{"status": "success", "data": {"description_md": "Let<br>\n$$ p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.<br>\n$$Suppose that<br>\\begin{align*}p(0,0) &= p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1) \\\\&= p(1,1) = p(1, - 1) = p(2,2) = 0.\\end{align*}There is a point $ \\left(\\tfrac {a}{c},\\tfrac {b}{c}\\right)$ for which $ p\\left(\\tfrac {a}{c},\\tfrac {b}{c}\\right) = 0$ for all such polynomials, where $ a$, $ b$, and $ c$ are positive integers, $ a$ and $ c$ are relatively prime, and $ c > 1$. Find $ a + b + c$.\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": "<p>Let<br/><span class=\"katex--display\"> p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.&lt;br&gt;</span>Suppose that<br/>\\begin{align*}p(0,0) &amp;= p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1) \\&amp;= p(1,1) = p(1, - 1) = p(2,2) = 0.\\end{align*}There is a point $ \\left(\\tfrac {a}{c},\\tfrac {b}{c}\\right)$ for which $ p\\left(\\tfrac {a}{c},\\tfrac {b}{c}\\right) = 0$ for all such polynomials, where $ a$, $ b$, and $ c$ are positive integers, $ a$ and $ c$ are relatively prime, and $ c &gt; 1$. Find $ a + b + c$.</p>&#10;<hr><p>Leading zeroes must be inputted, so if your answer is <code>34</code>, then input <code>034</code>. Full credit goes to <a href=\"https://maa.org/\">MAA</a> for authoring these problems. These problems were taken on the <a href=\"https://artofproblemsolving.com/\">AOPS</a> website.</p>", "hints_md": "", "hints_html": "", "editorial_md": "", "editorial_html": "", "flag_hint": "", "point_value": 6, "problem_name": "2008 AIME I Problem 13", "can_next": true, "can_prev": true, "nxt": "/problem/08_aime_I_p14", "prev": "/problem/08_aime_I_p12"}}