What is the probability that 3 randomly chosen vertices of a regular n-gon form an isosceles triangle?

Question

i.e, the points are the verticies of an isosceles triangle

The points are chosen without replacement. (No degenerate triangles allowed.)

Answer 1

Alright, this is my method:

Instead of treating the choosing of vertices like independent events (same way you'd calculate the probability of dice), I'm calculating all the possible isosceles triangles possible and dividing that by the total number of triangles you can make within the n-gon. This is where my issue with the equilateral triangles comes in.

To begin with, we have to set up two different cases; where n is even and where n is odd, because when creating triangles, even and odd n-gons behave differently.

Now here's my strategy: If you pick any particular vertex of the n-gon and declare it the apex of the triangle (the point between the two equal sides), you can determine how many isosceles triangles you can create that share the same vertex as the apex. Since no two isosceles triangles you can choose will have the same apex (unless it's equilateral, see below), if you do this for each vertex of the n-gon, you can come up with how many possible isosceles triangles there are in the n-gon. Finding the total number of triangles is easy: it's just nC3.

Odd:
Assume n is odd. If we pick a vertex to become the apex of our triangle, we'll notice that there is an even number of vertices left. Each vertex will have a unique partner that will create an isosceles triangle with our apex. Thus, the number of isosceles triangles for that apex is the number of remaining vertices divided by two. Multiply this by the number of vertices in the n-gon to get the total number of isosceles triangles:
n (n - 1) / 2

Even:
Same as with the odd, but this time, subtracting the apex in question leaves an odd number of remaining vertices to create triangles with. This is logical, because the vertex directly opposite the chosen apex will bisect the n-gon, thus creating a triangle with the chosen vertex as the apex would be impossible. So, instead of dividing the remainder by two, we must subtract another vertex (the opposite one) and then halve it. Again, multiply by n:
n (n - 2) / 2

Dividing the number of isosceles triangles by the total number of triangles (nC3 = n!/(3!(n-3)!) = n(n-1)(n-2)/6) will give us our probability:

Odd: (n(n-1)/2)/(n(n-1)(n-2)/6) = 3/(n-2)
Even: (n(n-2)/2)/(n(n-1)(n-2)/6) = 3/(n-1)

Hooray, we’re done! Or are we? Sadly, the work continues:

Equilateral triangles:
I said before that no two isosceles triangles in the n-gon will have the same apex. The only problem with this is when you have an equilateral triangle. An equilateral triangle has no definite apex, so using the apex method of counting the isosceles triangles means that if an equilateral triangle appears, we will be counting it three times (once for each vertex), which is twice too many. Here’s how we fix that. An equilateral triangle can only exist inside an n-gon where n is divisible by three. This is because between each vertex of the triangle around the n-gon, there must be an equal number of vertices. Since there are three vertices in a triangle, then the n-gon must be evenly divided into three portions. So, we can keep our equations from before, but they must be modified to accommodate for an n/3 polygon. In the case that n is divisible by three; we must subtract two triangles for each equilateral triangle present. There will be n/3 equilateral triangles in the n-gon, so if we subtract 2n/3 from our number of isosceles triangles we can fix the problem:

If n is divisible by three:
Odd: (n(n-1)/2 – 2n/3)/(n(n-1)(n-2)/6) = (3n-7)/((n-1)(n-2))
Even: (n(n-2)/2 – 2n/3)/(n(n-1)(n-2)/6) = (3n-10)/((n-1)(n-2))

To recap, if n is not divisible by three:
Odd: 3/(n-2)
Even: 3/(n-1)

This solves your problem with hexagons bpiguy. = )

Answer 2

Hmmm ... I never thought of this, and I don't know, so let's look for a pattern ...
For n=3, the probability is 1.
For n=4, the probability is also 1.
For n=5 it gets harder, but it turns out that the probability is still 1.
For n=6, the probability drops to 4 of 10 ==> 0.4
For n=7, you get 7 of 15 ==> 0.47
For n=8, it's 9 of 21 ==> 0.43
For n=9, it's 10 of 28 ==> 0.36
For n=10, it's 12 of 36 ==> 0.33

I'm tired, my brain is wearing out, so I'll stop here. There is a Fibonacci sequence developing in the number of combinations ... let me go back to develop the cases for n = 3, 4,and 5 more explicitly ...

