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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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There are \(\dbinom93\) total possible triangles that can be drawn in a regular nonagon. Of these, only \(10\) will contain the center of the circle. (I arrived at this by brute force.) Label the points on the circle, 1 to 9. If you pick 1 and 2, only 6 will form a triangle containing the center (1 choice). If you pick 1 and 3, you can use 6 and 7 (2 choices). If you pick 1 and 4, you can use 6, 7, and 8 (3 choices). If you pick 1 and 5, you can use 6, 7, 8, and 9 (4 choices). In total, you have \(1+2+3+4=10\). A certain pattern emerges if you use another odd-sided regular polygon. The number of triangles that contain the center takes the form of the triangular numbers. More generally, if you have \(2n+1\) points to choose from, you have \(\dbinom{2n+1}3\) total possible triangles. The number of triangles that contain the center is \(\displaystyle\large\sum_{k=1}^{\left\lfloor\frac{2n+1}{2}\right\rfloor}k\). (If what I knew of combinatorics was more fresh in mind, I'd have provided a much better explanation -_-)

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