{"problem":{"name":"Circle and Many Triangles","description":{"content":"There is a circle which radius is 1.   There are $N$ vertices on the circle's circumference.   The vertices divides into $N$ equal parts over the circumference.   ![image](https://atcoder.jp/img/s8pc","description_type":"Markdown"},"platform":"AtCoder","limit":{"time_limit":2000,"memory_limit":262144},"difficulty":"None","is_remote":true,"is_sync":true,"sync_url":null,"sign":"s8pc_3_e"},"statements":[{"statement_type":"Markdown","content":"There is a circle which radius is 1.  \nThere are $N$ vertices on the circle's circumference.  \nThe vertices divides into $N$ equal parts over the circumference.  \n\n![image](https://atcoder.jp/img/s8pc-3/4ed4ec92efd07ab59a34520d6e7f1c02.png)\n\n  \nYou can choose $3$ distinct vertices, and you can make a triangle.  \nThere are $\\\\frac{N(N - 1)(N - 2)}{6}$ ways choosing vertices. The question is: Calculate the area of $K$-th smallest triangle in $\\\\frac{N(N-1)(N-2)}{6}$ triangles.  \nIf the area is same, you can order in any order.  \n  \nIf $N = 4, K = 3$, the result is following:  \n\n*   If you select vertices $1$, $2$, and $3$, the area of triangle $= 1$.\n*   If you select vertices $1$, $2$, and $4$, the area of triangle $= 1$.\n*   If you select vertices $1$, $3$, and $4$, the area of triangle $= 1$.\n*   If you select vertices $2$, $3$, and $4$, the area of triangle $= 1$.\n\nAs a result, the 3rd smallest triangle's area $= 1$.\n\n[samples]","is_translate":false,"language":"English"}],"meta":{"iden":"s8pc_3_e","tags":[],"sample_group":[],"created_at":"2026-03-03 11:01:14"}}