Problem Analysis #3

To solve the problem, we consider a regular $n$-gon with $n = 9$. We are to compute the probability $p_9$ that a randomly selected set of four distinct vertices forms a quadrilateral containing the center of the polygon in its interior.

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Understanding the Geometric Condition

For a convex polygon inscribed in a circle, the center is inside the convex hull of its vertices if and only if the vertices are not all contained within any semicircle. This is a well-known geometric result for point sets on a circle.

In the case of a regular $n$-gon with $n$ odd (like $n = 9$), there are no diametrically opposite points. The maximum number of consecutive vertices that can lie within a semicircle (i.e., an arc of 180°) is:

$$\left\lfloor \frac{n}{2} \right\rfloor + 1 = \left\lfloor \frac{9}{2} \right\rfloor + 1 = 4 + 1 = 5$$

Thus, any set of 4 vertices that lies entirely within a block of 5 consecutive vertices does not contain the center, because these vertices are all on a single semicircle.

Conversely, if a set of 4 vertices is not contained in any such block, then the center lies inside the quadrilateral formed by them.

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Counting Total Subsets

The total number of ways to choose 4 distinct vertices from 9 is:

$$\binom{9}{4} = 126$$

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Counting Subsets Contained in a Semicircle

We now count how many of these 126 subsets lie entirely within some block of 5 consecutive vertices. Since the polygon is cyclic, there are exactly $n = 9$ such blocks (one starting at each vertex).

Each block of 5 consecutive vertices contains:

$$\binom{5}{4} = 5 \text{ subsets}$$

So, the total number of such subsets across all blocks is:

$$9 \times 5 = 45$$

However, this count includes some subsets multiple times. Specifically, subsets consisting of 4 consecutive vertices are counted twice, once in each of two adjacent blocks.

There are exactly $n = 9$ such subsets of 4 consecutive vertices.

Therefore, the number of distinct subsets contained in some semicircle is:

$$45 - 9 = 36$$

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Counting Favorable Subsets (Containing the Center)

A quadrilateral contains the center if and only if the 4 vertices are not all contained in any semicircle. Thus, the number of favorable subsets is:

$$126 - 36 = 90$$

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Computing the Probability

The probability $p_9$ is:

$$p_9 = \frac{90}{126} = \frac{5}{7}$$

This fraction is already in lowest terms.

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Final Answer

$$\boxed{12}$$