2008 AIME I 第 13 题

Apply the conditions in order: the first five force $a_0 = a_3 = a_5 = 0,$ $a_6 = -a_1,$ $a_9 = -a_2$

Eventually $p = a_1 f(x, y) + a_2 g(x, y);$ a point that works for every choice of $a_1$ and $a_2$ must be a common zero of $f$ and $g,$ which factor nicely

From $p(0,0) = 0$ we get $a_0 = 0.$ Adding and subtracting $p(1,0) = p(-1,0) = 0$ gives $a_3 = 0$ and $a_6 = -a_1;$ similarly $p(0,\pm 1) = 0$ give $a_5 = 0$ and $a_9 = -a_2.$ Then $p(1,1) = 0$ and $p(1,-1) = 0$ reduce to $a_4 + a_7 + a_8 = 0$ and $-a_4 - a_7 + a_8 = 0,$ so $a_8 = 0$ and $a_7 = -a_4.$ Now $p = a_1(x - x^3)$ $+ a_2(y - y^3)$ $+ a_4(xy - x^2y),$ and $p(2,2) = 0$ gives $-6a_1 - 6a_2 - 4a_4 = 0,$ i.e. $a_4 = -\frac{3}{2}(a_1 + a_2).$

Therefore $$\begin{aligned} p &= a_1\left[x - x^3 - \tfrac{3}{2}xy(1 - x)\right] \\ &\quad {}+ a_2\left[y - y^3 - \tfrac{3}{2}xy(1 - x)\right], \end{aligned}$$ and a point $(r, s)$ that is a zero for every choice of $a_1, a_2$ must kill both brackets. The first bracket factors as $r(1 - r)\left(1 + r - \tfrac{3}{2}s\right),$ so for a new point (with $r \ne 0, 1$) we need $s = \tfrac{2}{3}(r + 1).$ The second bracket is $\tfrac{1}{2}s(2 - 2s^2 - 3r + 3r^2);$ substituting $s^2 = \tfrac{4}{9}(r + 1)^2$ turns $2 - 2s^2 - 3r + 3r^2 = 0$ into $$\frac{19r^2 - 43r + 10}{9} = 0,$$ whose roots are $r = 2$ and $r = \frac{5}{19}.$

The root $r = 2$ reproduces the given point $(2, 2),$ so the new point has $r = \frac{5}{19}$ and $s = \frac{2}{3} \cdot \frac{24}{19} = \frac{16}{19}.$ Thus $(a, b, c) = (5, 16, 19)$ and $a + b + c = 40.$