all. R> well so, all the same, one area, or a little? Digit, the sister, digit! Well  stuck? There, where W - it is possible........... W................... W W.................. W W W................. W W W W................ W W W W W............... W W W W W W.............. W W W W W W W............. W W W W W W W W......... W W W W W W............... W W W W W................ W W W W................. W W W.................. W W.........

#### Re: Triangle from stick slices

Hello, Qulac, you wrote: Q> we Select on a stick a segment equal to its half are long. If fractures get to it a segment the triangle turns out. Probability 1/4 independent of position of this segment on a stick. Like so. Well, conceptually correctly confuses a statement vagueness "independent of position of this segment" a little.

#### Re: Triangle from stick slices

Hello, rg45, you wrote: R> Hello, Qulac, you wrote: Q>> we Select on a stick a segment equal to its half are long. If fractures get to it a segment the triangle turns out. Probability 1/4 independent of position of this segment on a stick. Like so. R> Well, conceptually correctly confuses a statement vagueness "independent of position of this segment" a little. Yes, itself noted. The graphic method is more evident.

#### Re: Triangle from stick slices

Hello, Qulac, you wrote: Q> Yes, itself noted. The graphic method is more evident. Actually, you too thought in the correct direction, but slightly did not finish  up to the end. Mentally we break our segment (stick) on two equal parts. Then for triangle formation simultaneous enough performance of two independent conditions is necessary also: 1) fractures should be allocated on different parts from the midpoint; 2) the distance between fractures should not exceed half of total length. The probability of each of conditions is separately equal 1/2 (it not difficult), therefore probability of joint event - 1/4.

#### Re: Triangle from stick slices

Hello, rg45, you wrote: R> the Direct stick in a random way break in two places. What probability of that from the received slices it is possible to add a triangle. Allocation of probability of position of fractures on length of a stick to consider as uniform. X ~ U (0,1), Y ~ U (0,1) P (|Y-X | <0.5) = P (Y-X <0.5 | Y> X) P (Y-X <0.5 | Y> X) = P (Y-X <0.5 | Y> X, X> 0.5) + P (Y-X <0.5 | Y> X, X <0.5) = 1/4 P (Y-X <0.5 | Y> X, X> 0.5) = 0.0 P (Y-X <0.5 | Y> X, X <0.5) = P (X <0.5) *P (X <Y <X + 0.5) = 1/2*1/2 = 1/4

#### Re: Triangle from stick slices

Hello, rg45, you wrote: and still it is possible to complicate and break in two steps: at first it is casual once, and then the second that remains I can not analytically count, ran monte-karlo, it turns out about 19 %

#### Re: Triangle from stick slices

Hello, rg45, you wrote: R> the Direct stick in a random way break in two places. <...> Allocation of probability of position of fractures on length of a stick to consider as uniform. IMHO, in this place complexity also sits. What is "is arranged uniformly"? If so that any of precisely square 11 is equiprobable, the obvious "graphic" decision. And if, for example, we break in a casual place, we select a casual half, and it is broken once again other answer can already be, and etc. There is a similar more known problem. There is a ruled paper with step between lines L, and there is a needle of length l. A needle casually throw on a paper, what probability, what a needle touches a line? To begin with it is possible to consider case L == l

#### Re: Triangle from stick slices

Hello, torvic, you wrote: T> Hello, rg45, you wrote: T> and still it is possible to complicate and break in two steps: T> at first it is casual once, and then the second that remains T> I can not analytically count, ran monte-karlo, it turns out about 19 % X ~ U (0, 1) Y ~ U (0, 1 - X) P (Y <0.5 and 1 - X - Y <0.5 | X <0.5) = P (0.5 - X <Y <0.5 | X <0.5) = integral (x / (1-x), 0, 0.5) = 0.1931477

#### Re: Triangle from stick slices

Hello, Erop, you wrote: E> IMHO, in this place complexity also sits. What is "is arranged uniformly"? It means that dP = dx/L and does not depend x. E> And if, for example, we break in a casual place, we select a casual half, and is broken once again other answer can already be, and etc. Is not present, about a choice of a casual half is told nothing. Consider that at first we define points, then we break.

#### Re: Triangle from stick slices

Hello, Erop, you wrote: E> And if, for example, we break in a casual place, we select a casual half, and it is broken once again other answer can already be, and etc. https://rsdn.org/forum/etude/7041129.1 the Author: novitk Date: 01.02 20:51

#### Re: Triangle from stick slices

Hello, rg45, you wrote: R> It means that dP = dx/L and does not depend from x. What is dx/L FOR PAIR? R> is not present, about a choice of a casual half it is told nothing. Consider that at first we define points, then we break. I from it and the beginnings, in a condition inaccurately am defined that for allocation means. If the point all is obvious, as though means uniformly arranged in a square 11... If something means another the answer depends that I knowingly about a needle resulted a problem. If it you know, should understand, if you do not know  and it. It too the cool.