For n=3, it's 1 of 1 ==> probability 1
For n=4, it's 3 of 3 ==> probability 1
For n=5, it's 6 of 6 ==> probability 1

The Fibonacci sequence is definitely there. As I did these, I saw some patterns ... if the first two points are adjacent, and if the third point is adjacent to either of the first two, you have symmetry (isosceles). If the first two points are adjacent and n is odd, then there's symmetry with the opposite center point, but if n is even, you don't have that.

For larger values of n, if the first two points are adjacent (if there's k points in-between, for example), then there is one or more symmetries if k points are in-between elsewhere, and the number of occurrences of this depends on whether n is even or odd.

I have the feeling that I could get the general formula, but it would still take a lot more work.

Oh -- for some reason, the hexagon (n=6) seems to act strange. I'm not sure why.

Very interesting and fun problem, in any event.

Answer 3

Consider a single vertex. A triangle through this vertex of the n-gon corresponds to a partition of n into three non-zero parts. The triangle is isosceles if and only if the partition contains a repeated part. E.g. if n=5 then the partition (1,1,3) corresponds to a triangle on the 1st 2nd and 5th vertices, 1, 1+1, and 1+1+3.

Notation: Let b be the number of partitions of n into three non-zero parts with at least one repeated part. Let i be the number of isosceles triangles within your n-gon. Let x(n) be an indicator variable, x(n)=1 if 3 divides n, and x(n)=0 otherwise.

By applying this to each vertex, it can be seen that i=b*n/3, since there are n vertices and each triangle will be counted 3 times.

It's straight forward to show that b=3*floor((n-1)/2) - 2*x(n), by simply counting the number of suitable partitions.

The total number of triangles is clearly (n choose 3), so the probability that a triangle, chosen uniformly at random, is isosceles is i/(n choose 3).

Answer 4

The desired probabilities for a regular n-gon, are as follows.
Note nC3 = n(n-1)(n-2) / 3x2 (Also for n= 3,4,5 the probability is 1)

Case 1: n is not divisible by 3:

If n odd then: n(n-1)/2 over nC3 = 3/(n-2)

if n even, then: n(n-2)/2 over nC3 = 3/(n-1)

Case 2: n is divisible by 3:

If n odd then: (3n-7)/(n-1)(n-2)

i.e (n(n-1)/2 - 2n/3) / nC3 = n(3n-7)/6 over nC3 =

(3n-7)/(n-1)(n-2)

If n even, then: (3n-10) / (n-1)(n-2)

(n(n-2)/2 - 2n/3) /nC3 = n(3n-10)/6 over nC3 =

(3n-10) / (n-1)(n-2)

Justification:

Case 1: n not divisible by 3

If we choose an arbitrary vertex V, then we can number the remaining vertices 1 to n-1, say clockwise from V. We get unique isosceles triangles (symmetric with respect to V) from the triplets (V, 1, n-1) (V, 2, n-2) etc. If n is odd, then n-1 is even, and we will have (n-1)/2 such triangles. However if n is even there will only be (n-2)/2 such triangles. Since n is not divisible by 3 there will be no equilateral triangles (and hence no 'double' counting) and since there are n vertices, the total number of such triangles will simply be n(n-1)/2 for n odd, and n(n-2)/2 for n even. To get the desired probabilities we simply divide by nC3 (the tolal number of unique triangles involving n vertices).

Case 2: n is divisible by 3

Here the only problem is there are equilateral triangles (which are also isosceles). If we count n, as above, then we are triple counting, and so we must subtract 2n/3 from the values derived in case 1, for n odd and even.

Answer 5

This is an interesting problem.



3 Sides = 1 of 1
4 Sides = 3 of 3
5 Sides = 6 of 6
6 Sides = 4 of 10
7 Sides = 9 of 15
8 Sides = 9 of 21
9 Sides = 10 of 28
n Sides = ? of (n-2)(n-1)/2

I just can't make out a pattern.

I think it has to do with the number of ways n-3 can be split into 3 pieces.
3 -> 0,0,3 or 1,1,1
4 -> 0,0,4 or 1,1,2 or 2,2,0
5 -> 0,0,5 or 1,1,3 or 2,2,1
6 -> 0,0,6 or 1,1,4 or 2,2,2 or 3,3,0

I think I'm going to give up on this one. Perhaps I think about it more tomorrow. This is a fun problem, and very difficult.