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I found the question here. The precise problem is

The unit interval is broken at two randomly chosen points along its length. Show that the probability that the lengths of the resulting three intervals are the heights of a triangle is equal to $$ \frac{12\sqrt{5}}{25}\log\left(\frac{3+\sqrt{5}}{2}\right)-\frac{4}{5}. $$ First off, is my understanding of the question correct? That the "heights" of a triangle really refer to its sides?

If this is the problem, I really have no clue how to resolve it, although I would be very interesting in seeing the answer. If it is not the problem, please set me straight.

EDIT - I have decided that the problem is really the perpendicular heights of a triangle, therefore - this has become a dual question, either show the above answer is true for the heights, or come up with a new answer interpreting the question using sides.

I will theorise a little on the "sides" problem.

I suppose that if $a$, $b$ and $1-a-b$ are the lengths of the segments, then if they do form a triangle then its area can be calculated using Heron's formula which says that $$ A = \frac{1}{4} \sqrt{(1-2a)(1-2b)(1+2a+2b)}. $$ An obvious restriction being that $a<1/2$, $b<1/2$. Perhaps we can compare this with the area calculated in the more traditional fashion?

Another obvious restriction is the triangle inequality, let's say $$1-a-b>a+b,$$ which implies that $a+b<1/2$?

I have found similar questions for a triangle and a quadrilateral, but none of the answers go into detail.

I envisage the answer comes from an expected value integral, although I have little knowledge of probabilistic methods.

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  • $\begingroup$ You should specify what you mean by splitting the unit interval into three random segments. This can be done in a lot of ways, resulting in different answers. Do you pick the two cutting points independently and with uniform distribution? $\endgroup$ – Dan Shved Nov 4 '13 at 6:32
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    $\begingroup$ Also, you should really decide which of the two questions you are asking. The question with sides is fairly easy (the answer is $\frac{1}{4}$ if cutting points have the uniform distribution and are independent). The question with heights seems more difficult, at least at first glance. $\endgroup$ – Dan Shved Nov 4 '13 at 6:33
  • $\begingroup$ Ok, let's say that they are independent, uniform distributions. Can you show me how the result is achieved for the sides case, the answer of $1/4$? $\endgroup$ – Bennett Gardiner Nov 5 '13 at 3:51
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OK, here is a brief solution for this variant of the question: two points are chosen uniformly and independently on the unit segment. Find the probability that the resulting three segments are sides of a triangle.

Let $a, b \in [0, 1]$ be the two random splitting points. Let $m = \min\{a, b\}$ and $M = \max\{a, b\}$. The three segments then have lengths $m$, $M - m$ and $1 - M$.

The necessary and sufficient condition for numbers $x, y$ and $z$ to be sides of a triangle is the three triangle inequalities: $x + y > z$, $x + z > y$ and $y + z > x$. In our case it boils down to $m < \frac{1}{2}$, $M > \frac{1}{2}$ and $M - m < \frac{1}{2}$.

So, we pick a point $(a, b)$ inside the unit square (it's just another way of saying that we pick two points on the unit segment). We want to find the probability that the minimum of $a$ and $b$ is less than $\frac{1}{2}$, the maximum is greater than $\frac{1}{2}$, and the difference between them is less than $\frac{1}{2}$. The figure below crudely shows which points inside the unit square satisfy all three of these conditions:

enter image description here

Now, since $a$ and $b$ are independent and distributed uniformly, probabilities of events are simply their areas on the diagram above. The black triangles occupy one fourth of the square, so the answer is $\frac{1}{4}$.

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  • $\begingroup$ Great answer - I love the idea of thinking of an augmented 2D solution space to find the area. One clarification, the diagonal line through the square is just to show where $b>a$, right? It's not an actual solution? $\endgroup$ – Bennett Gardiner Nov 5 '13 at 12:30
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    $\begingroup$ Right, it's just to show where $b>a$, and also to make it more obvious that black triangles occupy $1/4$ of the area. $\endgroup$ – Dan Shved Nov 5 '13 at 13:04
  • $\begingroup$ Awesome, thanks. $\endgroup$ – Bennett Gardiner Nov 6 '13 at 2:11

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