#### Re: Triangle from stick slices

Hello, Erop, you wrote: R>> It means, what dP = dx/L and does not depend x. E> That such dx/L FOR PAIR? R>> is not present, about a choice of a casual half it is told nothing. Consider that at first we define points, then we break. E> I from it and the beginnings, in a condition inaccurately am defined that for allocation means. A condition I formulated that is called, "on fingers", without the claim for mathematical accuracy, counting on intuitive understanding. And, to a campaign, all all is clear, to one you submit exact determination Well, I will try, specially for you. The uniform allocation of probability of a choice of points of a fracture is understood as that, at a casual choice of the next point of a fracture on segment L (not important what under the account - the first, the second, the fifth or the tenth) the probability of its hit in any section long l depends only on length of this section both is equal l/L, and does not depend from  this section on a segment (in a tail, a head, or somewhere else). Presence of already selected points of a fracture does not influence in any way a choice of new points. Very much I hope that this time I managed to satisfy your severity.

#### Re: Triangle from stick slices

Hello, rg45, you wrote: R> the uniform allocation of probability of a choice of points of a fracture is understood as that, at a casual choice of the next point of a fracture on segment L (not important what under the account - the first, the second, the fifth or the tenth) the probability of its hit in any section long l depends only on length of this section both is equal l/L, and does not depend from  this section on a segment (in a tail, a head, or somewhere else). Presence of already selected points of a fracture does not influence in any way a choice of new points. Very much I hope that this time I managed to satisfy your severity. I not about severity, and about an essence. That you described, it is possible to describe easier: uniformly distributed point of a square Where  - coordinate of the first break, and y - the second. But it is possible to select also other methods break points

#### Re: Triangle from stick slices

Hello, rg45, you wrote: R> the Direct stick in a random way break in two places. What probability of that from the received slices it is possible to add a triangle. Allocation of probability of position of fractures on length of a stick to consider as uniform. Here  very interesting approach to the decision - in trilinear coordinates: https://www.cut-the-knot.org/Curriculum … lity.shtml the Decision is based on that (the picture see) that the total of all perpendiculars (a, b, c), lowered of any point of an equilateral triangle on its sides, is a constant and is equal to its height for all points,  in this triangle. In the same way, as well as the total of lengths of slices on which we break our stick, too is a constant equal to length of a stick. Therefore we construct near to our stick an equilateral triangle with height to equal length of a stick and we argue. That from stick slices it was possible to add a triangle, it is necessary and enough that the length of each of slices did not exceed half of length of a stick: a <= L/2, b <= L/2, c <= L/2. In a triangle to all three conditions satisfies set of the points lying in a small equilateral triangle, designated by thin lines. Really, any of the points, lying outside of a small triangle has one of perpendiculars with long exceeding half of height of the big triangle (that is, half of length of a stick). The area of a small triangle is equal 1/4  the big triangle, as is the answer to the task

#### Re: Triangle from stick slices

Hello, Erop, you wrote: E> I not about severity, and about an essence. That you described, it is possible to describe easier: uniformly distributed point of a square Egor, well is not present any squares in the task. The square is already your interpretation. I understand that you mean but to include such description in a statement of the problem, at first it is necessary to give the formal description of this approach. Fairly, I do not see sense something to complicate. Well it is clear, what meant?

#### Re: Triangle from stick slices

Hello, rg45, you wrote: R> Yes there are in the task no squares. How is not present? The pair of points on a segment [0, 1] is, and the Cartesian product of segments is not present? I stated not the claim, and thought, simple enough that the most interesting place of this task - the formal interpretation of the informal description "uniformly distributed pair of points of a segment". The formal interpretation gives at once the answer. But interpretation it is possible to offer a little IMHO, in this place there is an interesting place in this task. You about a needle solved the task earlier? If is not present, it too interesting, approximately the same place.

#### Re: Triangle from stick slices

Hello, Erop, you wrote: E> You about a needle solved the task earlier? If is not present, it too interesting, approximately the same place. Well did not solve while, but the decision for me looks clear enough - through integral on a needle angle of rotation: int (cos (a), da);

#### Re: Triangle from stick slices

Hello, rg45, you wrote: R> Well did not solve while, but the decision for me looks clear enough - through integral on a needle angle of rotation: int (cos (a), da); Round what point?.

#### Re: Triangle from stick slices

Hello, Erop, you wrote: R>> Well did not solve while, but the decision for me looks clear enough - through integral on a needle angle of rotation: int (cos (a), da); E> Round what point?. I you at all do not understand Something today. The corner between lines and a needle is round what point? (Is more true, a corner between a needle and an axis, perpendicular to lines, so far as I under integral wrote a cosine, instead of a sine. But not an essence).

#### Re: Triangle from stick slices

Hello, Erop, you wrote: E> I not about severity, and about an essence. And unless the essence consists not what from the infinite number of points there is no method to select in a random way at least one point?