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Return a single integer only. The problem has been revised so that no final answer is an irrational number, a parameterized expression, a range, a proof, a drawing, a graph, a locus equation, or a region description.
Let . Choose four distinct vertices of a regular -gon uniformly at random. Let be the probability that the quadrilateral whose vertices are the four chosen points contains the center of the polygon in its interior. If in lowest terms, find .
For a regular n-gon with n odd, no diameter exists, so I need to count quadrilaterals containing the center.
Total ways: C(9,4) = 126.
A convex polygon (the 4 points form a convex quadrilateral) contains the center if and only if no semicircle (open half-plane through center) contains all four points. Equivalently, the center is inside iff the four points are NOT all contained in some half-plane through the center.
Let me count quadrilaterals that do NOT contain the center: those where all 4 points lie in some open half-plane through center.
For odd n, count subsets of 4 points lying within a half-plane. Standard approach: count subsets contained in an arc of less than half the circle (semicircle).
The number of 4-subsets that lie in some semicircle: For each point as the "first" point (most counterclockwise start), count subsets where all points lie within a semicircle starting at that point.
For odd n, a semicircle spans (n-1)/2 = 4 other positions on one side. Actually let me use the formula: number of k-subsets of n points on a circle all lying within some semicircle.
For n points, the number of k-subsets lying in a semicircle (when no two points are diametrically opposite, n odd) is n·C(m, k-1) where m is the number of points strictly within a half-circle starting clockwise from a chosen point.
For each point, consider it as the most clockwise point. The semicircle starting at this point covers the next (n-1)/2 = 4 points (since n odd, exactly floor(n/2)=4 points fit). The remaining k-1 points must be chosen from these 4 points.
So number of "bad" subsets = n · C((n-1)/2, k-1) = 9 · C(4, 3) = 9 · 4 = 36.
So quadrilaterals NOT containing center: 36.
Quadrilaterals containing center: 126 - 36 = 90.
Probability p_9 = 90/126 = 15/21 = 5/7.
So p=5, q=7, p+q = 12.
\boxed{12}
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