Let $P$ be a $30$-sided polygon inscribed in a circle. Find the value of $\frac{N}{100}$.

Question

Let $P$ be a $30$-sided polygon inscribed in a circle. There are $N$ number of triangles whose vertices are the vertices of $P$ such that any two vertices of each triangle are separated by at least three other vertices at $P$. Find the value of $\frac{N}{100}$.

What I Tried: This is more like a Combinatorics problem rather than a geometry problem, so here is what I think.

First, fix a point of a triangle. The next point can be chosen in $23$ ways. But I am not sure how to choose the $3$rd point, as for choosing the $2$nd point there are slight variations as well, which dosen't follow the rule.

I thought before of fixing one point, and then the next $2$ points can be chosen in ${23}\choose{2}$ ways, but then I realised that is wrong since those $2$ points might not have a $3$ point gap, and I couldn't get on how to progress on this.

As usual, I also know that the number of triangles on an $n$-sided polygon with no shared sides is given by the formula :- $$\rightarrow\frac{n(n-4)(n-5)}{6}$$ So the total number of triangles is $3250$, but I am not sure on how this fact will help in this problem.

Can anyone help me? Thank You.

Answer 1

Choose any point and call it $A_1$. Label the points in counterclockwise manner $A_2,\ldots,A_{30}$ .

Second vertex can be any from $A_5$ to $A_{27}$.

When second is $A_5$, third vertex can be any from $A_9$ to $A_{27}$. That's $19$ ways.

When second is $A_6$, third vertex can be any from $A_{10}$ to $A_{27}$. That's $18$ ways.

And so on. Number of triangles $= 19+18+17+\ldots+1$

We could start on any point as first vertex, so desired is $$\dfrac{19\cdot20}{2} \cdot \dfrac{30}{3}$$

If we were to leave atleast $k$ points between adjacent vertices, by the same logic we'll get $$\dfrac{n(n-3k-1)(n-3k-2)}{6}$$

for appropriate $k$. Since $3k+2$ number of points are left out first when second vertex is $A_{k+2}$.

Answer 2

An alternative approach is to use the stars and bars method.

We can generalize and consider instead of triangles, $k$-sided polygons. Also let $d$ be the minimum "distance" among vertices of those $k$-sided polygons, where "distance" is the number of inner vertices plus one. In our case we have $k = 3$ and $d = 4$. So the problem becomes finding the number of solutions of:

$$ x_1 + x_2 + \ldots + x_{k-1} + x_k = n$$

where $x_i, i=1,\ldots,k$ are the "distances" among vertices of the $k$-sided polygons, with the constraint:

$$x_i \ge d, i=1,\ldots,k$$

We can define $y_i = x_i+d, i=1,\ldots,k$, and then the first equation becomes:

$$y_1 + y_2 + \ldots + y_{k-1} + y_k = n-kd$$

with $y_i \ge 0, i=1,\ldots,k$. Therefore, by the stars and bars method, the solutions for each vertex are:

$${n-kd+k-1 \choose k-1}$$

and there are $n$ vertices, but every $k$-sided polygon is in common with $k$ of them, so the final solution is:

$${n-kd+k-1 \choose k-1}\frac{n}{k}={30-3\cdot4+3-1 \choose 3-1}\frac{30}{3}={20 \choose 2}\frac{30}{3}=1900$$