{"status": "success", "data": {"description_md": "For $\\pi\\leq\\theta<2\\pi$, let

\n$$ P=\\dfrac12\\cos\\theta-\\dfrac14\\sin2\\theta-\\dfrac18\\cos3\\theta+\\dfrac1{16}\\sin4\\theta+\\dfrac1{32}\\cos5\\theta-\\dfrac1{64}\\sin6\\theta-\\dfrac1{128}\\cos7\\theta+\\ldots
\n$$ and
\n$$ Q=1-\\dfrac12\\sin\\theta-\\dfrac14\\cos2\\theta+\\dfrac1{8}\\sin3\\theta+\\dfrac1{16}\\cos4\\theta-\\dfrac1{32}\\sin5\\theta-\\dfrac1{64}\\cos6\\theta+\\dfrac1{128}\\sin7\\theta
+\\ldots $$ so that $\\tfrac PQ = \\tfrac{2\\sqrt2}7$. Then $\\sin\\theta = -\\tfrac mn$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.\n___\nLeading zeroes must be inputted, so if your answer is `34`, then input `034`. Full credit goes to [MAA](https://maa.org/) for authoring these problems. These problems were taken on the [AOPS](https://artofproblemsolving.com/) website.", "description_html": "

For \\pi\\leq\\theta<2\\pi, let

P=\\dfrac12\\cos\\theta-\\dfrac14\\sin2\\theta-\\dfrac18\\cos3\\theta+\\dfrac1{16}\\sin4\\theta+\\dfrac1{32}\\cos5\\theta-\\dfrac1{64}\\sin6\\theta-\\dfrac1{128}\\cos7\\theta+\\ldots<br> and
Q=1-\\dfrac12\\sin\\theta-\\dfrac14\\cos2\\theta+\\dfrac1{8}\\sin3\\theta+\\dfrac1{16}\\cos4\\theta-\\dfrac1{32}\\sin5\\theta-\\dfrac1{64}\\cos6\\theta+\\dfrac1{128}\\sin7\\theta<br>+\\ldots so that \\tfrac PQ = \\tfrac{2\\sqrt2}7. Then \\sin\\theta = -\\tfrac mn where m and n are relatively prime positive integers. Find m+n.

Leading zeroes must be inputted, so if your answer is 34, then input 034. Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.

", "hints_md": "", "hints_html": "", "editorial_md": "", "editorial_html": "", "flag_hint": "", "point_value": 6, "problem_name": "2013 AIME I Problem 14", "can_next": true, "can_prev": true, "nxt": "/problem/13_aime_I_p15", "prev": "/problem/13_aime_I_p13"}}