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anonymous
 one year ago
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anonymous
 one year ago
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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There 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 oddsided 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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