Question

令 $\begin{aligned} p(x, y) &= a_0 + a_1x + a_2y \\ &\quad {}+ a_3x^2 + a_4xy + a_5y^2 \\ &\quad {}+ a_6x^3 + a_7x^2y \\ &\quad {}+ a_8xy^2 + a_9y^3 \end{aligned}\text{。}$ 假设 $\begin{aligned} p(0, 0) &= p(1, 0) = p(-1, 0) \\ &= p(0, 1) = p(0, -1) \\ &= p(1, 1) = p(1, -1) \\ &= p(2, 2) = 0 \end{aligned}\text{。}$

存在一点 $\left(\frac{a}{c}, \frac{b}{c}\right)$ ，使得对所有这样的多项式都有 $p\left(\frac{a}{c}, \frac{b}{c}\right) = 0$ ，其中 $a$ 、 $b$ 、 $c$ 为正整数， $a$ 与 $c$ 互质，且 $c \gt 1$ 。求 $a + b + c$ 。

Let $\begin{aligned} p(x, y) &= a_0 + a_1x + a_2y \\ &\quad {}+ a_3x^2 + a_4xy + a_5y^2 \\ &\quad {}+ a_6x^3 + a_7x^2y \\ &\quad {}+ a_8xy^2 + a_9y^3. \end{aligned}$ Suppose that $\begin{aligned} p(0, 0) &= p(1, 0) = p(-1, 0) \\ &= p(0, 1) = p(0, -1) \\ &= p(1, 1) = p(1, -1) \\ &= p(2, 2) = 0. \end{aligned}$

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 $c \gt 1.$ Find $a + b + c.$

答案

答案：40

知识点：多项式方程组因式分解

难度评级：3160

小提示：

依次应用条件：前五个条件迫使 $a_0 = a_3 = a_5 = 0$ 、 $a_6 = -a_1$ 、 $a_9 = -a_2$

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

大提示：

最终 $p = a_1 f(x, y) + a_2 g(x, y)$ ；对任意 $a_1$ 和 $a_2$ 都成立的点，必须是 $f$ 与 $g$ 的公共零点，而它们可以漂亮地分解

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

解答：

由 $p(0,0) = 0$ 得 $a_0 = 0$ 。将 $p(1,0) = p(-1,0) = 0$ 相加和相减，得 $a_3 = 0$ 且 $a_6 = -a_1$ ；同样由 $p(0,\pm 1) = 0$ 得 $a_5 = 0$ 且 $a_9 = -a_2$ 。接着 $p(1,1) = 0$ 与 $p(1,-1) = 0$ 化为 $a_4 + a_7 + a_8 = 0$ 和 $-a_4 - a_7 + a_8 = 0$ ，所以 $a_8 = 0$ ， $a_7 = -a_4$ 。此时 $p = a_1(x - x^3)$ $+ a_2(y - y^3)$ $+ a_4(xy - x^2y)$ ，而 $p(2,2) = 0$ 给出 $-6a_1 - 6a_2 - 4a_4 = 0$ ，即 $a_4 = -\frac{3}{2}(a_1 + a_2)$ 。

因此 $\begin{aligned} p &= a_1\left[x - x^3 - \tfrac{3}{2}xy(1 - x)\right] \\ &\quad {}+ a_2\left[y - y^3 - \tfrac{3}{2}xy(1 - x)\right] \end{aligned}\text{，}$ 一个点 $(r, s)$ 若使该式对所有 $a_1, a_2$ 都为零，就必须使两个括号都为零。第一个括号分解为 $r(1 - r)\left(1 + r - \tfrac{3}{2}s\right)$ ，所以对新点（ $r \ne 0, 1$ ）需要 $s = \tfrac{2}{3}(r + 1)$ 。第二个括号为 $\tfrac{1}{2}s(2 - 2s^2 - 3r + 3r^2)$ ；代入 $s^2 = \tfrac{4}{9}(r + 1)^2$ ，将 $2 - 2s^2 - 3r + 3r^2 = 0$ 化为 $\frac{19r^2 - 43r + 10}{9} = 0\text{，}$ 其根为 $r = 2$ 和 $r = \frac{5}{19}$ 。

根 $r = 2$ 重现了已给点 $(2, 2)$ ，所以新点有 $r = \frac{5}{19}$ ，且 $s = \frac{2}{3} \cdot \frac{24}{19} = \frac{16}{19}$ 。因此 $(a, b, c) = (5, 16, 19)$ ， $a + b + c = 40$ 。

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 $a + b + c = 40.$

2008 AIME I 第 13 题

答案：40

难度评级：3160

小提示：

大提示：

解答：

其他年份的第 13 题