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  1. anonymous
    • one year ago
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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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