People Sitting Around Triangular Tables 
What is the ratio of number of arrangements of $9$ people sitting
around an isosceles triangular table and the same $9$ people sitting
around an equilateral triangular table with $3$ people sitting on each
side if the table in both the cases?

My solution approach :-
Let's consider the case of isosceles triangular table.
Number of ways in which people can be arranged on the unequal side = $^9C_3 \times 3!$
Number of ways in which people can be arranged on the equal side = $^6P_6 = 6!$
Total number of ways = $^9C_3 \times 3! \times ^6P_6 = 9!$
But when I come to the case of equilateral triangular table I do not have fixed point to start with. That's where I am getting stuck. How to count the number of ways in case of equilateral triangular table? 
One side question what can be the number of ways in which $n$ people can sit around a $x-$ sided polygon? I asked this question because it will help me generalize the cases for any regular polygon.
Thanks in advance !!!
 A: It's not necessary to count the number of arrangements around either table; the ratio is simply $3$, because each of the equilateral arrangements corresponds to three isosceles arrangements, letting the "bases" of the isosceles triangles be the three different sides of the equilateral triangle. (This assumes "isosceles" means "isosceles but not equilateral.")
Remark: I was initially just going to post this as a comment, but decided to make it a full fledged answer to stress a point: If you play close attention to exactly what a question asks for, you can sometimes get away with doing a lot less work than you might otherwise anticipate. In this case, actually counting the numbers of arrangements isn't terribly difficult, but it does take some effort. Since the question only asks for the ratio, it suffices to observe that the isosceles arrangements come in groups of three. A more advanced problem might pose a daunting challenge to count the arrangements but still have a simple way to arrive at the ratio.
A: Three people on each side of a table in the shape of an isosceles triangle:  Pick one of the seats as a starting point (it does not matter which seat you choose).  There are nine ways to choose the person who sits in that seat.  The remaining eight people can be arranged in $8!$ ways as we proceed clockwise around the table from that person.  Up to rotation, there are $9 \cdot 8! = 9!$ such arrangements.  However, there are three equivalent rotations of the table, so we have counted each such arrangement three times.  Hence, there are
$$\frac{9!}{3}$$
arrangements of nine people around a table with the shape of an equialateral triangle if three people sit on each side.
If $n = kx$, where $k$ is an integer, the number of ways $n$ people can sit around a table in the shape of a regular polygon with $x$ sides is
$$\frac{n!}{x}$$
since there are $n!$ ways to arrange the people around the table and $x$ equivalent rotations.
A: If we draw a triangle on a piece of paper and put 3 slots on each side, the number of ways to place people on the 9 slots is $9!$. But, there is a symmetry present, where if we were to rotate the placement on each side to the next side, the seating would look the same to everyone.
I.e., if you have, persons A,B,C on side 1, persons D,E,F on side 2, and persons G,H,I on side 3, this is equivalent to persons G,H,I on side 1, persons A,B,C on side 2, and persons D,E,F on side 3, with the same order of people in each side. There are a total of 3 equivalent placements for this configuration. There are no other equivalent placements. Say you are person A sitting at the table; on your side, you know that B is to your right, and C is to B's right, and on the next side, it is D,E,F, etc. This forces these 3 to be the only placements corresponding to this configuration.
Therefore, the total number of ways to seat the 9 people is $9!/3.$
A: I am assuming that there are three people on each side, symmetrically seated
If the seats are numbered, the number of arrangements will be $9!$ for both.
If they are not numbered, then for the isosceles triangle  it will remain $9!$, but that for the equilateral triangle will become $\frac{9!}{3}$ because you can't distinguish between the three vertices. [For sitting equally spaced in a circle, it would be $\frac{9!}{9}$ because you couldn't distinguish between any of the $9$ points]
