# Difference between revisions of "2008 AIME I Problems/Problem 13"

## Problem

Let

$$p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.$$

Suppose that

$$p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1)\\ = p(1,1) = p(1, - 1) = p(2,2) = 0.$$

There is a point $\left(\frac {a}{c},\frac {b}{c}\right)$ for which $p\left(\frac {a}{c},\frac {b}{c}\right) = 0$ for all such polynomials, where $a$, $b$, and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c > 1$. Find $a + b + c$.

## Solution

\begin{align*} p(0,0) &= a_0 = 0\\ p(1,0) &= a_0 + a_1 + a_3 + a_6 = a_1 + a_3 + a_6 = 0\\ p(-1,0) &= -a_1 + a_3 - a_6 = 0\end{align*}

Adding the above two equations gives $a_3 = 0$, and so we can deduce that $a_6 = -a_1$.

Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5 = 0$ and $a_9 = -a_2$. Now, \begin{align*}p(1,1) &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9\\ &= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 = a_4 + a_7 + a_8 = 0\\ p(1,-1) &= a_0 + a_1 - a_2 + 0 - a_4 + 0 - a_1 - a_7 + a_8 + a_2\\ &= -a_4 - a_7 + a_8 = 0\end{align*} Therefore $a_8 = 0$ and $a_7 = -a_4$. Finally, $p(2,2) = 0 + 2a_1 + 2a_2 + 0 + 4a_4 + 0 - 8a_1 - 8a_4 +0 - 8a_2 = -6 a_1 - 6 a_2 - 4 a_4 = 0$ So $3a_1 + 3a_2 + 2a_4 = 0$.

Now $p(x,y) = 0 + x a_1 + y a_2 + 0 + xy a_4 + 0 - x^3 a_1 - x^2 y a_4 + 0 - y^3 a_2$ $= x(1-x)(1+x) a_1 + y(1-y)(1+y) a_2 + xy (1-x) a_4$.

In order for the above to be zero, we must have

$x(1-x)(1+x) = y(1-y)(1+y)$

and

$x(1-x)(1+x) = 1.5 xy (1-x).$

Canceling terms on the second equation gives us $1+x = 1.5 y \Longrightarrow x = 1.5 y - 1$. Plugging that into the first equation and solving yields $x = 5/19, y = 16/19$, and $5+16+19 = \boxed{040}$.