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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.

Albus Dumbledore
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Anonymous
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    Note that the formula is for "number of triangles with no edges on the polygon". Do you know how that formula is derived (ideally combinatorially)? If so, you can modify it slightly to deal with your case. It will look like $ n (n-k)(n-k-1) / 6$. – Calvin Lin Dec 05 '20 at 12:33
  • @CalvinLin Yes, I saw it on https://math.stackexchange.com/questions/1446438/how-many-triangles-can-be-formed-by-the-vertices-of-a-regular-polygon-of-n-sid/1446470 and understood the answers as well. – Anonymous Dec 05 '20 at 12:46

2 Answers2

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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 $A_{k+2}$.

cosmo5
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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:

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

Fabius Wiesner
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